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Speed of a projectile launched from the moon?

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A projectile launched vertically from the surface of the Moon rises to an altitude of 370 km. What was the projectile's initial speed?

    2. Relevant equations

    h = 370,000 m
    g moon = 1.62 m/s^2
    G = 6.67x10^-11 N m^2/kg^2
    Mass moon = 7.35x10^22 kg
    Radius moon = 1.74X10^6 m

    vf^2 = vi^2 - 2gh

    3. The attempt at a solution

    I calculated g on the moon to be 1.62 m/s^2. I then used kinematics (vf^2 = vi^2 - 2gh) and I get:

    vf = 0
    vi = ?
    g = 1.62 m/s^2
    vf^2 = Vi^2 - 2gh
    solving for Vi = sqroot of Vf^2 + 2gh

    I get initial velocity to be 1095.57 m/s. The answer is not correct, and I really don't understand what I'm doing wrong.
     
  2. jcsd
  3. Nov 15, 2009 #2

    mgb_phys

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    Two methods
    Conservation of energy
    1/2mv^2 = mgh
    Or motion equations:
    v^2 = u^2 + 2gs (where final velocity is zero)

    ps. g = 1.62m/s^2 for the moon is correct. Check your calculations
     
  4. Nov 15, 2009 #3
    Thanks for your reply. I tried the equations again and I'm still getting the same answer. It is wrong and I don't understand why :(

    Based on the conservation of energy equation you mentioned:
    1/2mv^2 = mgh
    1/2v^2 = gh
    v^2 = 2gh
    v = sqroot (2gh) = sqroot [(2)(1.62m/s^2)(370000m)]
    v = 1095 m/s
     
  5. Nov 15, 2009 #4

    mgb_phys

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    Sorry the altitude is 370km so you can't assume that g is constant.
    You need to do gravitational potential energy at the surface and at 370km - the difference is the KE

    (Energy = GMm /r)
     
  6. Nov 15, 2009 #5
    Ok, here is what I got using your advice:

    GPE at the surface:
    GMm/r = (6.67x10^-11 N m^2/kg^2)(7.35x10^22 kg) / 1.74x10^6 m = 2817500 J

    GPE at 370km:
    GMm/r = (6.67x10^-11 N m^2/kg^2)(7.35x10^22kg) / 370000m + 1.74x10^6 m = 2323436 J

    2817500 J - 2323436 J = 494064 J = KE

    Am I correct so far? Do I plug this number into 1/2mv^2 ?
     
  7. Nov 15, 2009 #6

    mgb_phys

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    Almost - you can't have Joules because you don't know the mass of the object (essentially you have assumed a 1kg mass) but that doesn't matter since you can also assume 1kg in the KE
     
  8. Nov 15, 2009 #7
    I plugged in 494064 into KE = 1/2 mv^2

    494064 = 1/2 (7.35x10^22kg) v^2
    v = 3.66x10^-9

    Does this sound correct?
     
  9. Nov 15, 2009 #8

    mgb_phys

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    No, the 'M' in that equation is the mass of the object you are throwing - not the mass of the moon.
     
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