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Radius of earth

  1. Aug 15, 2005 #1
    A rock is dropped from outer space (initial velocity=0)
    at a radius R from Earth's center. It is recorded moving at a
    speed 8368 m/s when it strikes the surface of the Earth. What
    was R? (in m) (Ignore the air resistance felt during the last
    few miles of the approach to the planet) R earth = 6.38*10^6
    m, M earth = 5.98*10^24. (in m)

    Not sure about what equation to even start with. Was thinking about
    g = (G*M earth)/r^2, but don't know what g is, and I don't see how I can find out without being given a time frame. Anyone know? thanks a lot
     
  2. jcsd
  3. Aug 15, 2005 #2

    Galileo

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    What you know is that the force acting on the rock is F=GMm/r^2.
    From this you could try setting up the equation of motion.
    Or you could take an energy approach, which is probably easier since you don't have a time variable to worry about. Just set up the total energy (potential + kinetic) at the beginning and at the end when the rock hits the earth.
     
  4. Aug 15, 2005 #3

    Fermat

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    You would start with,

    F = GM1 M2 / r^2

    where F is the gravitational force of attraction between two masses, M1 and M2 separated by a distance r between their centres of mass.

    Then use newton's 2nd law to work out an eqn of motion.
     
  5. Aug 15, 2005 #4

    Fermat

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    Can you do that? If the acceln isn't constant ?
     
  6. Aug 15, 2005 #5

    HallsofIvy

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    Of course. Energy is still conserved. Find the potential energy at radius R (kinetic energy is 0). Find the potential energy at the earth's surface. The difference is the kinetic energy of the rock as it hits the earth.

    (This is, of course, ignoring air resistance which would probably vaporize the rock before it hit.)
     
  7. Aug 15, 2005 #6

    Fermat

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    Is this it?

    KE = PE
    ½mv² = mgh
    v² = 2gh
    =======

    but shouldn't g be a constant value here?
     
    Last edited: Aug 15, 2005
  8. Aug 15, 2005 #7
    I agree....shouldn't g be constant? I don't see how can I use 9.81 m/s^2 in this situation, and I still have to determine g regardless. You guys are helping me understand it a lot, but I'm at a standstill with how to determine F, let alone r
     
  9. Aug 15, 2005 #8
    And wouldn't the potential energy at the earth's surface be 0?
     
  10. Aug 15, 2005 #9
    The acceleration due to gravity is different at different distances from Earth's center. You know F = ma = GMm/r2 where M is the mass of the Earth. Solve for a to get acceleration as a function of radius.
    On the other hand, the gravitational potential energy is -GMm/r (integration of force over distance), so instead of going through that acceleration formula, just use the above to find change in potential energy. Note r is with respect to the center of the Earth, although it doesn't really matter with potential energy where you set your 0.
     
  11. Aug 15, 2005 #10

    Andrew Mason

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    Your approach using change in potential energy is correct. I am not sure what you mean by "where you set your 0", though. If you use potentials, then you have to use [itex]U_{\infty} = 0[/itex] if you want [itex]U_{R} = -GM/R[/itex].

    AM
     
  12. Aug 16, 2005 #11
    Good point. I just wanted to reassure the person who assumed potential energy was 0 at the Earth's surface that it was okay to do so, as long as he added a sufficient constant of integration to the potential energy function. :) To the OP, the constant becomes irrelevant as it cancels itself in any change of potential energy equation.
     
    Last edited: Aug 16, 2005
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