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An Interesting Gravitation Problem

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Two satellites are launched at a distance R from a planet of negligible radius. (Yes, that's what the problem says...) Both satellites are launched in the tangential direction. the first satellite launches correctly at a speed [tex]v_0[/tex] and enters a circular orbit. The second satellite, however, is launched at a speed [tex]\frac{1}{2}v_0[/tex] . What is the minimum distance between the second satellite and the planet over the course of its orbit?

    2. Relevant equations

    3. The attempt at a solution
    I thought about using energy. The two satellites both start out with the same potential energy but different kinetic energy.

    So satellite one's TME: [tex]-\frac{GMm}{R} + \frac{1}{2}mv_0^2[/tex]

    where as satellite two's TME:[tex]-\frac{GMm}{R} + \frac{1}{8}mv_0^2[/tex]

    And the second satellite's minimum distance is when its potential is the least...
    Somehow I think this problem might relate to angular momentum... L = mvr, but not exactly sure.

  2. jcsd
  3. Jan 20, 2009 #2


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    You can use the fact the first satellite goes into a circular orbit to figure out M in terms of R and v0 using the centripetal acceleration of an object in circular motion. Let R1 and v1 be the radius at the lowest point. At the lowest point v1 is again tangenential to the planet. So conservation of angular momentum gives you one equation in R1 and v1 and conservation of energy gives you another. Solve them simultaneously.
  4. Jan 20, 2009 #3
    Ah, i c~

    and i got it right =D

  5. Dec 31, 2009 #4
    How can you prove that the lowest point does indeed occur when the velocity is tangent again? And why do we need to know that it is tangent? Is something not conserved otherwise?
    Last edited: Dec 31, 2009
  6. Jan 1, 2010 #5


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    The lowest point occurs when the object is travelling 'horizontally' i.e. perpendicular to the radial vector connecting the object to the planets center. Draw a picture. Knowing the angle between those vectors makes it easy to compute angular momentum. Try it.
  7. Jan 1, 2010 #6
    haha I forgot that angular momentum is a cross product
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