Astronaut Problem: Homework Equations and Solution Attempt

[blackboy]

Homework Statement

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.

Homework Equations

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.

 

[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

 

[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.

 

[glueball8]

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

 

[LowlyPion]

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.

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?

 

[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.

 

[blackboy]

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

 

[LowlyPion]

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

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

 

[blackboy]

Ok yea dum error. v=1684?

 

[ideasrule]

1.So vt=1092100. t also satisfies y=(gt^2)/12. Then I get y=(9.74*10^11)/(v^2).

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.

 

[LowlyPion]

Ok yea dum error. v=1684?

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

 

[nealh149]

LowlyPion, what do you mean when you say "the centripetal 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.

 

[ideasrule]

LowlyPion, what do you mean when you say "the centripetal 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.

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.

 

[nealh149]

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