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Dynamics of Uniform Circular Motion Quesiotn

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A satellite has a mass of 5700 kg and is in a circular orbit 4.2 multiplied by 10^5 m above the surface of a planet. The period of the orbit is two hours. The radius of the planet is 4.25 multiplied by 10^6 m. What is the true weight of the satellite when it is at rest on the planet's surface?

    2. Relevant equations

    G = gravitational constant
    Me = Mass of planet
    r1 = radius from planet to satellite
    r2 = radius of planet
    v = velocity of satellite in orbit
    Fc = centripetal force
    T = Time

    Fc = msat*v^2/r

    G*msat*Me/r^2 = msat*v^2/r

    v = Sq. Root(G*Me/r)

    3. The attempt at a solution

    r1 = 4.67e^6
    r2 = 4.25e^6

    T = 7200seconds

    so v1 = 2*pie*r1/T = 4075.343803 m/s

    so Me = v^2 * r1/G = 1.1628389e^24

    not sure how to find mass of satellite on planet from here...
  2. jcsd
  3. Oct 7, 2009 #2


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    The mass of the satellite on the planet is the same as everywhere else, 5700 kg. That's only part of your problem. When you say

    G*msat*Me/r^2 = msat*v^2/r

    what did you use for r? It'd better be r1+r2, the distance from the center of the planet not the surface.
  4. Oct 7, 2009 #3
    Yeah for r1 I did the radius of the planet plus the radius of the satellite.

    So it should be 5700 kg up in orbit and still 5700 kg on the planets surface? Edit: I guess the answer is supposed to be in "N"
  5. Oct 7, 2009 #4


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    Don't confuse mass with weight. Mass is the same everywhere, but "weight" is the force with which the planet attracts the satellite near its surface. Mass is expressed in kilograms and weight in Newtons. Do you know how to find the weight?
  6. Oct 7, 2009 #5
    I am not sure how to find weight...? With the given problem that is
  7. Oct 8, 2009 #6


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    You can find the mass of the planet when the satellite is in orbit from the given quantities. Then put the satellite at rest on the surface of the planet and find the gravitational attraction at that point. That is the weight.
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