Orbital Mechanics
Recovered historical discussion from Orbital Mechanics.
Regarding
this post
at another board:
Am I off base here, or is the direction quite irrelevant? The post makes it sound like a huge amount of energy is required to move from an earth orbit to a non-orbit (either a parabolic or hyperbolic trajectory), because the direction the craft needs to move is perpendicular to the direction it is moving. But that just seems wrong to me. As long as the trajectory doesn't take the craft into the atmosphere, it seems to me the direction is quite irrelevant, get enough velocity, and you're gone.
Force of gravity is
where
is the mass of the spacecraft,
is the mass of the earth,
is the gravitational constant, and
is the distance between the centres of mass (assuming the objects can be treated as point masses, shell theorem, that sort of thing). Then the potential energy of an object at distance
from the centre of the earth is
where the zero-level is the potential energy of an object at "infinity" distance. Then an object in circular orbit in the x-y plane with an orbital radius of
moves according to
for some
Taking the second derivatives, and calculating the magnitude of the resulting acceleration vector, we get the acceleration has magnitude
. But Newton's second law is
, so
which means
The magnitude of the velocity vector is
, so
Kenetic energy is then
So if you are in a circular orbit at distance of
from the centre of the earth, you have kinetic energy of
, and you need energy of
to escape to "infinity". So the kinetic energy you have, is half the energy you need, to get out of town. The craft started with some kinetic energy (because the earth is rotating), but that is probably close to negligible; so it looks to me like if you got into orbit, you're more than half way to leaving earth for good.
Sound about right?
Archived source: https://web.archive.org/web/20121203223829/http://www.spacetimeandtheuniverse.com:80/general-physics/5495-orbital-mechanics.html
Replies
grapes
I'm not sure which post at the "other" board. The one linked doesn't seem to say anything about direction.
You're right as far as it goes. Orbital velocity (at the surface of the Earth, which is not that far from ISS or Shuttle orbit) is 1/sqrt(2) of the escape velocity, in general, so since kinetic energy is as velocity-squared it's half.
But remember how much stuff it took to get that Shuttle into that orbit ("half way"), a huge set of boosters, and a lot of fuel. To get past the next "half" you'd need similar rockets and boosters
in orbit
. To get
those
in orbit is an immense task--basically, you start with an even smaller satellite instead.
Coelacanth
You are correct, my link should have been to the previous post, by
WayneFrancis
.
Also good point, there is no requirement that your exhaust also reach escape velocity
But my main point is, the direction you are currently moving in, shouldn't affect how hard it is to escape. If you are in a circular orbit, then the requirement is the same as if you were at the same altitude moving at the same speed directly away from the earth. Sound right?
grapes
The direction is immaterial, almost. A velocity directly away from Earth would be perpendicular to an orbit velocity, and
parallel
, for different points of the orbit.
The main point of the post is that you'd need another amount of energy the same as you expended in getting to orbit, pretty much what you have calculated.: