# Astronaut Problem

1. Jun 16, 2009

### blackboy

1. The problem statement, all variables and given/known data
An astronaut standing on the Moon fires a gun so that the bullet leaves the barrel moving in a horizontal direction. What must be the muzzle speed of the bullet so that it travels completely around the Moon and return to its original location? How long does this trip take? Assume that the the free fall acceleration on the Moon is one-sixth that on Earth.

2. Relevant equations

3. The attempt at a solution
Say the height of the guy is y. I got the circumference as 1092100m. So vt=1092100. t also satisfies y=(gt^2)/12. Then I get y=(9.74*10^11)/(v^2). I don't know where else to go with this.

2. Jun 16, 2009

### LowlyPion

Basically the bullet needs to go into orbit at the radius of the surface doesn't it?

So centripetal acceleration needs to match moon gravity.

m*v2/r = m*gmoon = m*g/6

v = (r*g/6)1/2

3. Jun 16, 2009

### blackboy

No but this is in motion in 2d. I don't want to bring m into it yet. Is there a way to do it with pure 2d Motion stuff.

4. Jun 16, 2009

### glueball8

But m is canceled out. Do you mean that you don't want to use a=v^2/r ??

5. Jun 16, 2009

### LowlyPion

When the centripetal acceleration outward is the same as gravity inward ... shouldn't it go on and on, unless it hits a crater wall taller than where the astronaut is standing?

Why do you think it is not in 2D motion?

6. Jun 16, 2009

### blackboy

Oh I overreacted when I saw the m. I know it was 2d motion, but I didn't want to use m or any dynamic stuff yet.

7. Jun 16, 2009

### blackboy

I had a few errors in the first post. Anyway I got y=1.23 and v=2838733. Thanks

8. Jun 16, 2009

### LowlyPion

I hope you are missing a decimal point. That's speed of light territory.

9. Jun 16, 2009

### blackboy

Ok yea dum error. v=1684?

10. Jun 16, 2009

### ideasrule

You have a bullet in a circular orbit, so what does "y" represent? The equation y=(1/2)at^2 only applies if the acceleration is of constant magnitude and constant direction. In this case, the direction of the acceleration constantly changes to point to the Moon's center, so you can't use that equation.

11. Jun 16, 2009

### LowlyPion

Close enough I suppose. I get 1660 m/s scratching it out roughly.

12. Jun 16, 2009

### nealh149

LowlyPion, what do you mean when you say "the centripital acceleration outward"? There is only one force on the bullet. Here's what I got.

$$mv^2/r = GmM/r^2 v = sqrt(GM/r) = sqrt(1/6gr)$$

Plugging in values I get 532 m/s.

Last edited: Jun 16, 2009
13. Jun 17, 2009

### ideasrule

That's because you are assuming that the circumference the questioner gave, 1092100 m, is right. It's not; it corresponds to a radius of 173.8 km whereas the actual radius is 1738 km, one decimal place off. 1684 m/s is right if 1738 km is used.

14. Jun 17, 2009

### nealh149

Okay, that's true, I did assume the OP's circumference was correct.

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