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Orbits


Why don't satellites just fall down?

May people believe that spacecraft stay up because there is no gravity in
space. This is not true. The space shuttle, for example, orbits less than
1000 km up. Gravity at this altitude is only slightly weaker than at the
Earth's surface.
The force F of gravity between two objects of mass m and M, whose centres
are a distance r apart, is given by Newton's famous equation:
[pic],
where G = 6.67(10-11 N m2 kg-2, and F is measured in Newtons. So, the force
in orbit Fo is weaker than the force on the Earth's surface F0 by the
ratio:
[pic]
where the radius of the Earth is 6,400 km, and the orbit is 1000 km above
the Earth's surface. So gravity is only 25% weaker up at the Space
Shuttle's orbit. This cannot be why satellites stay up.
In reality, they stay up because they are moving sideways. If a satellite
wasn't moving sideways, it would fall straight back down to Earth. There
are two ways to see this:

Free Fall.

Consider a tower-block 1000 km high, protruding above the Earth's
atmosphere. Imagine that you are dropping something off the top of this
tower. If you just drop it (ignoring the rotation of the Earth), it will
fall straight down, burning up in the atmosphere near the base of the
tower.
But now, give it a sideways push as you drop it. As it falls, it will
continue to move sideways, until it burns up. The harder you push it, the
further away from the base of the tower it will land.
If you push it hard enough, it will miss the Earth altogether - by the time
it's fallen 1000 km, it will have moved so far sideways that the Earth is
no longer below it. If you're clever, you can get it moving in a circle
around the Earth - perpetually falling but never hitting the bottom (see
picture below).
[pic]

Centrifugal Force

Another equivalent way of seeing the same thing is to use the concept of
centrifugal force. When an object of mass m moves in a circular path of
radius r at a velocity v, it will experience an outward centrifugal force
given by the equation:
[pic]
For a satellite to stay in orbit, this centrifugal force must balance
gravity. Thus,
[pic].
Simplifying, we find that
[pic],
and hence that
[pic].
Thus for a satellite 1000 km above the Earth (and hence 7,400 km from the
centre of the Earth) travelling in a circular orbit,
[pic]
(the mass of the Earth is 5.967(1024 kg).




Stability of Orbits

So, if we go at exactly the right speed, centrifugal force balances gravity
and we can stay up. But are we safe up there, or is this a precarious
balance? Consider the Earth in orbit around the Sun. Centrifugal force
balances gravity at a speed of 30 kms-1, and indeed, that is the speed of
the Earth.
But what would happen if the Earth slowed down, even slightly, for some
reason? Say an asteroid hit the Earth head on. We would be travelling ever
so slightly slower, so the centrifugal force (which depends on the square
of the velocity) will decrease. Gravity, however, remains just as strong.
Surely we would fall inwards. And as we move inwards, gravity gets stronger
still. So wouldn't we spiral inwards to our doom? Why not?
Luckily, the law of conservation of angular momentum saves us from frying.
The angular momentum of an object moving in a circular orbit is l=mvr. This
is conserved, so
v=l/mr, ie. v is inversely proportional to r. Now centrifugal force is
proportional to v2/r. Substituting for v, we thus find that F( 1/r-3. So
the centrifugal force is proportional to the inverse cube of the radius.
The gravitational force is only proportional to the inverse square of the
radius.
So, as an object with a given angular momentum moves inwards, gravity
increases, but the centrifugal force increases faster. Thus at some point,
the centrifugal force will once again balance gravity.
So - orbits are stable - if you nudge them, they will come back into
balance.


Energy Considerations

Another way to see all this is to consider the energy of an object in
orbit. Break its velocity v into two perpendicular components - the radial
velocity (towards or away from the thing it is orbiting - vr, and the
tangential velocity vt.
[pic]
The kinetic energy of this object is KE = Ѕ mv2 = Ѕ m(vr2+vt2) (by
Pythagorus' theorem). Its angular momentum l = mvtr, and is conserved. What
is its potential energy?
Take an object at distance r from the source and push it outwards against
gravity until it reaches infinity (we choose to define the potential energy
at infinity as zero). The energy used for each change in radius is just the
force times the distance, and the force is given by Newton's law, so we
just have to integrate this out.
[pic]
The total energy is thus:
[pic]
As angular momentum is conserved, we can substitute vt=l/mr. Everything
except the radial velocity component is thus a function of radius.
[pic]
This equation is actually motion in one dimension only - all we have is the
radial position and velocity. It actually looks just like the normal 1D
energy equation, with a KE term that depends on the square of the velocity,
and a PE term that depends only on position. So we can see our orbiting
object as moving in one dimension, with a rather strange pseudo potential
energy given by the term in brackets above.
What does this pseudo-PE term look like? When r is very small, the 1/r2
term will dominate, so the pseudo PE will be large and positive - indeed
the pseudo-PE becomes infinite as r goes to zero. When r is large, the 1/r
term dominates, so the pseudo-PE will be small and negative. As r goes to
infinity, the pseudo-PE goes to zero. So, a graph of pseudo-PE against
radius goes something like this:
[pic]
The graph has a minimum (exactly where depends on the angular momentum l),
which is why orbits are stable. If an object has just enough energy to sit
at this minimum, vr=0 (there's no spare energy for radial motion), so it
will be in a perfect circular orbit.
If the energy of an object is greater than this minimum, but still
negative, the object will oscillate in radius between rmin and rmax. This
is an elliptical orbit! If the energy is positive, the object will fly out
into space (it has achieved escape velocity) and will never come back (see
the picture below).
[pic]

An object with negative energy, but more energy than that needed for a
stable orbit, will undergo approximately simple harmonic motion in this
potential well. If and only if gravity follows the inverse square law, the
period of this oscillation in radius will exactly equal the orbital period
- this is why orbits are ellipses. If gravity obeyed any other law, the
periods will not match, and the orbits will not repeat themselves, but will
look more like flower petals.
This pseudo-PE curve is very informative - it tells us how hard it is to
send spacecraft to various destinations. For example, notice that it takes
more energy to go to r=0 than it takes to go to r=(. Thus it is easier to
send a spacecraft to Alpha Centauri than it is to drop it into the Sun!
This same curve applies to orbits about black holes (almost - see below).
This gives the lie to most science fiction novels - it is actually very
hard to fall into a black hole. Unless you have zero angular momentum,
falling into a black hole requires vast amounts of effort on your behalf.
This has long been a puzzle in astrophysics. Objects like Cataclysmic
Variables and Quasars are believed to be powered by matter falling into
black holes. But how can it get rid of enough angular momentum to fall in?
We'll come back to this later.
One complication - this calculation has used only Newtonian physics. Close
to the event horizon of a black hole, we have to use relativity, and this
changes things quite a bit. It turns out that if you are really close to
the black hole (the exact distance depends on the angular momentum of the
black hole, but is at most a few times the Schwartzschild radius), there is
no such thing as a stable orbit, and without the use of rockets you will
inevitably fall in. This basically occurs because as you get close, you
would need to go very fast to fight off the severe gravity. But as your
speed approaches the speed of light, your mass increases, so you get sucked
in harder still.