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Finding altitude on the moon

  1. Dec 18, 2007 #1
    Finding altitude on the moon (urgent)

    1. The problem statement, all variables and given/known data
    A projectile is launched vertically from the surface of the Moon with an initial speed of 1700 m/s. At what altitude is the projectile's speed 27% of its initial value?


    2. Relevant equations
    Kinematics, Conservation of energy


    3. The attempt at a solution
    I tried using First kinematics and solved for t, using t I plugged it into the 2nd kinematics, but the solution was wrong because of change in altitude. Since altitude changes, the gravity will also change as altitude increases. I then tried using Vf = square root 2Gm/Xr solving for x, but this came to a small increase in altitude which would not make sense as an answer. Any ideas or how to find the solution to this problem? I've tried every method I could think of, but none of the solutions work.
     
  2. jcsd
  3. Dec 18, 2007 #2

    Dick

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    Your second approach (conservation of energy) looks correct. But why doesn't it make sense as an answer, i.e. what DO you get? What numbers are you putting in?
     
  4. Dec 18, 2007 #3
    ok instead of using my numbers, I was using a friends because they had a small initial speed which gravity did not have a significant change and they got the right answer. Their speed was 600m/s and their percentage was 55%

    I did Vf = square root 2Gm/Xr so I took 55% of the initial as Vf. The equation came out to be 330m/s = sqaure root (2 x 6.673e-11 x 7.35e22kg) / x1.74e6. 7.35e22kg is mass of moon and 1.74e6 is the radius of the moon. Solving for x I got 51.768m and the answer they wanted was in km which would mean this answer would equal 0.0517km. The right answer was 76.87km.....any ideas?
     
  5. Dec 18, 2007 #4

    Dick

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    The concept you want is that the change in kinetic energy (1/2)*mv^2 is equal to the change in the potential energy between the moon's surface and the final height. The latter is something like GMm/(R)-GMm/(R+h), where m is the mass of the object, M is the mass of the moon, R is radius of the moon and h is height. Right? Notice m cancels.
     
  6. Dec 18, 2007 #5
    That means you're left with GM/(R)-GM/(R+h). Which is equal to what? We didn't get on the topic of using kinetic and potential energy so I'm not certain of how to exactly do this problem using kinetic and potential energy. If possible could you explain please?
     
  7. Dec 18, 2007 #6

    Dick

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    It's equal to (1/2)*(vi^2-vf^2). Take the difference of (1/2)mv^2 between initial and final velocities and divide by m. Where vi and vf are initial and final velocities. BTW, I don't think 76.87km is correct either. 1700m/s is a substantial fraction of the moon's escape velocity. It should go much higher. Are you sure about that?
     
  8. Dec 18, 2007 #7
    76.87km was the answer to my friend's because his initial speed was 600m/s and they wanted it to know the height at 55% of that speed. The difference of (1/2)mv^2 divide by m? I wasn't given a m of anything....I'm getting confused. This is what it would be GM/(R)-GM/(R+h) = (1/2)*(vi^2-vf^2). You want me to do ((1/2)mv^2)/m but I'm not given an m.
     
  9. Dec 18, 2007 #8

    Dick

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    If you are doing (1/2)mv^2/m, you don't need an m. It cancels. And yes, the rest is right. You should look at potential and kinetic energy and conservation thereof. In the case of your friends problem you can get away with ignoring the change in gravity, in your's you can't. But you knew that, right?
     
  10. Dec 18, 2007 #9
    Yes, I knew that in my friends problem change in gravity could be ignored. My equation would be GM/(R)-GM/(R+h) = (1/2)*(vi^2-vf^2)^2? Since you said to take the difference so I'm assuming the (1/2)*(vi^2-vf^2) = (1/2mv^2)/m where the (vi^2-vf^2) = the (v^2)
     
  11. Dec 18, 2007 #10

    Dick

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    Yesss. Again, try to figure out why that is correct. When you get to kinetic and potential energy. I don't know why you were given this problem if you haven't already covered this.
     
  12. Dec 18, 2007 #11
    So using my numbers my equation would be (6.673e-11 x 7.35e22)/(1.74e6) - (6.673e-11 x 7.35e22)/(1.74e6 + h) = 1/2(1700^2-459^2)^2?? When I solved up to 2.81877e6 - (6.673e-11 x 7.35e22)/(1.74e6 + h) = 3.58938e12m/s I end up with -4.90465e12 = 6.24552e18 + 3.58938e12m/s(h) which would mean I end up with a neg answer. Any idea which step I messed up on?
     
  13. Dec 18, 2007 #12

    Dick

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    Yes, "1/2(1700^2-459^2)^2". Why did you square that? Actually, I know that. It's because you responded to "(1/2)*(vi^2-vf^2)" with "(1/2)*(vi^2-vf^2)^2" and I didn't correct you because I didn't notice. That's the advantage to actually knowing what you are doing instead of blindly following instructions.
     
  14. Dec 19, 2007 #13

    Dick

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    BTW, tracking what the units are would have made you realize that was wrong pretty quickly. Try it!
     
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