- #1

Brad2007

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## Homework Equations

## The Attempt at a Solution

I don't really know where to start there is so little information given, I thought it might be an escape velocity problem, but I don't know where to even start.

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- Thread starter Brad2007
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- #1

Brad2007

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I don't really know where to start there is so little information given, I thought it might be an escape velocity problem, but I don't know where to even start.

- #2

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- #3

Jimmy

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If anyone knows the answer to this, please PM me.

- #4

ideasrule

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- #5

Jimmy

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I didn't want to give the answer away in case the OP comes back needing more help. That probably won't happen but you never know...

I calculated the change in kinetic energy:

[tex]\Delta{}E_k = \frac{m(v_0^2 - v_f^2)}{2}}[/tex]

Next, I calculated the gravitational energy at the surface:

[tex]E_g = \frac{GMm}{r}[/tex]

subtracted the change in**E**_{k} from **E**_{g} and calculated r based on the difference:

[tex]r = \frac{GMm}{E}[/tex]

The altitude of course is**r - radius**_{moon}

I used 1kg for the projectile mass but any mass should result in the same answer. Because of that, I'm pretty sure there's a simpler way to do this*sans* the projectile mass(m)- if I'm doing this correctly at all...

I calculated the change in kinetic energy:

[tex]\Delta{}E_k = \frac{m(v_0^2 - v_f^2)}{2}}[/tex]

Next, I calculated the gravitational energy at the surface:

[tex]E_g = \frac{GMm}{r}[/tex]

subtracted the change in

[tex]r = \frac{GMm}{E}[/tex]

The altitude of course is

I used 1kg for the projectile mass but any mass should result in the same answer. Because of that, I'm pretty sure there's a simpler way to do this

Last edited:

- #6

ideasrule

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Your method is correct. If you substitute E=m(v0^2-vf^2)/2 into your equation for r, you'll see that the mass cancels out.

- #7

Jimmy

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I was fairly confident it was correct, but was convinced my solution was overly complicated. Thanks for the input.

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