# Stone falling from moon's orbit

1. Jan 19, 2009

### weltza

I have a bunch of ideas how to solve this problem, but I cannot figure out which one is right...

A stone is released from rest at the Moon’s orbit and falls towards the Earth. What is the stone’s speed when it is 96,000 km from the center of the Earth?

Thanks,

2. Jan 19, 2009

### skeptic2

3. Jan 19, 2009

### weltza

well, I was thinking of using taking the work earth does on the stone and subtracting the kenetic energy of the stone based on the orbital velocity. but, I'm not sure if the problem wants me to take into account the gravity of the moon.

also, I could use gravitational potential of earth and again, use TE=KE+PE

I'm trying to sift through fifty pages of text to figure this out.

4. Jan 19, 2009

### Helios

A way to think of these things is to pretend the stone is in orbit, but traveling by an ellipse with a huge eccentricity, a pencil-thin ellipse. Equipt with just Keplerian orbit equations, this should be an easy problem.

5. Jan 19, 2009

### weltza

ok, maybe you could give me a little more guidance. This is my first week of astronomy.

Is the ellipse is pencil thin because the stone has such a small mass compared to earth? And can I then just find the change in KE based on the change in PE? Or am I still missing something?

6. Jan 19, 2009

### LURCH

I think what Helios was trying to say was that you should use the pencil-thin elipse as a good aproximation of falling straight down. So, the elipse is that thin only because you set it that way, not because any property of the bodies in question causes it to be. Use Kepler's equations to figure out the velocity of the rock at any point along its "orbit," which is almost stationary at aphelien and falls almost straight down.

7. Jan 20, 2009

### weltza

by huge eccentricity do you mean just less than 1? because e=1 is a parabola and >1 is hyperbola. I am not quite sure what equation I should be using.

8. Jan 20, 2009

### Staff: Mentor

There are three different cases where e=1. One is a parabola (zero total energy, non-zero angular momentum). Another is a degenerate parabola (zero total energy, zero angular momentum). The third is a degenerate ellipse (negative total energy, zero angular momentum). Some jokingly call this case an "orthogonal orbit". The rock falling from at rest represents an example of this third case.

Think of it in terms of a limit. Suppose that instead of starting with zero velocity at the Moon's orbital distance, the rock starts with zero radial velocity and a tiny but non-zero tangential velocity. The rock will follow an elliptical orbit with apogee at the Moon's orbital distance. In the limit that that tangential velocity goes to zero, the eccentricity will go to one. As Kepler's equations become singular at this limit, you cannot use Kepler's equations directly. However, you can use Kepler's equations to solve the problem for any non-zero angular momentum. Take the limit of this solution as angular momentum goes to zero.

These machinations would have come in handy had you been asked to determine the time needed to fall from 384,400 km to 96,000 km. However, you were not asked that. You were just asked to determine the velocity. All that takes is conservation of energy considerations. Use the approach to which you alluded in post #3.

Last edited: Jan 20, 2009
9. Jan 20, 2009

### weltza

gotcha, thanks!