Thread: 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?

Originally Posted by Coelacanth

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.

I'm not sure which post at the "other" board. The one linked doesn't seem to say anything about direction.

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?

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.

Originally Posted by grapes

I'm not sure which post at the "other" board. The one linked doesn't seem to say anything about direction.

You are correct, my link should have been to the previous post, by WayneFrancis.

Originally Posted by grapes

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.

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?

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.:

You'd almost have to use up as much power to kick it out of Earths orbit as you did putting it into orbit and even then you aren't removing the possibility that it won't impact at a later date.