No, because to an excellent approximation the sun is in an inertial frame
(at rest or moving with constant velocity) and the earth is an accelerating frame because of its orbit.
There are also innumerable other examples which could not be understood
(like the motions of the planets in the sky) by an earth-centric model.
Sorry, but the mass of Mercury is almost 20 times smaller than earth and
its radius is almost 3 times smaller. You can calculate the acceleration
due to gravity at the surface, g=MG/R2 where
G=6.67x10-11 N∙m2/kg2. The data for
Mercury (earth) are: mass M=3.3x1023 kg (6x1024
kg); radius R=2.4x106 m (6.4x106 m);
gravitational acceleration at surface g=3.7 m/s2 (9.8
m/s2).
QUESTION:
I'm wondering about whether it would be possible for human activity, given current technology, to move a moon such as Europa or, better yet, Ganymede out of Jupiter's orbit and into a solar orbit (down the road, ideally it would be nice to have it impact with another planet with water resources for, you guessed it, terraforming).
The parameters of this situation are as follows. Assume cooperative human action which would allow the use of nuclear weapons on the far side of the moon to push it away from Jupiter, but at regular enough intervals to keep it from exploding. Assume also any "tractor" technologies, such as a large, solar-powered spacecraft which uses (some of) the electricity created by the solar panels to create lasers which push it away from this moon.
Am I right in thinking that either or both of these techniques this could possibly move an "earth-like" moon into the habitable zone, eventually?
ANSWER:
I am not going to address any of your hypothetical scenarios for delivering
the necessary energy because, as I shall show, this is not going to work
for any "current technology" you wish to put into play. The energy
necessary to remove Ganymede (MG≈1.5x1023
kg) from its orbit (R≈109 m) around Jupiter (MJ≈2x1027
kg) is E=GMJMG/R≈2x1031
J, where G=6.67x10-11 J∙m/kg2 is the
universal gravitational constant. Suppose that you wanted to achieve
this over a period of T=100 years≈1010 s. The required
power would be P=E/T≈2x1021 W=2,000,000,000 TW. The
current average power consumption of the entire earth is about 15 TW.
(By the way, Ganymede is not really "earth-like" at all since the
acceleration due to gravity at its surface is only about 1/5 of earth's.)
QUESTION:
What causes the birth of a black hole?
ANSWER:
When a star has exhausted most of its hydrogen fuel, it begins
to collapse under its own gravity. The next thing that happens is that it
explodes, a supernova. Then, providing that the star was massive enough, it
will continue collapsing until it has essentially collapsed into zero
volume, a black hole.
QUESTION:
I was at a training and learned that the Andromeda Galaxy revolves around a black hole. Is that true of all galaxies?
ANSWER:
Astrophysicists believe that nearly all large galaxies have a
supermassive black hole at their centers.
QUESTION:
If one stood on a typical neutron star and turned on a flashlight pointed horizontally, would the light bend down to the surface much like water coming out of a garden hose here on earth?
ANSWER:
The density of a typical neutron star is about 5x1017
kg/m3 and the radius is about 12 km, so the mass is about M=5x1017x4xπ(12x103)3/3=3.6x1030
kg. To find the acceleration due to gravity, g=GM/R2=6.67x10-11x3.6x1030/(12x103)2=1.7x1012
m/s2. So your weight on the surface would be about a trillion
times what it is on earth. You would be crushed totally flat. We can
estimate how far light would "fall" in such a gravitational field. The time
it would take to go 100 m would be about t≈100/(3x108)=(1/3)x10-6
s. The distance it would fall would be y=�gt2≈10 m.
On earth the speed of something which dropped 10 m in 100 m would be about
71 m/s≈160 mph, quite a bit faster than a garden hose. But qualitatively,
yes the light would fall quite a lot.
QUESTION:
Can you please tell me the difference between dark matter and dark energy.
ANSWER:
There are many instances where astronomical objects behave as if
there were more mass present than we can see. The best-known example is the
dynamics of galaxies where the speeds of stars far from the centers of the
galaxies are much bigger than expected on the basis of the mass that can be
seen; see the figure to the left. The conclusion usually drawn is that there
is much more mass present than we can see and this hypothesized stuff is
called dark matter. Nobody knows what it
is nor has it yet been detected directly. A second anomalous feature of the
universe is that it has recently been discovered that the expansion of the
universe is actually accelerating; the conclusion usually drawn is that
there is a repulsive force not heretofore observed which is "winning" a
tug-of-war with the attractive force of gravity. This is referred to as
dark energy.
QUESTION:
I understand you don't answer questions about stars, but im wondering about the nuclear dynamics involved in nuclear fission. In a star with 250 solar masses or more there is something called photofission, where high energy gamma rays cause elements as light as tin to go through nuclear fission. I was wondering, how much energy would be required to cause fission to occur in the element tin?
ANSWER:
Right, this is certainly more nuclear physics than astrophysics,
so I can do some rough estimates for you. I can tell you approximately how
much energy is released in the symmetric fission Sn112
�>2Mn56. The binding energies of the tin and manganese are 953.5
MeV and 489.3 MeV respectively. So the energy released in the fission would
be 2x489.3-953.5=25.1 MeV. What I cannot tell you is what energy photon
would be required to cause the fission. This depends on the structure of the
Sn112
and how it interacts with the photon. Fission might be induced by a much
lower energy photon than 25 MeV. After all, uranium can fission with no
energy whatever added to it (spontaneous
fission).
QUESTION:
What is the Cosmological Constant and why it couldn't be described perfectly by Einstein?
Was only because of it that Einstein thought the universe was static?
ANSWER:
When Einstein proposed the theory of general relativity, around
1918, it was generally believed that the universe was static. General
relativity is the theory of gravity, and if gravity is the main interaction
among stars and galaxies, this is not possible; with only gravity, the
universe would have to be either expanding and slowing down or compressing
and speeding up. Therefore Einstein had to introduce something to balance
the universal gravitational force; this something was called the
cosmological constant. It was later discovered that the universe is
expanding and he later denounced the cosmological constant as "�my greatest
blunder�" Interestingly, several years ago it was discovered that the
expansion of the universe is actually speeding up implying some kind of
repulsive force, akin to the cosmological constant, often referred to as
dark energy; this has reignited interest in Einstein' "greatest (or maybe
not) blunder". I do not know what you mean by "�couldn't be described
perfectly�"
QUESTION:
If I shine a light it emits a light particle outwards. Assuming it is travelling in vaccum all the way out and hits no light absorbing medium.
The farther it travels it starts losing energy and shifts to the red. Now how long in Earth years will it take for that photon particle to lose "ALL" its energy. How far will it travel in distance (light years) and what happens to that particle at the end of it all.. i.e it just cannot "disappear" or become void or null.
Or perhaps it will all be converted to some other form of energy but then what would it lose its energy to given it was hypothetically travelling in a vaccum/void all the way out?
ANSWER:
Let's first consider the particle to be a ball. If you throw it
up it will begin immediately losing energy and eventually stop. You can now
think of the earth as having acquired the energy the ball had, now in the
form of potential energy. Being a ball, it then falls back and extracts all
its energy back. However, if you throw the ball hard enough it will keep
going forever and the farther away it gets, the less energy it loses,
essentially moving with constant speed when it has gotten pretty far away.
The same is true of a photon, as you note, except for two important
differences: it cannot stop and it has no mass, just pure energy. The red
shift you refer to is the gravitational red shift and as the photon becomes
redder it is, as you stated, losing energy (also to the mass it is leaving)
but not speed. For the earth (and almost every object in the universe) the
effect of gravity is insufficient to ever cause the photon to lose all its
energy. Only black holes can rob a photon of all its energy; if a photon is
inside the event horizon of a black hole, as it tries to escape it gets
"redder and redder" until it has finally lost it all and it disappears. The
energy it had will now be found in a slightly increased mass of the black
hole.
QUESTION:
In looking at the photgraphs from the Rosetta comet mission, I notice what seems to be "loose boulders" on much of the surface. How much gravitational attraction does that comet have? Is it enough to keep the boulders on the surface as it would on Earth, or are the loose parts of the comet simply in a coinciding orbit?
ANSWER:
If they remain in the same place relative to the main body of the comet they are resting on the surface. If they were moving alongside, there would be an attractive force which would bring them together;
otherwise, to remain separated, the boulder would have to be orbiting the
main body. I once answered a question which surprised me�two
dice, in empty space and separated by 10 cm, take only about 8 hours to come
together.
QUESTION:
If I am at a point in space where very, very, very little gravitational pull exists, what happens to the time clock on my spaceship as opposed to clocks on earth?
ANSWER:
A clock in a gravitational field runs slower.
FOLLOWUP QUESTION:
What I am asking, if you were at a point in space, where the gravitational forces were almost 0, would your clock, in your space craft be running extremely fast as observed from earth?
I was thinking about a spot in the universe were the gravitational effect
would be almost opposite of a black hole. Like the trampoline example you used on the web site. I set 5 or 6 bowling balls on the trampoline so each makes a dent. I then use a marble and find a spot to set it where it is not affected by the bowling balls. Now I know my ship is moving always towards some mass but there has to be a spot where gravitational effects all almost 0. Maybe that spot in space would be inaccessible to my ship, kind of an anti event horizon where I would not have enough power to get too. a spot where Light can not even enter.
I Know that my question is flawed, but I can't help to think about a black hole as a gravity pit, and that makes me think of a spot in space where there is a gravitational mountain (Anti-Black Hole, with an anti-Event horizon). In this spot time runs so fast as not to exist....
ANSWER:
First of all, all your thrashing around with black holes etc.
trying to find a spot with small gravitational field is totally unnecessary;
all that is required is that you need to be far away from any mass and
nearly the whole universe satisfies that criterion. We generically call such
places intergalactic space. So, choose some place, maybe halfway
between here and the Andromeda galaxy, our nearest galactic neighbor. For
all intents and purposes the field there is zero and a clock would run at
some rate. Now, take that clock and put it on the surface of a sphere of
radius R and mass M. The
gravitational time dilation formula you need to compute how much slower
the clock would tick at its new location is √[1-(2MG/(Rc2))] where G=6.67x10-11
N∙m2/kg2 is the universal constant of gravitation and
c=3x108 m/s is the speed of light. Putting in the mass and
radius for the earth, I find that the space clock runs about 7x10-8
% faster than the earth clock, certainly not "extremely fast"! The
bottom line here is that, except very close to a black hole, gravitational
time dilation is a very small effect. An interesting fact, though, is that
corrections for this effect must made in GPS devices because they require
extremely accurate time measurements to get accurate distance measurements.
QUESTION:
How fast would a more massive planet than Earth, say twice as massive, with the same diameter have to spin to produce an Earth-like sense of gravity, due to centripetal force, at its equator? Might there exist a massive Earth-like planet, environmentally speaking, that we could still live on because its spin negated its gravitational pull enough to produce an Earth-like sense of gravity that wouldn't crush us?
ANSWER:
Let's just do it in general, the pseudo-earth being N
times more massive than our earth. The apparent weight at the equator would
be Nmg-Fc=mg
where Fc=mv2/R=mRω2 is
the centrifugal force. Solving, ω=√[(N-1)g/R]
where ω is the angular velocity in radians per second. For example,
if N=2, ω=√[g/R]=√[9.8/6.4x106]=1.24x10-3
s-1=2π/T where T is the period. So, T=5.08x103
s=1.41 hours.
QUESTION:
Suggest that we have constructed a model of the solar system by
decreasing the sizes of all orbits by n = 0.005. We have kept the average
density of the planets and that of the Sun the same as in the original solar
system. How will the periods of rotations of planets change in our model?
ANSWER:
Is this homework?
Are the sizes of the planets also scaled by n?
QUESTION:
No this is not homework-it was in a past paper and I was curious as to how to answer it!
And the question does not say, although I assume so!
ANSWER:
The period T of an elliptical orbit with semimajor axis
a around a body of mass M which is much bigger than the mass
of the orbiting body is given by T=2π√[a3/(GM)]
where G is the universal gravitational constant 6.67x10-11
N∙m2/kg2. So the answer to the question depends on how
the masses are scaled. If all the masses stay the same, the new periods
T' are simply scaled by √0.0053=3.54x10-4. If the
densities remain the same and the sizes of the sun and planets themselves
are also scaled down by 0.005, then the masses will scale by the factor M'=0.0053M
because the mass is the density (unchanged) times the volume
(proportional to the cube of the radii). Therefore the periods would remain
unchanged,
T'=2π√[(0.005∙a)3/(G∙0.0053∙M)]=T.
QUESTION:
I have a question that been thats been unanswered from quite some time now. I even tried asking that to a person in NASA houston and he couldn't even understand. Hope you can help.
We know that space dust and debris keep on falling on earth. And I have read that its many tons a day. Considering mass of earth is increasing every day for millions of years now, how does it affect its revolution, speed and axis. Or does it even affect that.
ANSWER:
One estimate is that the earth gains 40,000 metric tons (4x107
kg) per year. So, in a million years that would be a gain of 4x1013
kg. The mass of the earth is 6x1024 kg. The moment of inertia
will be proportional to the mass times the square of the radius, so,
assuming the change in radius is negligible, the fractional change in moment
of inertia over a million years would be approximately 4x1013/6x1024≈10-9=10-7
%. Since angular momentum (moment of inertia times angular velocity) is
conserved, this would mean that the angular velocity would decrease by about
10-7 %, an increase in the length of a day of about (24 hr)(3600
s/hr)x10-9≈10-3
seconds in a million years!
QUESTION:
This might seem dumb, but i cannot understand the laws of thermodynamics applied to gravity.
If a moon orbits a planet in a stable orbit, it still pulls on the planet, and the planet pulls on the moon. This seems to generate heat (the heatsource on europa) and motion (tidalwaves on earth). How can this not violate the laws of thermodynamics?
ANSWER:
The first law of thermodynamics, which seems to be the crux of
your question, is nothing more than conservation of energy. Energy
conservation is true only for isolated systems, so let us think about a
single planet and a single moon isolated from everything else. In addition
to the simple central force of gravity, the two interact with each other via
tidal forces which result
from the sizes and relative masses of the two. Just what the effects of the
tidal forces are depends a lot on the relative sizes and distances of the
two, but, as you note, often appear to violate energy conservation. So,
let's look at that more closely. Soon after the moon was created, probably
by a cataclysmic collision with an asteroid, it was much closer to the earth
than today and it rotated on its axis at a much faster rate than it does
today (about 28 days, which is why it always shows the same side to us). The
tidal force caused what is called "tidal locking", the same side always
facing the earth. It would seem that this would represent a loss of energy
since it is not spinning as fast as it used to be, but as this happened, the
moon moved farther and farther away from the earth which represents an
increase in energy. So, the energy of the whole system stayed the same.
Today, the tidal effect is mainly due to the ocean tides on the earth and
now it is the earth which slows down its rotation�the earth is losing
rotational kinetic energy and tending toward being tidally locked with the
moon. This is an extremely small effect�a day gets about 2 milliseconds
shorter during one century. But, just as was the case when the moon was
slowing down its rotation, the way energy is being conserved is by the moon
moving farther away, resulting in an increase in the earth-moon energy
keeping the total energy constant. If the tidal forces are great enough to
cause significant frictional heating because of the tidal force distortion
of the whole moon (your Europa example), energy is lost because some of the
heat is radiated into space. So the first law would be violated, but you
would expect that since some energy is escaping from the planet-moon system.
You can see that your error was to assume a "stable orbit".
QUESTION:
This might seem dumb, but i cannot understand the laws of thermodynamics applied to gravity.
If a moon orbits a planet in a stable orbit, it still pulls on the planet, and the planet pulls on the moon. This seems to generate heat (the heatsource on europa) and motion (tidalwaves on earth). How can this not violate the laws of thermodynamics?
ANSWER:
The first law of thermodynamics, which seems to be the crux of
your question, is nothing more than conservation of energy. Energy
conservation is true only for isolated systems, so let us think about a
single planet and a single moon isolated from everything else. In addition
to the simple central force of gravity, the two interact with each other via
tidal forces which result
from the sizes and relative masses of the two. Just what the effects of the
tidal forces are depends a lot on the relative sizes and distances of the
two, but, as you note, often appear to violate energy conservation. So,
let's look at that more closely. Soon after the moon was created, probably
by a cataclysmic collision with an asteroid, it was much closer to the earth
than today and it rotated on its axis at a much faster rate than it does
today (about 28 days, which is why it always shows the same side to us). The
tidal force caused what is called "tidal locking", the same side always
facing the earth. It would seem that this would represent a loss of energy
since it is not spinning as fast as it used to be, but as this happened, the
moon moved farther and farther away from the earth which represents an
increase in energy. So, the energy of the whole system stayed the same.
Today, the tidal effect is mainly due to the ocean tides on the earth and
now it is the earth which slows down its rotation�the earth is losing
rotational kinetic energy and tending toward being tidally locked with the
moon. This is an extremely small effect�a day gets about 2 milliseconds
shorter during one century. But, just as was the case when the moon was
slowing down its rotation, the way energy is being conserved is by the moon
moving farther away, resulting in an increase in the earth-moon energy
keeping the total energy constant. If the tidal forces are great enough to
cause significant frictional heating because of the tidal force distortion
of the whole moon (your Europa example), energy is lost because some of the
heat is radiated into space. So the first law would be violated, but you
would expect that since some energy is escaping from the planet-moon system.
You can see that your error was to assume a "stable orbit".
QUESTION:
Isn't the fact that Andromeda galaxy (being the galaxy supposedly going through a
collision with the Milky way in millions of years) defying the dynamics of
dark energy? Shouldn't it be moving AWAY?
ANSWER:
The expansion of the universe is the behavior on average, not
some kind of absolute rule. Imagine that you drop a stone�shouldn't it move
away from the earth? Local gravitational fields and initial conditions often
trump the overall expansion properties. The Andromeda and Milky Way galaxies
are attracted to each other gravitationally and since they never have had a
large velocity moving away from each other, they are doomed to fall into
each other.
QUESTION:
ok, two stones. Both spherical and same mass and
density evenly spread in each stone. Set each about 1/2 the distance to the
moon. One leading the earth's orbit and one following the earth's orbit.
Both not moving relative to the earth, yet the same speed as the earth as it
moves around the sun. No tangent or orbital speed, the stones are starting
in freefall. Which gets to earth's surface first, neglecting
air drag.
ANSWER:
Since you stipulate that the stones are not
orbiting, the stones are at rest with respect to the sun and the earth is
not. Therefore, the stone on the leading side of the orbiting earth will win
the race because the earth is moving toward it and away from the other when
the stones start dropping.
BETTER
ANSWER:
I see that I misread this question. I guess you meant there is no orbital
speed around the earth. To make this manageable at all I will neglect the
influence of the moon and assume a
spherically-symmetric mass distribution of the earth. In the figure
above, I show your two stones and the forces (blue arrows) on them. The
down-pointing forces are the from the sun (keeping them in orbit) and the
horizontal forces are the weights making them want to fall toward the earth.
The distances and forces are not drawn to scale; when the stones are about
30 earth radii away from the earth center (the moon is about 60 earth radii
away), the weight forces are about 2 times larger than the sun forces. The
big blue arrow shows the direction everyone is orbiting the sun. So, the
leading stone slows down its orbital speed and so will slightly fall toward
the sun as it falls toward the earth; the trailing
stone increases its orbital speed and so will slightly fall away from the
sun as it falls toward earth. These deflections are shown (probably quite
exaggerated) by the red arrows. Given the symmetry of the situation, I
would expect the two to be at the same distance from the center of the earth
at any given time�which is the
crux of your question, I think. To actually do this more quantitatively,
though, would be very hard because as the stones were deflected the weight
force would change direction now having a vertical component in the figure.
QUESTION:
What effect (if any) would there be on the solar system if the sun were to lose the mass of Jupiter? This question is based on various schemes to power a "warp drive" by converting mass about the size of the planet Jupiter into pure energy. Since vaporizing Jupiter is probably not a good idea, it might make more sense to get the necessary mass by skimming off a chunk of the sun's surface. Assuming mankind might have the technology to do this someday, would the sun be appreciably different if it lost the mass of Jupiter? Would the orbits of the planets change to a significant degree?
ANSWER:
I usually do not answer astronomy questions, but will
take a stab at this one since I have
earlier talked about effects
which Jupiter has on other objects in the solar system. If Jupiter
disappeared it would have a negligible effect on other planets. It would,
though, have a large impact on many asteroids whose orbits are controlled by
Jupiter's gravity (see that earlier
answer). This would release a swarm of asteroids into earth-crossing
orbits considerably increasing the likelihood of catastrophic collisions
with earth. The mass of Jupiter is only 0.1% that of the sun, so taking that
much mass from the sun would probably cause fairly minor effects on solar
system orbits. "Skimming" that much mass from the sun could have pretty
serious short-term effects on the sun's radiation. Oh, by the way, there is
no such thing as warp drive!
QUESTION:
Something
ridiculous I thought of, if the Moon suddenly stopped moving and began to
fall toward the Earth, how long would it take to impact? I'm stumped as to
how to calculate this, as the force on the Moon gradually increases as it
falls, and the Moon also pulls the Earth toward it, and the radius of each
object would have to be included.
ANSWER:
I guess I am going to have to put
questions like this one on the
FAQ page. You should read the details of these
earlier questions since I do
not want to go over all the details again. It is tedious and uninstructive
to try to do this kind of problem precisely. I, being a great advocate of
"back of the envelope" estimates, use Kepler's laws to solve this kind of
problem; I have found that a very excellent approximation to fall time can
be found this way. I note that the mass of the moon is only about 1% of the
mass of the earth, the period of the moon is about 28 days, and the moon's
orbit is very nearly circular. The trick here is to use Kepler's third law
and recognize that a vertical fall is equivalent to the very special orbit
of a straight line which is an ellipse of semimajor axis half the length of
the line. Kepler's third law tells us that (T2/T1)2=(R2/R1)3
where Ti is the period of orbit i and Ri
is the semimajor axis of orbit i. Now, T1=28 and R2=R1/2
and so T2=T1/√8=9.9 days. But this is
the time for this very eccentric orbit to complete a complete orbit, go back
out to where it was dropped from; so, the time we want is half that time,
4.9 days.But this is not what you
really wanted since I have treated the earth and the moon as point masses.
What you really want is when the two point masses are separated by a
distance of the sum of the earth and moon radii, 6.4x106+1.7x104≈6.4x106
m. To see how much error this causes, I can use the equation for the
velocity v at the position r=6.4x106 m if dropped
from r=Rmoon-orbit=3.85x108 m which I derived
in one
of the earlier answers: v=√[2GM(1/6.4x106-1/3.85x108)]=1.1x104
m/s. It would continue speeding up if the collision did not happen, but even
if it went with constant speed the time required would be about t=R/v=6.4x106/1.1x104=580
s=9.6 min. This is extremely small compared to the 4.9 day total time, so,
to at least two significant figures, 4.9 days is the answer to your
question.
An important part of doing physics, or any
science, is knowing when to eliminate things which are of negligible
importance!
QUESTION:
If our solar system was formed from a cloud of dust, created by numerous super nova explosions, where we get various elements, and the sun is comprised of the greater part of this cloud, how is it that the sun is comprised of hydrogen burning to make helium? How did the sun separate all the other elements?
ANSWER:
Again, I am not an astronomer/astrophysicist. However, I
was just reading in the April 2014 Scientific American an article
about stars early in the universe. When the universe was young, it was
composed almost entirely of hydrogen with only a tiny amount of helium and
lithium. When stars form, they contract gravitationally and heat up as they
do so. But, heating up, the pressure increases and keeps the star from
collapsing further until it can shed some of the heat so that it can
continue contracting. Eventually, a core dense and hot enough forms where
the fusion can ignite. It turns out that hydrogen is not very good at
getting rid of the heat and so more and more hydrogen accumulates;
eventually ignition occurs but the typical early star, because of the
inabilty of hydrogen to cool, is hugely bigger than the sun, anywhere from
100 to 1,000,000 solar masses. These giant stars now eventually die after
burning much hydrogen and creating lots of heavier elements up to iron; they
explode in supernovae and fall back into black holes, scattering the heavy
elements (including heavier than iron made in the supernovae) into space.
Now, to answer your question, the sun has lots of the heavier elements in
it, just a smaller fraction than the planets. It turns out that these
heavier elements are much more effective in cooling the protostar as it is
forming which allows much smaller stars like the sun to form.
QUESTION:
What is the moon's orbit around the earth? I was wondering if you could send me pictures and diagrams or whatever you could about the orbit.
ANSWER:
I am not sure what you want. The moon has a nearly circular
orbit of radius about 385,000 km. It takes about 28 days to go around earth
once (which is how it came to be that a month was a standard time
measurement). The same side of the moon always faces earth which means it
also takes about 28 days to rotate once on its axis. The picture above is
drawn to scale. For more detailed information, see the
Wikepedia entry.
QUESTION:
The number of known Mars Meteorites on Earth at last count I know of was greater than the number of Lunar meteorites. This is the opposite of what you would expect: The moon is 140 times closer than Mars with a weaker gravitational field to recapture the meteorites and no atmosphere to slow material blasted off the planet. The number of Lunar meteorites on Earth should be hundreds of times greater than Mars Meteorites.
ANSWER:
As always, a disclaimer that I am not an astronomer. The
origins of either lunar or Martian meteorites are major asteroid or comet
impacts; after impact, some of the debris has enough velocity to escape the
gravitational field. First of all, the cross sectional area of the moon is
about 4 times smaller than Mars, so all things being equal, the probability
of a major impact is 4 times greater for Mars. Second, Mars is closer to the
asteroid belt and would therefore, I presume, be more likely to suffer an
asteroid impact. Most major impacts occurred early in the history of the
solar system when embryonic planets were "sweeping up" material in their
orbits and there was a much greater potential of major impacts. These
early meteorites from the moon which landed on earth would have been eroded
away or taken under the crust by tectonic action; those from Mars would have
gone into orbits around the sun and land on the earth over a much longer
period, probably still landing today.
QUESTION:
During a discussion with my 6th grade class about the Law of Inertia and space travel, a student asked: "If a spacecraft leaves a solar system, for example Voyager 1, will its velocity increase due to the lack of gravity from the sun." Fairly certain the answer is no; however, it did spark a rather interesting debate. Can you explain?
ANSWER:
Perhaps the key is to note that "leaves the solar system"
does not mean that there is no longer any gravity from the sun.
Rather there is a boundary where the solar wind, particles like protons and
electrons, stops; this is called the heliopause and has been definitely
observed by the Voyager 1 spacecraft as shown in the graph to the right.
This shows the amount of solar wind the Voyager detected as it moved during
the months of 2011-2. In September 2012 it dropped to near zero. This is
where we define the edge of the solar system to be and it is about 50
Astronomical Units (AU) from the sun; the earth is one AU from the sun. But
the gravity from the whole solar system is still present and the craft will
continue slowing down, but ever so slightly since as you get farther away
from a mass the gravity gets weaker. This craft has enough speed that, if it
never encountered any other mass it would keep going forever. I did a rough
calculation and found that the acceleration of Voyager is about a=-0.01
mi/hr2 which means that it loses about 1/100 mph per hour; but
the speed is about 40,000 mph, so I think we could agree that it is moving
with an almost constant speed. As it gets farther away, the acceleration
will get even smaller (physicists call slowing down negative acceleration,
not deceleration). Until it gets close to something else, like some other
star, it will keep going with an almost constant speed. When it does
approach another star it will begin speeding up. Your students should
appreciate that the only thing which can change the speed of something is a
force, a push or a pull.
QUESTION:
Isnt the gravitational pull of an object determined SOLELY on the mass of that object and NOT its size or density? obviously distance from said object plays a role just as size and density affect mass.... however, if what i believe is true, there can be no such thing as black holes, which even today their existence is only theoretical, not proven... anyway my point is this... i believe that if our sun was the size of a basketball but still had the same mass it has today, (giving it almost infinite density)..(what many scientists today consider a 'black hole') but the Earth was still the same 93,000,000 miles from the actual surface as it is today, the orbit of the Earth and the planets would remain, and the sun would not be considered a 'black hole', nor would it behave like one, (sucking in all its orbiting masses, and even light itself).. it would be much smaller, but given the exact same mass and distances surface to surface, little would change as far as orbits go... could this be correct theoretically?
ANSWER:
If the object is spherically symmetric and you are totally
outside the mass distribution, then you are right�only the total mass
matters. But, this does not mean that black holes do not exist. It is
dramatic to say that nothing can escape a black hole, not even light, if you
are inside a critical distance called the Schwartzchild radius, but that
does not mean that objects could not orbit the black hole, either inside or
outside that radius. If the sun were a black hole, its Schwartzchild radius
would be
R=2GM/c2=2x6.7x10-11x2x1030/(3x108)2=3,000
m; so all the planets would be outside and would orbit just fine; they could
even be dragged away if you wanted. Only objects inside 3000 m would be
"trapped" but even they could still orbit the black hole. The figure to the
left shows some orbits of stars around the supermassive black hole at the
center of the galaxy which have been observed by astronomers.
QUESTION:
I'm not a physicist or mathematician, but could there be a better explanation to the expansion of the universe without assuming so much dark matter? The galaxies accelerating away from one another seems to be analogous to the same way a sudden release of gas molecules would disperse across a large room, or the way a drop of liquid would disperse across a vast body of water.
ANSWER:
First, let's get some things clear. Dark matter is a proposed
type of matter which exerts an attractive force (gravity) on all other
matter and therefore works against the expansion of the universe because it
is trying to pull the rest of the universe toward it. The acceleration of
the universe is the result of some proposed repulsive force trying to push
matter away, and this is called dark energy. Even without dark energy or
dark matter, we can talk about the expansion of the universe. Imagine there
is an object, say a planet, in the middle of empty space which suddenly
explodes in to numerous pieces. There are essentially two possible ultimate
outcomes. If a relatively small amount of energy was added by the explosion,
the pieces will all move away from each other slowly but, because they are
all attracting each other gravitationally, they will all eventually turn
around and come back to recoalesce into the planet. However, if enough
energy is added by the explosion, the pieces will have such a large velocity
that they will never come back, they will just keep expanding forever. (This
is analogous to giving a rocket a high enough velocity, the escape velocity,
so that it will never fall back to earth.) One of the most important goals
of astrophysics over the last century has been to determine which is the
case for our universe, the big bang being the explosion which started it
all. Now, the problem which prompted astrophysicists to hypothesize dark
matter was not the expansion of the whole universe, but the way that
galaxies move; the way the stars in a galaxy revolve around the center
cannot be understood as resulting only from gravitational interactions among
all the matter in that galaxy. The more recent discovery of the observed
acceleration of very distant galaxies is what requires something like dark
energy, the source of something pushing them apart, "beating out" the
gravity which is trying to slow them down. Neither of your two examples have
any mechanism which would accelerate the expansions. My own feeling, totally
not mainstream, is that perhaps these problems are simply indicative of our
not understanding gravity as well as we think we do.
QUESTION:
How far from Earth would you have to travel before the constellations changed shape?
ANSWER:
Of course, this is impossible to answer in a quantitative was
because different constellations are made up of different stars which are
different distances away from us. Also, you need to have some criterion for
"changed shape". In some sense, you already can observe a tiny change in
shape due to moving around the sun as shown in the picture to the left. When
the earth is at position 1, a nearby star (yellow) is observed at one
location relative to very distant stars (orange and blue) and 6 months later
at position 2 it will be observed to have apparently moved, thereby changing
the shape of this 3-star constellation. But by how much? The nearest star to
us is about d=4.22 ly (light years) from us and the radius of the
earth's orbit (called an astronomical unit, AU) is about B=1
AU=1.6x10-5 ly, so the angle p (called the parallax angle)
is about 1.6x10-5/4.22=3.8x10-6
radians=0.000220≈1" (one arcsecond, 1/3600 of one degree). You
would probably not consider that to be much of a change! So, suppose we now
asked how big the earth's orbit would have to be to have a parallax angle of
100=0.175 radians. In that case we would have B/4.22=0.175
or B=0.74 ly. So finally we can conclude that if you want the
position of a star in the sky to change by about 100, you will
need to move laterally about 20% of the distance to that star. Be sure to
note that most stars are enormously more distant than a few light years from
us.
QUESTION:
Is there something wrong with the theory that says, if there is enough dead matter from dying stars that comes together, gravity will take over and eventually produce a new star from this dead matter?
I am aware that stars form from hydrogen, but what if 999 trillion tons of the heaviest known elements got together all in one place? Most say a star would not form because there is no hydrogen, but�what then? It would appear that gravity, being indiscriminate, will continue crunching this big iron ball until critical mass is surpassed, regardless of what the result might end up being. What would that result be?
ANSWER:
As I state on the site, I usually do not do
astronomy/astrophysics/cosmology questions, so take my answer here with a
grain of salt. But, sometimes the question is sufficiently interesting to me
that I attempt an answer. I would say that a "normal star" might be defined
as an object which makes energy by
nuclear fusion, fusing light nuclei to heavier nuclei. However, when you
reach iron on the periodic table you no longer gain energy by fusing nuclei
and therefore the energy production ceases. Before the whole star becomes
iron, though, other instabilities cause the star to, in one way or another,
end its life, the details depending on the mass of the particular star. The
most dramatic end is a supernova (of which there are several types, again
depending on details), a very dramatic explosion of the whole star; the
energy from such an event is what is generally thought to be responsible for
creation of heavier elements beyond iron. One possible end following a
supernova explosion is called a neutron star. The remnants of the star,
mostly heavy elements like you refer to, will undergo gravitational collapse
and, in essence, electrons will be pushed into protons such that the star is
made mostly of neutrons and is extraordinarily dense (it is essentially a
giant nucleus). With sufficient mass, that is what I believe would happen to
your collection of heavy elements. However, your number, about 1018
kg, is way too small for the requisite mass. Neutron stars have about
1.4-3.2 solar masses and the mass of the sun is about 2x1030 kg.
QUESTION:
If we can talk about "escape vel. for light ,(in case of black holes)"
that mean light is affected by gravitational fields and thus like other
masses, must be "decelerated", while moving away from a heavy planet.
Thus, can we slow down light to near zero speed? I know the ans is 'no',
but WHY? if light is affected by gravity then we should be able to slow
it down? Also if light is affected by gravity , then what's it's mass??
ANSWER:
As light travels upward in a gravitational field, it loses
energy. If light were a classical object like a baseball, you would say
that, since its energy is �mv2 and the mass is constant,
the velocity must decrease. However, the energy of light, since it has no
mass, is not �mv2 but rather each photon has an energy
hf where h is Planck's constant, and f is the frequency of
the light. So, as it goes higher its frequency gets smaller�the wavelength
gets longer; that is why it is called gravitational red shift. In the
case of a black hole, eventually the energy of the light is zero and the
black hole has gained all the energy which the light had. Similarly, if
light is aimed down in a gravitational field it does not speed up but rather
increases frequency as it goes down, gravitational blue shift. Light
does not have mass but it is affected by gravity. Newton's universal law of
gravitation is not the last word; the modern theory of gravity is general
relativity and if you go to my FAQ
page you can learn about how gravity affects massless objects.
QUESTION:
If energy can neither be created nor destroyed, how can people who believe in the bing bang believe what they believe since we have energy now so they are saying that it was created by something in the natural world which we know is not possible.
ANSWER:
The law of conservation of energy applies only to our
universe after its creation. Nobody knows where the original energy came
from for the "bing bang". I should also point out that energy conservation
can be violated in nature; the Heisenberg uncertainty principle allows
energy to appear out of nothing but only for short times.
QUESTION:
i am a university senior student. my department is elementary science education. I know that every atom that is in our planet was once created in stars. I know that first H used to create He and all the way to iron. However, iron is not the last element in periodic table how other elements are formed? Thank you for your answer.
ANSWER:
As you note, iron is the heaviest element created in stars;
the reason is that iron is the most tightly-bound nucleus and therefore any
heavier elements will require that energy be added rather than energy being
released by fusion (which is what stars do). Heavier elements are created in
very energetic events like novae and supernovae.
Recently it has been found that collisions between neutron stars also
play an important role, particularly for the element gold.
QUESTION:
From what I understand, the vast majority of the large bodies in the solar system rotate and orbit in a counterclockwise fashion relative to their magnetic north pole. Is this true for most/all solar systems in this galaxy/the observable universe, or is it different every time? Why is this?
ANSWER:
First, you have a misunderstanding. Most rotations in our
solar system are counterclockwise as viewed from the North Star
(Polaris). The magnetic pole of any particular body has nothing to do with
this rule. In fact, the earth's NS poles have swapped positions many times
in the distant past. Also, many of the bodies in the solar system (like our
moon) have no or negligible magnetic field. So, to state the rule you need
to specify whether you are viewing the solar system from "above" (which I
will call looking down on our north pole) or "below" (which I will call
looking down on the south pole); in the latter case, the rule would say all
rotations are clockwise. The reason is that in the current best theory for
formation of the solar system says that a huge cloud of gas and dust
collapsed to form it. That cloud will have an angular momentum (net
rotation) which remains and speeds up as the cloud collapses. To see more
detail, see an earlier answer. Any other
planetary system in the galaxy will have a similar rule with most objects
spinning the same direction, but not likely in the same direction as ours.
Other galaxies do not have the same rotation as our Milky Way galaxy because
when we see them, some are seen edge on and others are seen face on. (Both
of those pictured are spiral galaxies which spin like a big plate.)
QUESTION:
i am currently working on a book that deals a lot with space. while it is technically fiction, it's not heavily science fiction (no aliens or new planets or anything of the like) and i want it to be as accurate as possible. my question is, is it possible for a satellite/spacecraft to be able to have and elliptical orbit around the earth and moon, with the two bodies being the elliptical centers. the main problem i see with this idea is that the moon, well, moves, which could wreak havoc on the orbit, but i'm still wondering if it is still theoretically possible.
ANSWER:
If, by "elliptical centers" you mean foci, the answer is
absolutely no. The
three-body problem has no general analytical solution. In your special
case, one approximate solution would be that if the satellite/spacecraft
were at all times very far away compared to the earth-moon distance, the
orbit would be approximately an ellipse with the center of mass of the earth
moon system at one focus of the ellipse. Another interesting case is if the
moon-earth system is approximated to have circular orbits (about their
center of mass) and the satellite/spacecraft moves in the same plane. In
that case the satellite will have a closed orbit (called a
horseshoe orbit)
or even be stationary points (called
LaGrange points)
as seen from the (rotating) earth-moon system. The picture to the right
shows some possible orbits around the sun-earth system, but would be similar
for the earth- (replacing the sun) -moon (replacing the earth) system; also
shown are the LaGrange points (L1-L5). The picture to the left shows the
asteroid belt (white dots) along with other asteroids (green dots) which
orbit with Jupiter near the L5 and L4 LaGrange points and still other
asteroids (orange points) which are referred to as Hildas; for a more
detailed discussion of the Hildas which have elliptical orbits, see the
article by
John Baez. An animation of the Trojans, Greeks, and Hildas is shown
below; keep in mind that this is in the frame of Jupiter and the sun which
is approximately stationary but the picture from outside would have Jupiter
going around the sun counterclockwise. I
think these are the most interesting solutions for your situation, but
there are numerous other special analytical situations of the three-body
problem which you can research starting with the
Wikepedia article.
QUESTION:
Im sitting here watching the science channel, which I love! And they are talking about the core of Jupiter, which is by popular discussion they believe it is mostly metallic hydrogen. Now my question is, "if you had a 1 inch cube of metallic hydrogen, and took away the extreme pressure, would it stay metallic or dissolve into back into a gas?
ANSWER:
Metalic hydrogen is atomic hydrogen (not H2,
molecular hydrogen) which behaves like a conductor and which occurs at high
pressures. The phase diagram to the left shows that metallic hydrogen cannot
occur at pressures less than about 100 GPa≈1000 atmospheres or temperatures
above about 2000 K. Lower temperatures, say around 300 K (near room
temperature), require pressures more like 350 GPa≈3500 atmospheres. To the
right is a phase diagram for H2 at much lower temperatures and
pressures and you can see that
hydrogen will be a gas of H2 for normal conditions. Note that
atmospheric pressure is about 1 on the pressure scale and the temperatures
are far below room temperature So, the
answer to your question is that putting 1 in3 of metallic
hydrogen in an environment of normal pressure and temperature would cause it
to change to a gas of H2 molecules (hugely expanded in volume).
NOTE
ADDED:
I found this "Jupiter cake". Cake depicting the layers of Jupiter! The center (mudcake) is the theoretical rock/ice core, then comes the (almond butter) layer of liquid metallic hydrogen, and finally the (colored vanilla) liquid molecular hydrogen.
QUESTION:
What causes matter to rotate/spin/orbit? All I can find is the
statement that in space particles of dust/gas/matter contract into a
spinning disk due to gravity (to form stars, solar systems, galaxies
etc.), with no explanation as to why the spin. I see a lot about the
conservation of angular momentum, but these discussions all presume that
the 'spin' already exists. What caused the spin in the first place?
Shouldn�t gravity simply attract particles of dust, gas or matter
together along a straight path till they collide, as a magnet does to a
paper clip? The magnet does not make the paper clip revolve around it,
and if I fall off of a building, I don�t spin around the earth. I fall
in a straight path till I collide with the earth. What am I missing?
ANSWER:
A star and its planets form from a cloud of dust and gas
which is hugely bigger than the size of the final star-planet system.
Imagine that this cloud is a million times bigger than the final system.
Each dust particle and molecule is moving with a random velocity; labeling
each member of the cloud by the number i, the angular momentum ℓi
of each particle of mass mi and speed vi
is miviθri=ℓi
where ri is the distance to the center of mass of
the cloud and viθ is the component of the velocity which
is perpendicular to the vector ri. Notice that if
at large distances out into the cloud all the viθ do not
add up to zero exactly, even if the sum is a very tiny number (the cloud is
almost not rotating) the angular momentum can be appreciable because
ri is very large. Probably the whole cloud has an
imperceptibly small spin but it still can have a large angular momentum. Now
the cloud starts collapsing because of its gravity. There is a very
important law in physics called conservation of angular momentum: the
total angular momentum of any isolated system must remain constant.
Here, isolated means that there is nothing which exerts torques on the
system and since this cloud is far from other large objects, it is
essentially isolated. As the cloud collapses, ri for each
particle gets smaller and so viθ for each particle
must get bigger by the same factor to conserve momentum. So, if the cloud
shrinks by a factor of a million, it will end up with a million times
greater spin. Also, see an earlier
answer.
QUESTION:
I am a teacher in New Zealand, and am much more familiar with Chemistry than with Physics. I have been trying recently to understand magnetism better, in general, and as applied to Earth and Space Sciences. When a permanent magnet is created, what is actually lining up to create the magnetic field? I have heard it described as atoms, or pockets of charge, or micromagnets, but what does this chemically look like? Are they polar molecules within the solid? My understanding of bonding of metals doesn't include any polarity, they are positive nuclei in a 'sea of electrons', so how could a metal be aligned by charge? Is it an alteration of electron orbitals? Am I over-thinking this?
ANSWER:
An electron is, itself, like a tiny bar magnet, called an
electric dipole moment. In most materials, the electrons point in random
directions resulting in no bulk magnetism. In a ferromagnetic material like
iron, there is a quantum mechanical effect which causes the electrons
responsible for bonding to neighbors in the crystal to align their dipole
moments. An ordinary piece of iron is normally not a magnet because the
aligning of electron magnets happens only below a certain temperature called
the Curie temperature and when the iron cools down the alignment occurs
in local small volumes, each of which is pointing in a random direction;
these are called domains and are hugely bigger than atoms but still
microscopic. To magnetize the iron, you put it in a strong magnetic field
and the domains already aligned with the field will grow at the expense of
those not aligned with the field. You can also cool iron in a strong
magnetic field and then most domains will form being aligned with the
external field.
CONTINUED
QUESTION:
Also, how does the movement of charged particles in the Earth's liquid outer core, or in the Sun's convective zone, actually produce a magnetic field? I have read about this effect many times, but I just can't picture chemically what this would look like. Is it that the general movemnet of ions within a fluid tends to align in the same direction over time, thus influencing other ions to do the same, and this spreading the force?
ANSWER:
(In future, please follow site
groundrules and submit single
questions!) Any electric current causes a magnetic field. So a simple
model of the earth's field would be to imagine a huge ring of current inside
the earth. The magnetic field of a current ring is shown to the left.
Compare this to the field of the earth shown to the right. Clearly a model
where the core is thought of as a collection of current loops gives rise to
a reasonable description of the earth's field. You are barking up the wrong tree
by trying to understand it chemically. I should also note that the details
of the earth's field mechanism are not fully understood.
QUESTION:
So if black holes have so much mass that light can't even escape and gravity has almost an infiniate range of influance, when our universe is at its end and black wholes have swallowed up everything will the black holes start to move toward each other and then come together? If so would this singularity be so heavy that it would rip space time and then "big bang" again?
ANSWER:
Again, I emphasize that I usually do not answer
astronomy/astrophysics/cosmology questions. I believe, though, that the
current best model of the universe has, because of dark energy, the most
likely fate of the universe not recollapsing (big crunch) but ever expanding
and black holes eventually all evaporating via
Hawking radiation.
QUESTION:
Is there a factor in Newtonian Gravity similar to the factor for precession as in GR(ie., the factor in the Schwarzschild equation for planet precession)?
ANSWER:
If you are wondering whether an orbit precession is possible
in Newtonian gravity, the answer is yes. For example, if the sun were not
perfectly spherical (which it is not), orbit precession is possible.
QUESTION:
I�m not sure if you would consider this an �off the wall� question (it seems like a �deceptively simple� type to me), but here it goes: what would be the relative density (to water�s 1 (1000 kg/m^3) say) of something with a theoretical �complete density,� i.e. something with no empty space between mass components (atoms, quarks, etc.)�a celestial goo, if you will. Would this be a singularity material or would this be less than singularity? Some sources list a singularity as being infinite density, so, if that is the case, I�d guess relative comparisons might be pointless.
Part of what I am trying to figure out is how deceptive the term "density" actually is; I am thinking it will end up being something like the differential between the Fahrenheit scale and Kelvin. I would also like to know the how much �more� surface area a 1 meter by 1 meter square of this material would have than the traditional one square meter (of water I guess) of any traditional material. In that case, I figure ewe have always assumed complete density with surface area calculations, but, for some calculations I need to do, I need to know what �true� surface area is.
ANSWER:
Well, you are pretty close to "off the wall" here! First, a
singularity has infinite density and zero surface area. A black hole is
thought to be such a thing. I could quit there since this is certainly the
limit. However, maybe you want something somewhere between the density of
normal matter and infinite density, something which does not depend on what
the material is. If we think about all matter as being composed of
electrons, neutrons, and protons, we can imagine somehow compressing them
all together to make some "primal stuff". Since electrons have very small
mass compared to neutrons and protons, the number you seek is the density of
nuclear matter, that is, the mass of a typical nucleus divided by its
volume; that number is about 2x1017 kg/m3; so, that is
2x1014 times more dense than water. There are
actually, in nature, objects with this density.
Some stars, late in their lifetimes, collapse under gravitational attraction
to this density and the enormous pressure causes electrons to be "forced" into the protons to make
neutrons; they are called
neutron stars.
Regarding your question about area, it really has no meaning because you
cannot have a square meter of something, it always has a volume as well. If
you had a sphere of nuclear matter of radius 1 m, it would have a surface
area of 12.6 m2
and a mass of 8.4x1017 kg. I guess you could imagine a "sheet of
neutrons" with this density and ask what its area was. Taking the radius of
a neutron to be about 10-15 m, the volume of a sheet of area
1 m2
would be 2x10-15
m3
so its mass would be 400 kg. Taking the radius of a water molecule to be
about 10-10 m, a sheet of water of mass 400 kg would have an area
of about 4x109 m2.
QUESTION:
With all the space junk circling our planet, I wonder how much mass do we have to lose before it affects our orbit around the sun?
ANSWER:
First of all, we have not lost that mass, it is still
attached to the earth. Second, the orbit of an orbiting body is independent
of the mass; a 1 gram earth, given the same orbit and speed, would take one
year to go around. For example, there are hundreds of communications
satellites in orbits high above the equator and all have a period of 24
hours, although they have many different masses.
QUESTION:
If all the galaxies are accelerating away then is nothing an
intertial reference frame? If so, why do Newton's laws work the way they
do?
ANSWER:
I am not, as I emphasize on the home page, a cosmologist, so
take what I say with a grain of salt. I do believe, though, that there is no
such thing as an inertial frame of reference and it really has nothing to do
with acceleration. General relativity posits the equivalence of accelerating
frames and frames in a gravitational field, so any frame where there is any
gravitational field is not an inertial frame. Because there is no place in
the universe where there is truly zero field, there are no inertial frames.
Therefore, Newton's laws are false! However, there are many frames which are
very close to inertial and for which they are almost not false. On earth, a
patently noninertial frame, Newton's laws work remarkably well in many
situations. For much of celestial mechanics (calculating orbits of
satellites and planets, for example), Newton's laws work almost perfectly.
Special relativity originally overthrew Newtonian mechanics which does not
work, even in an inertial frame, at very high speeds without serious
modification of definitions of things like linear momentum and energy.
QUESTION:
which takes more fuel- a voyage from earth to moon or from moon to earth?
ANSWER:
You can get a pretty good idea by just looking at the escape
velocities from the earth and from the moon. From the earth, ve=11.2 km/s
and from the moon, ve=2.4 km/s.
Since kinetic energy is proportional to ve2,
it would take about 23 times as much fuel to escape from the earth as to
escape from the moon.
QUESTION:
We've all been reading about dark matter and dark energy for some years now and I believe you've said that you (among many others) are not yet persuaded that dark energy and dark matter exist. If matter and energy, as traditionally and conventionally understood, comprise only a very small part of the substance of the Universe, does it follow that classical mechanics, thermodynamics, relativity and quantum theory, etc., correspondingly apply only to that (seemingly) tiny aspect of the world? Is there any reason to think that the laws and theories of physics that humanity has discerned to date would apply also to dark matter and dark energy?
ANSWER:
First, a disclaimer: as I state on the site, I am not an
expert in astrophysics, astronomy, or cosmology, so you can take my opinion
with at least a grain of salt or ignore it altogether! I would not say "many
others"! Most astrophysicists, astronomers, and cosmologists talk about dark
matter as if it is surely there but just not directly observed yet. My own
point of view is that I need to see some direct evidence before I accept
that such a thing really is there; there is lots of indirect evidence of
dark matter―the dynamics of galaxies, the time when galaxies first began to
form, to name a couple―but it is altogether possible that we do not
understand gravity as well as we assume that we do. The best theory of
gravity, general relativity, makes many assumptions which are not
necessarily true over really large distances. If this were the case, maybe
dark matter is the 21st century equivalent of the lumeniferous
�ther (see the following answer) and is something we are looking for in vain
because there is no such thing. There are lots of good ideas about what dark
matter might be (including WIMPS, for which some
evidence has recently been observed in the observed excess of
high-energy positrons) and I will be happy to accept experimental evidence
when it happens. Dark energy is a different matter in that it has not caused
a search for some "stuff". There is already a place for dark energy in
general relativity, known as the
"cosmological
constant"; in other words, many cosmologists do have the point of view
that dark energy does result from an incomplete theory of gravity.
QUESTION:
why can nothing escape event horizon when all it needs to escape
is just to exceed the gravitational force and not to reach the escape
velocity-speed of light in this case- the way rockets on earth do as far
as I know they do not cross the escape velocity
ANSWER:
The escape velocity Vescape is defined to
be the minimum speed something must have to escape an object of mass M
to a distance infinitely far away and from a distance R away from
M. This is easily shown in classical physics to be given as Vescape=√(2GM/R)
where G is the universal gravitational constant. Suppose that an
astronomical object exists such that M is large enough and R
can be small enough that Vescape=c, c being
the speed of light. Then R=2GM/c2≡Rs,
this being the Schwartzchild radius. Since no object with mass can travel as
fast as the speed of light, anything inside this radius cannot escape. Also,
light itself inside this radius (for which Vescape>c)
could not escape either since the speed of light is always equal to the
speed of light. Technically, the calculation of
Vescape is not correctly calculated using classical
physics for either photons or masses going at relativistic speeds, but this
is one of those cases where a correct relativistic calculation gives the
same results as the classical calculation.
QUESTION:
Even though mars weighs one ninth of earth why does is have 38 percent of earth's gravity?
ANSWER:
For a spherically symmetric mass, the gravity at the
surface is proportional to the mass of the planet and inversely
proportional to the square of its radius. For Mars the radius is RM=0.533RE
and the mass is MM=0.107ME, so
MM/RM2=(0.107/0.5332)ME/RE2=0.38ME/RE2.
There is your 38%.
QUESTION:
Are the eliptic orbital planes of all our solar systems planets on the same 2d plane, or are there angles between the planes of the planets orbital elipses' ?
ANSWER:
Orbits of the planets are nearly coplanar. Inclination, measured
relative to the plane of the earth's orbit, are all less than 3.50
except for Mercury which is about 70. Pluto, no longer considered
a planet, has an inclination of 170.
QUESTION:
Could you explain the origin of the inverse square law that occurrs in nature? Why can't Newton's law of gravitation vary as R^3 or R^5?
How did Newton deduce this so long ago?
Could it be because of the geometry of space when something radiates from a central point source...as a sphere gets larger & larger radii?
ANSWER:
Let us start with the historical question of how Newton could deduce the
inverse square law for gravity and that will take care of your questions of
why it cannot possibly be some other power law. There were data available to
Newton which described the motion of the planets; these were
Kepler's three laws. These can only be described by an inverse square
law. So once again we see that, at its heart, physics is an experimental
science and data dictate theory. Is there a deeper reason? As I see it, that
is part of your question too. The electrostatic force is also an inverse
square law (Coulomb's law) and in this case the theory of quantum
electrodynamics correctly predicts the 1/r2 dependence of
the force. One reason is that the quanta which transmit the field,
photons, are massless;
forces like the nuclear strong interaction are mediated by particles with
mass and the force is a much shorter-ranged force. For gravity, it is, as
you suggest, related to the geometry of space (the area surrounding a point
increases like r2 as the distance r increases) as best we
understand. General relativity, our best theory of gravity, correctly
predicts the force to be inverse square law. Although there is no theory of
quantum gravity, the mass of a graviton, if it exists, is expected to be
zero because the force law is what it is.
QUESTION:
Why the orbit of Mercury around the Sun cannot be all
accouted from the Newton's Law of Gravity, because there is a disprency
between observation and Newton's Law.
ANSWER:
The orbit of Mercury is an elipse and the semimajor axis of this elipse
precesses in time and this should not happen in Newtonian physics. However,
this is predicted by the theory of general relativity and is one of the
conerstone tests of that theory.
QUESTION:
I'm doing a report for my physics class on gravity and I was
wondering if the force of gravity from the sun and the size of the
planets determines the distance that the planets are from the sun.
ANSWER:
The size of the planets has nothing to do with it. The orbital period is
independent of the mass. For example, any object, regardless of its mass,
which orbits the earth near the earth's surface (like the shuttle) has a
period of about an hour and a half. The distances of the planets from the
sun is purely accidental. The distance does determine the period of the
orbit (the length of a "year" on that planet). If the orbit is circular
(which is approximately true for all planets), then the square of the period
is proportional to the cube of the distance to the sun; this is Kepler's
third law. For example, if a planet is a distance twice as far from the sun
as the earth, then its period T may be found from (1/2)3=(1/T)2
, so T=√8=2.83 years. To learn about
Kepler's laws, see the Wikepedia entry.
QUESTION:
Hey dude,
With the endless amount of stuff floating around in space, how come none of it hits earth?
And to be more specific, how come nothing so massive and unstoppable by earths defensive capabilities ever comes towards earth and destroys it?
ANSWER:
Hey Dude back at ya! It happens all the time�what
do you think shooting stars are? And, tons of space dust hit the earth every
year. In the early solar system, there were a lot more pieces of stuff
"floating around" and there were many more collisions with bigger objects
than there are today. One collision with an asteroid-sized object is thought
to have caused a major extinction of life millions of years ago. But most
remaining objects are in reasonbly stable orbits now. And, you just have no
idea how mostly empty space is�just statistically it is immensely improbable
that collisions will occur. Imagine two guys with rifles 5 miles apart
firing their rifles; what is the likelihood that two of their bullets will
collide? There was one major collision in 1994 when Comet Shoemaker-Levy hit
Jupiter. Many asteroids have highly variable orbits which could possibly
result in a collision with earth, but it is a very low probability event.
QUESTION:
Galaxies recede from us at a velocity proportional to their distance. Why then is Andromeda on a collision course with our beloved Milky Way?
ANSWER:
The overall average behavior is that the universe is expanding.
However, at distances much smaller than the size of the universe there is no
fundmental reason why everything has to be moving apart. For example, the
sun is not moving away from us. And, there are attractive gravitational
forces which can be important between nearby galaxies. Astronomers have
observed many examples of galaxies colliding. The Milky Way and Andromeda
galaxies will collide in about 5 billion years. You can see a simulation of
that collision at
this link.
QUESTION:
If a black hole pulls material in doesn't it gain mass and therefore slow down?
ANSWER:
Yes, it gains mass. But, it gains the mass by interacting with
something else. So, to find out what the black hole is doing after the
collision, you have to conserve momentum of the event when it gained that
mass. It is pretty much like the collsion of two putty balls�you
start with two and end with one. For example, suppose the black hole is at
rest and the to-be-captured object is far away and has a velocity v
directly toward the black hole and a mass m. So the momentum of the
system is mv. After the collision, the mass must still be mv (that is
what momentum conservation means). So, if the mass of the black hole were
999m, the speed after the collision would be V=mv/(1000m)=v/1000.
You could go through the whole thing for other initial conditions, for
example if the black hole were moving with a speed v and the object
were at rest, the speed after the collision would be V=999mv/(1001m)=0.998v,
slower, as you suggested, but not really for the reason you suggested. (I
have done this nonrelativistically, but it conveys the idea.)
QUESTION:
If the Big Bang is the result of everything in the universe is rushing away from each other then why is the Andromeda galaxy hurtling toward the Milky Way galaxy? Should they move in opposite directions? Did something more massive than the Andromeda galaxy fling it our way?
ANSWER:
The fact that the universe is expanding does not mean that
everything is moving away from earth. For example, sometimes Venus moves
in a direction toward the earth, sometimes away, depending on the
positions we are in our orbits. Almost everything very far away moves
away from us, but things closer, like other stars in our galaxy or
nearby galaxies, may move toward us. This can be understood in terms of
initial conditions and forces. For example, if two galaxies form
reasonably close together initially at rest with respect to each other,
their mutual gravitational attraction will pull them together.
QUESTION:
I read that the Andromeda Galaxy will collide with the Milky Way Galaxy in about 3 billion years. How could this occur unless the Big Bang allowed for a change of direction, OR there was more than one Big Bang?
ANSWER:
I do not normally answer astrophysics questions, but I can handle
this one. Just because the universe is expanding does not necessarily
mean that everything is moving away from everything else. Imagine two
galaxies which are originally moving away from some point in space but
fairly close to each other. Then, if they keep moving like this, each
will see the other moving away. But, they are gravitationally attracted
to each other. Therefore, as they move away from the point, they will
begin moving toward each other; so looking at one from the other, we
will see it approaching. We know that the universe is made up of local
clusters of galaxies which are gravitationally bound to each other and
therefore orbit each other, collide with each other, etc. But
if you look at that cluster, you will find that the average of all those
motions is that the cluster moves away from us.
QUESTION:
I read the other day that Tycho Brahe didn't think the earth could be rotating partly because if it did, then a cannonball fired in the direction of the earth's rotation should go farther than if it were fired in the opposite direction.
I'm pretty sure I understand the idea of relative motion to understand why it doesn't do so, but what I don't understand is why he thought it would travel further if it were shot in the direction of the earth's rotation. It seems if I were going to be confused about that topic, I would think the cannonball would travel further if the ball were fired in the opposite direction of the earth's rotation. So, I imagine I'm not understanding his confusion. Can you help me understand why he was confused
ANSWER:
In the simplest terms, the cannonball should go the same distance
either way because it is fired relative to the earth which is
moving. If you are in a moving train and throw a ball forward or
backward, it will go the same distance in the train. However, if someone
watches from outside the train, the ball goes faster when you thow it
forward than backward. That is why most satellites are launched in an
easterly direction�to get the extra
boost of the earth's rotational speed. Brahe was right in thinking about
this issue, though, because things behave strangely in rotating
coordinate systems and do not behave the same as in stationary systems.
Most notable is the motion of projectiles due to the
Coriolis force where cannonballs curve left or right. But the effect
was too small to be observed for cannons of his time and, since he was
born 20 years before Galileo, these effects were unknown.
QUESTION:
We know the Earth's rotational speed is 24 hours approximately. Using T= sqr (4 x pi^2 x R^3/GM), where R=radius of earth= 6400km, M=Mass of earth= 6x10^24 kg, I always get 1.41 hours and not 24 hours. Whats wrong with my concept?
ANSWER:
This is a perfect example of using a perfectly good equation for a
situation where it is not valid. The equation you state is Kepler's
third law for the period of a circular orbit of a mass mm going
around a mass M if m<<M. For example, if you set R
equal to the radius of the earth's orbit and M equal to the mass
of the sun, T=1 yr. Or, if you set R equal to the radius
of the moon's orbit and M equal to the mass of the earth, T≈1
month. What you are doing is calculating the orbit of a satellite which
orbits near the earth's surface (like the International Space Station)
which does, indeed, orbit the earth about every hour and a half. If R=6.6
RE, RE being the radius of the
earth, T=24 hr; this is called a geosynchronous orbit where a satellite
over the equator appears to remain fixed in space and is where
communication satellites are placed.
QUESTION:
This lady lost her legal battle to have the Large Hadron Collider shut down. She believed that the atom smasher could create a black hole and suck up the earth. The courts sided with the scientists. The scientists said it's not possible, and even if it did create a black hole, it would be a micro black hole and collapse in on itself. That seems to go against the laws of physics and quantum physics. We now know that matter sucked up by a black hole is permenantly imprinted on the black holes surface and the size of the black hole increases. Did I miss something?
ANSWER:
I do not normally answer astrophysics questions, but I can
deal with this one, I think. First, some cosmic rays (radiation which
strikes earth from space) have energies much greater than the proton
energies in the LHC and if a black hole could be created and have sucked in
the whole earth, that would have happened long before we evolved. Second,
you did miss something�Hawking
radiation whereby a black hole can radiate energy. For something to be
"sucked into" the black hole, it must come within the
Schwartzchild
radius which is 2Gm/c2=6.67x10-11m/(3x108)2≈1.5x10-27m
where m is the mass of the black hole. The maximum kinetic energy of
each proton in the collider is 7 TeV which is about equal to the mass energy
of 7000 protons, so the heaviest black hole they could make would have a
mass of about 14,000 protons, about 2.3x10-23 kg; the
corresponding Schwartzchild radius is 3.5x10-50 m! It seems to me
that even if this black hole never evaporated, it could go a really long
time before it got close enough to anything to suck it in.
QUESTION:
Why is everything spinning? Moons, planets and stars all rotate on their axes, moons orbit planets, planets orbit stars, spiral galaxies rotate around super-massive black holes, etc. What force is causing all this spinning?
ANSWER:
First, the fundamental fact: if something is spinning it
takes a force (technically a torque) to change that spinning, nothing is
required to keep it spinning. Imagine that you are in the middle of empty
space and a mass moving with a constant velocity goes by you; no force is
necessary to keep it moving and it will move like that forever as long as no
forces act on it. (This is called conservation of linear momentum.) Now,
imagine that you are in the middle of empty space and there is some object
in front of you spinning; then, if you do nothing to it, it will keep
spinning like that forever. (This is called conservation of angular
momentum.) Take the solar system, for example. The sun and all the planets
all formed around five billion years ago when a huge cloud of gas and dust
in space collapsed because of gravitation. If that cloud had any rotation at
all before collapse, then the collapsed cloud (aka solar system) has to
retain that angular momentum. Although the collapsed cloud rotates much
faster than the original cloud did, it still has the same amount of angular
momentum, just like a spinning
figure skater
pulling in her arms spins faster.
QUESTION:
Is there a mechanism that create Hydrogen? It
seems to me that over billions of yrs all of the H would be burned up
in stars. Where does it get converted back to H after its been used
up???
ANSWER:
This is the nature of our
universe. When all the hydrogen is burnt up (that is, converted to
heavier elements), that will be the end. After more than 14 billion
years, there is still quite a bit left.
QUESTION:
Hydrogen is the most abundant element in the
universe. A star, such as the sun in our solar system can produce all
of the naturally occurring elements on the periodic table. Are all of
the naturally occurring elements on the periodic table created by
adding hydrogen atoms together under extreme temperatures and extreme
pressures? (Ex: 1 hydrogen atom equals hydrogen, 2 hydrogen atoms added
together equals helium, 3 hydrogen atoms added together equals lithium,
and so on.)
ANSWER:
Starting with hydrogen,
stars fuse nuclei together to get heavier elements, but not all heavier
elements. The net effect is to make helium from two hydrogen, but the
process is a little circuitous. Heavier elements are made by fusing
lighter elements, but not three hydrogen for a lithium, rather
something like a helium and a hydrogen; for example, see this link to see how
carbon, oxygen, and nitrogen are made.. After iron, energy is lost when
fusion to heavier nuclei occurs so it will not happen. The elements
heavier than iron are made when the star explodes (supernova) and the
energy from the explosion can be used to fuse heavier elements.
QUESTION:
Does everything in the universe rotate in the same
direction?
ANSWER:
No. However, most things in the solar system
do; if viewed from the north star, most objects in the solar system
rotate counterclockwise.
QUESTION:
As a shotgun fires, the bb's
make a pattern that gets wider and less dense with more distance. So
how can light leave a small source and fill all the void in between for
what seems could be an indefinate distance? i.e. the sun
ANSWER:
Because a shotgun is more like a flashlight
than the sun. The light from a flashlight is pretty well collimated and
spreads a little. The light from a source like the sun goes out in all
directions. However, the intensity of the light from this kind of
source gets smaller as you get farther away quite rapidly--if you go 10
times as far away, the intensity of the light gets 100 times smaller.
That is why, of course, star appear much dimmer than the sun even
though the sun is a pretty average star.
QUESTION:
OK. I have been thinking
about this for a very long time, however since i do not possess the
intimate knowledge that a well-versed person in the ream of physics
might have. My question is this: From my understanding, there are four
main forces which keep the universe together. One of them is gravity.
Gravity affects the heavens and all that it encompasses even light.
Light is bended by the gravity exerted from a massive object. From what
ive learned, this is one way black holes are observed. As is commonly
known, nothing can escape the gravitational pull of a black hole, not
even light. So what we have are two extremes: in one extreme, a beam of
light would travel (theoreticlly) in a straight line if no force of
gravity was imposed on it and in the other extreme, light is bended so
far that it is incapapble of escaping when it enters a supermassive
graviational field. The question i have is is there a medium between
the extremes? Would it be possible for a beam of light to orbit an
object of a specific mass, trapping it in orbit?
ANSWER: (answer
provided by L. A. Magnani)
Yes, light can orbit around a black hole of
any given mass. The key question is, how far
from the singularity can the light maintain a circular orbit? It turns
out that, according to General Relativity,
light can maintain a circular orbit at 1.5 times the Schwarzschild
radius from the
singularity (the Schwarzschild radius is 2GM/c2
where M is the mass of the black hole, G is the
gravitational
constant and c is the speed of light. This region is sometimes
called the photon sphere. Inside 1.5 times the Schwarzschild
radius, no circular orbits exist; all orbits are unstable and all
objects moving inside of this distance
fall into the black hole.
QUESTION:
I was told to ask this question to a phyicist, so here goes. Where did
air come from?
ANSWER:
Well, how far back do we want to go? All heavy elements
(essentially heavier that hydrogen) were produced in stars and then,
when the star was "all burnt up", it exploded and sent all the heavy
elements flying into space and then they eventually come together again
to form planets, etc. (Scientists like to say that we are
all made of "star dust". Then, depending on the chemistry of the
planet, its temperature, and other factors, some of the planet will
become an atmosphere, i.e. gases will escape from the surface
somehow In some cases (like the moon) the gravity is not strong
enough to hold the atmosphere and it eventually "leaks" off into
space. In the case of the earth, there is virtually no hydrogen
or helium in the air because it has all leaked off. The detailed
composition of the atmosphere depends on chemistry and biology.
For example, it is thought that originally the earth had much more
carbon dioxide in its air but that evolution of green plants resulted
in there being much more oxygen now.
QUESTION:
Do gravitons exist?Or are they only a theory?
ANSWER:
There is no working theory of quantum gravity and so a
graviton is a purely hypothetical particle. A graviton has never
been detected experimentally. Trying to unify general relativity
(the working theory of gravity) with quantum mechanics (the working
theory of everything else) is one of the holy grails of theoretical
physics.
QUESTION:
Given that the neutron has a magnetic moment - is this why
neutron stars have very strong magnetic fields? Are all of the neutrons
within the extreme density of the star aligned along their magnetic
fields?
ANSWER:
The simple answer to your question is no! The reason is not
simple but is related to the fact that a neutron star has lots of
neutrons but it also has lots of other stuff. An astronomer here
in the department is going to consult with friends of his who are
neutron star experts and then we can get back to you with a more
complete answer.
Have a
look at
http://www.astro.umd.edu/~miller/nstar.html#internal.
ANOTHER ANSWER: (Provided
by M. C. Miller)
(I referred the questioner to the above site. He
emailed the author of that site, Professor Cole Miller of the
University of Maryland who then provided the following answer.)
"The
origin of neutron star magnetic fields is not, by any means, well
understood. The existing field is supported by normal currents in
normal fluids (superconductors exclude magnetic fields), but why it got
so strong is an evolutionary question. Let me start by giving you the
pat answer you'll find in many textbooks, then I'll explain why it's at
best misleading and tell you a bit about current thinking.
It's often "explained" like this. The Sun has a magnetic field, with an
average strength of about 100 Gauss. If you were to compress the Sun to
the size of a neutron star and conserve magnetic flux, the strength of
the field would be about a trillion Gauss, which is about the strength
inferred for young pulsars. Voila, that's why you expect such a strong
field.
The problem is that the Sun will not become a neutron star. Stars that
do become neutron stars (massive O and B stars) have relatively little
convection in their interiors, meaning that they don't generate a field
via a dynamo. If you were to take their central cores and compress them
to the size of a neutron star, you'd get a rather weak field.
So, people think that the field is generated during the collapse of the
core that initiates the supernova. Somehow, with all the associated
turbulence, you stir things up and produce a dynamo that generates the
observed field. Again, the field itself isn't especially mysterious --
it's standard currents -- but how it got there is not well known. One
problem is why it stops at a trillion Gauss for many or most neutron
stars. In principle, the dynamo could generate something a good hundred
thousand times stronger, so why doesn't it? Maybe there's something
preventing such growth, or maybe it's just that such a star would spin
down so fast we usually wouldn't see it. There is a class of objects
thought to have magnetic fields up to 1000 times stronger, so perhaps
such stars are out there."
QUESTION(1):
Today there is evidence that the big bang occurred. But do
scientists have any theories as to WHY it occurred? Or is it,
that they just don't know for sure and are working on it? Also, any
theories are not testable right?
ANSWER(1): (Provided
by L. A. Magnani)
Right now,
there is no consensus on why the Big Bang happened. Modern physics
breaks down in conditions typical of the Plank Time (10-43
seconds after the Big Bang) so the safest thing to do is to discuss
what happens after the Plank Time. However, people who work on Grand
Unified Theories and string theory feel that when these theories are
fully developed we will be able to discuss what happens before t=10-43
seconds. People talk about vacuum energy fluctuations and phase
transitions, but, at the moment this is just speculation backed up by
theories which are in the developmental and not the testing stage.
String theories and GUTs may be testable in the years to come, but they
are not testable now. However, the next few decades could lead to
breakthroughs in this area.
QUESTION(2):
Oops! I have another question. According to this article: http://www.space.com/scienceastronomy/generalscience/physicists_bigbang_000209_wg.html
Does this mean that at the most we can have quarks come out of this
"recreation" of big bang conditions? Or do scientists think they
can make other matter, like trees, rocks, gold, etc. through this
matter, or do they think that it only occurred through the big
bang?
ANSWER(2): (Provided
by L. A. Magnani)
I hadn't
heard of this result and I would like to know more about it. But,
I think the answer to the student's question is that the conditions in
the early Universe which led to quark production subsequently led--as
the Universe expanded and cooled--to the production of protons and
neutrons. Hydrogen, Helium, Lithium, and Beryllium followed. But by
then, the Universe had cooled to the point where nucleosynthesis could
no longer proceed and the abundances of those elements and their
isotopes were frozen in (some elemental and isotopic compositions
changed with time because of processing of original material in stars).
The rest of the elements, necessary for trees, rocks, gold, etc. were
synthesized in stars.
QUESTION(3):
Hello, I have one last question!!! Do scientists believe
that the big bang explosion came from a single atom, a glob of matter,
or they just don't know and are working on it?
ANSWER(3):
See ANSWER(1).
YET ANOTHER
FOLLOW-UP QUESTION:
Prof. Magnani had said that "elements, necessary for trees,
rocks, gold, etc. were synthesized in stars." What does
he mean by this? I mean, you never see gold, or a tree, or a rock or
plants, or a coral reef coming out of a star or inside a star? How
could these things come to earth through a star? Also, how are stars
and the Sun formed? I couldn't find this on the Web!!! I mean, stars
are balls of gas, and the sun is too right? So could we just put
hydrogen and helium together to make a mini-sun?" How do
scientists think the earth is formed? I have a feeling that the
professor will tell me that stars were created through the matter by
the big bang, but what about the earth? You'll never find the earth
coming out of those balls of gases, right?
ANSWER:
As Dr. Magnani said, after the big bang the universe
consisted entirely of the very lightest elements, mostly
hydrogen--nothing you could make a tree out of. The reason is that what
was created were the "elementary" particles (protons, neutrons,
electrons) and to make a heavy element like gold you have to put
together a bunch of protons which takes a lot of energy to push those
positively charged protons together which want to keep apart. Therefore
when things have settled down after the big bang you have what is
essentially a universe full of hydrogen. And that would be the end of
it if the universe had been homogenous (perfectly the same in all
directions). But there were lots of places where the density of H gas
was bigger than neighboring regions and the gas started to contract
because of the gravitational attraction. Eventually the gravitational
force has compressed this H gas so much that it gets real hot, that is
the atoms are moving around with a very large velocity. Now the H
starts "burning". It isn't really burning in the usual sense; rather it
is colliding with other H atoms and causing nuclear fusion to occur.
The main thing that happens in a normal star in the "prime of its life"
is that H burns to Helium and the result is the release of huge amounts
of energy. That is the energy that keeps us all alive. As the star gets
older and begins to exhaust its H, things get complicated and things
start happening relatively fast. Basically the He starts combining to
form heavier elements, the star starts cooling down, so that it starts
to contract under gravity again. In many cases the star at some point
just explodes (a supernova) throwing all its matter out into space. Now
we have in the instellar space material other than H. But there is
still a lot of H around. So now a new star will form but when it does
there is stuff other than H there. What apparently happens is that
there is an "accretion disk" which forms around the new young star
which has much of the heavier elements in it (sort of like Saturn's
rings which are sort of an accretion disk) and the "stuff" in the disk
condenses into lumps which become planets. So our sun must have formed
long after the big bang or else there wouldn't have been any of the
heavier elements around. A favorite poetic saying astronomers like to
use is that "we are all made of star dust".
The
answer to your question "So could we just put hydrogen and helium
together to make a mini-sun?" is yes! That is what a hydrogen
bomb is--a mini-sun. That is what scientists trying to harness
fusion energy are trying to do is to make a controlled H bomb as an
efficient power source. It is a very difficult job still far from
finished.
You need
to get yourself a good book on elementary astrophysics and one which
addresses stellar evolution. There were things called books and
libraries back before the internet and they actually still exist!
Seriously, though, you clearly have a very inquiring mind and you
should read a good book or take a course so that you can get the big
picture. It will be better for you than surfing the web!
