# Gravitational satellite question.

1. Aug 11, 2006

### al_201314

Hi guys I need help with this after quite some time

A piece of rock far out in space is relative to the earth. Under the influence of the earth's gravitational attraction, it begins to fall towards the earth along a straight adial line. With what speed will the rock hit the earth? (Given radius of earth = 6.4 x 10^6m)

I couldn't find the radius of orbit of the rock. I suppose since it is at rest relative to earth it must have the same angular speed? How do I find the radius from here? I figured that its period is not the same as the earth's as the T^2 = k(r^3).

Thanks a lot!

2. Aug 11, 2006

### Päällikkö

Falling in straight radial line means that the rock will fall straight down ie. it will not orbit the earth.

Think about the conservation of energy principle.

3. Aug 11, 2006

### al_201314

The total energy the rock has is -GMm/r^2 during the orbit where r is radius of orbit am I right? So when it drops straight all these energy will be converted to KE of the rock? So -GMm/r^2 - GMm/(6.4 x 10^6)^2 = 1/2mv^2? However I still can't solve from here as mass of earth is not given as well as r?

Thanks!

Last edited: Aug 11, 2006
4. Aug 11, 2006

### Päällikkö

Well although the formula for potential energy you have is correct, you must confirm this as the rock is not orbiting the earth. (You stated that using the formula would be correct as the rock is orbiting the earth, but this is not the case)

Mass of earth can be assumed known (Google for the precise numerical value). For r: The rock starts out falling from really, really far away. How would you therefore approximate its potential at the beginning of the drop?

Last edited: Aug 11, 2006
5. Aug 11, 2006

### al_201314

Since it is stationary relative to the earth, wouldn't it be orbiting just like a geostationary satellite in orbit?

The potential is 0 at infinity?

Anyway the given answer is 1.1 x 10^4 ms^-1

6. Aug 11, 2006

### Päällikkö

No. Looking from an inertial earth-centered frame, the rock does not have any velocity at the beginning (you earlier, correctly, assumed that the kinetic energy of the rock was 0). This would not be the case if the rock were orbiting (as it would mean that anglular momentum L = 0, which would in turn mean that the rock weren't orbiting).

Yes.

True. And you should get it from the energy conservation equation you had there a few posts ago.

Last edited: Aug 11, 2006
7. Aug 11, 2006

### WigneRacah

The formula for the potential energy which you were using is not correct.
Go it over before you go on in your calculations.

Last edited: Aug 11, 2006
8. Aug 11, 2006

### Päällikkö

I cannot believe that went totally unnoticed to me. Worse than that, I called it correct. Thank you!

9. Aug 11, 2006

### DaveC426913

Unless I'm completely mistaken, this question is unsolvable. The initial distance of the rock is not given. We don't know if by "far out in space" they mean 10 thousand miles or 10 million miles.

Now, there seems to be some suggestion that "at rest relative to the Earth" is to be interpreted as "in geostationary orbit". I believe this is an erroneous interpretation. I believe "at rest relative to the Earth" simply means "no velocity towards (or away from) Earth".

10. Aug 11, 2006

### Päällikkö

Usually such phrase is interpreted (modeled, approximated) as infinitely far, which applied to this particular problem does give the wanted 11 km/s (escape velocity).

Last edited: Aug 11, 2006
11. Aug 11, 2006

### al_201314

Ahh thanks guys I got the answer in just one step. It is basically just the equation for the escape velocity. I guess all my wrong interpretations lead me to think otherwise and I can't believe at my level I got the GPE equation wrong!

Thanks anyway!

12. Aug 11, 2006

### al_201314

Allow me to ask another one I couldn't really get the answer as well.

Earth has a radius R and value of acceleration of free fall at its surface is g. A planet in another system has same average density as earth but free fall at its surface is 3g/2. Whats the radius of this planet?

g = Gm/R^2 for earth.

g' = GM/x^2.

They didn't mention anything about the mass how can I go about it? Equating this 2 gave the wrong answer of (2^1/2)R/3.

EDIT: typo.

Last edited: Aug 11, 2006
13. Aug 11, 2006

### Päällikkö

Are you missing a word or two there?

Start by relating the mass to something given in the problem (density).

14. Aug 11, 2006

### al_201314

Sorry for the mistake.

And I just managed to solve it. Sorry for the trouble and your tip on the mass helped.

Thanks!