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Gravitational Potential Energy of a planet

  1. Jun 12, 2006 #1
    I'd be grateful if someone can help me with this problem -

    Zero, a hypothetical planet, has a mass of 4.4 × 10^23 kg, a radius of 3.2 × 10^6 m, and no atmosphere. A 2.4 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of 7.4 × 10^7 J, what will be its kinetic energy when it is 4.8 × 10^6 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.9 × 10^6 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

    I started out by using the equation for energy which is E = KE + U (kinetic energy plus potential energy).

    So I got 7.4x10^7 = KE + (mGR). But I have 2 main problems: one is, do I use G = 6.67x10^-11? The second question is, I know there is something I have to do with the radius, but I don't exactly know what. Do I do mass of probe/R? I did that but still got the answer wrong...I know I'm doing something wrong with the radius.

    Thanks a lot.

    *If I get part a, I'm sure I can get the second part by myself.
     
  2. jcsd
  3. Jun 12, 2006 #2
    i think by G you mean g which is the gravitational field strength or the acceleration due to gravity, in earths case it is [tex]9.81ms^{-2}[/tex]

    also, the gravitational potential energy is [tex]V=\frac{GMm}{r}[/tex]

    because the potential decreases with 1/r. M is the mass of the earth and m is the mass of the probe in this case G is [tex]6.67x10^{-11}[/tex]

    hope this helps

    newo
     
  4. Jun 12, 2006 #3
    PS G=6.67x^-11 the [tex]x[/tex] was meant to be a multiplication sign sorry.
     
  5. Jun 12, 2006 #4
    Thanks for the gravitational potential energy equation.

    Yet I'm still confused about what to use for r:

    [​IMG]

    I don't know what the new radius would be. Does the height of the rocket matter? :frown:
     
  6. Jun 12, 2006 #5

    Doc Al

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    Staff: Mentor

    In the equation for gravitational potential energy, r is the distance of the probe to the center of the planet:
    [tex]U = -\frac{GMm}{r}[/tex]
    (note the minus sign)
     
  7. Jun 12, 2006 #6
    For some reason I'm not getting the right answer (I got part B though, for some unknown freaky psychotic reason).

    I did: E = KE - GMm/R which became:

    7.4 × 10E7J = KE - (6.67E-11)(4.4 × 10E23kg)(2.4kg)/4.8 × 10E6m

    Then got the KE, which was the wrong answer.

    I'm still thinking I have to do something with the radius. :frown:
     
  8. Jun 12, 2006 #7

    Doc Al

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    Staff: Mentor

    You need to consider the change in potential energy as it moves from its initial to its final position.
     
  9. Jun 12, 2006 #8
    got it, thanks
     
  10. Jun 13, 2006 #9

    oooops yeah i forgot that. lol
     
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