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Gravitational pull on asteroid/Period it takes for one orbit

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A bit of asteroid material of mass 2.5 x 10^6 kg orbits around the earth at an altitude of 2800km. How long does it take the asteroid to complete one orbit.

    M asteroid = 2.5 x 10^6 kg
    altitude of asteroid = 2800 km
    radius asteroid = 200m
    radius earth = 6.37 x 10^6 m


    2. Relevant equations
    Fnet = mv^2/r
    C = 2∏r
    t = d/v



    3. The attempt at a solution
    r = rEarth + rAsteroid + altitude
    r = 9.2 x 10^6 m

    C = 2∏r
    C = 5.78 x 10^7
    C = d

    Fg = mv^2/r
    mg = mv^2/r
    9.8 = v^2/r
    9.8 (9.2 x 10^6) = v^2
    v = 9500 m/s

    t = d / v
    t = 5.78 x 10^7 / 9500
    t = 6084 s
    t = 1.7 h

    My friend got 0.29 and the teacher had told him he was right. I didn't have a chance to see what I did wrong so I'm wondering if anyone here can tell me where I made my mistake.
     
  2. jcsd
  3. Mar 1, 2012 #2

    gneill

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    Staff: Mentor

    Use Newton's law of gravitation to find the gravitational force when the object is not very close the the surface of the Earth; f = m*g just won't cut it here.
     
  4. Mar 1, 2012 #3
    That would be

    F = G * M * m/r if I'm not mistaken?

    Would F be mg still? Also how does that help me find the velocity?
     
  5. Mar 1, 2012 #4

    gneill

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    Staff: Mentor

    The r should be squared in your formula.

    No, F is not mg. You just wrote the equation for the force due to gravity. Use that where you used mg in your work.
     
  6. Mar 1, 2012 #5
    I think I know what you mean, would it be like this.

    mv^2/r = (G)(M)(m)/r^2

    v^2/r = (G)(M)/ r^2

    v^2 = (G)(M)/r

    Before I substitute values in, how do I know what to put for M and what m would be? I know that the two m would cancel out and that was the mass of the asteroid, so then M should be the mass of the earth. Generally speaking though, how do I know what mass to use?

    Edit:

    I tried substituting values in and I got 2.4 h which is still wrong apparently.

    Work:

    v^2 = GM/r
    v^2 = (6.67 x 10^-11)(5.96 x 10^24) / 9.2 x 10^6
    V = 6573

    t = d/v
    t = 5.78 x 10^7 / 6573
    t = 8794 s
    t = 2.4 h
     
    Last edited: Mar 1, 2012
  7. Mar 1, 2012 #6

    gneill

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    Staff: Mentor

    That looks better. The small m, which was also used in the expression for the centripetal force, represents the mass of the orbiting body. The large M is the mass of the primary, in this case the mass of the Earth.
     
  8. Mar 1, 2012 #7
    Ah, I editted the other post not thinking you had seen my previous post. Anyways..

    I tried substituting values in and I got 2.4 h which is still wrong according to what I was told.

    Work:

    v^2 = GM/r
    v^2 = (6.67 x 10^-11)(5.96 x 10^24) / 9.2 x 10^6
    V = 6573

    t = d/v
    t = 5.78 x 10^7 / 6573
    t = 8794 s
    t = 2.4 h
     
  9. Mar 1, 2012 #8

    gneill

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    Staff: Mentor

    Your answer looks fine to me.
     
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