Projectile launched from the Moon

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  • #1
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A projectile is launched vertically from the surface of the Moon with an initial speed of 1380 m/s. At what altitude is the projectile's speed two-thirds its initial value?


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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.
 

Answers and Replies

  • #2
kuruman
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The initial speed is too high for the kinematic equations to be applicable because they only work at constant acceleration. The projectile is likely to reach a height at which the acceleration of gravity is appreciably smaller from its value on the surface of the Moon. Look up the mass and radius of the Moon and use energy conservation. The potential energy will be (of course) zero at infinity. If the total energy on the surface is positive, then you know that the 1380 m/s is larger than the escape velocity. Your initial hunch about escape velocity is good.
 
  • #3
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If anyone knows the answer to this, please PM me.
 
  • #4
ideasrule
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We don't PM solutions; we post them here for other students, who might have similar problems, to see. How far have you gotten in solving the problem?
 
  • #5
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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 Ek from Eg and calculated r based on the difference:

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

The altitude of course is r - radiusmoon

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...
 
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  • #6
ideasrule
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Sorry Jimmy, I thought you were the OP.

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
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No problem, It's probably better this way.

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

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