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Homework Help: Centripetal force or escape velocity?

  1. May 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A certain spherical planet which is not rotating has a radius of 6.36 x 105 m and a mass of 1.89 x 1021 kg. At what minimum speed would a vehicle traveling along its surface just begin to leave the ground?

    2. Relevant equations
    FC = mv2/r
    FG = GMm/d

    3. The attempt at a solution
    My problem is I don't know how to set the initial equation up: Is it,
    FC = GMm/d2 = mv2/d
    Or is it,
    1/2mv2 = GMm/d?
    They will differ by a factor of root 2.
    If it was vertical velocity I would know it needs to be KE = PE, but since it's horizontal I'm confused. Can someone explain?
  2. jcsd
  3. May 25, 2010 #2


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    Homework Helper

    Well the planet is not rotating, so there is not centripetal force.

    So you will need to equate KE to PE as in your second equation.
  4. May 25, 2010 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    You're looking for the speed required for a stable circular orbit at radius "R" where R just happens to be the radius of the planet itself. At speeds higher than this, gravity won't be able to provide enough centripetal force to keep the car moving in an orbit at that radius, and it will move into a higher orbit. So it is indeed the first equation you wrote that is applicable. The condition |KE| = |PE| is required for the car to escape to infinity. For the car to remain gravitationally bound to the planet in a closed orbit of radius R, it is sufficient that |KE| = (1/2)|PE|, where |PE| is the potential energy of the system when the car is in an orbit of radius R. You know that this last statement is true because it comes from equating the centripetal and gravitational accelerations. But you might be wondering why it must be true from an energy standpoint. This last statement comes from the virial theorem, which is a more general statement about the energy balance in systems that have achieved some sort of dynamical equilibrium. More generally the virial theorem says that for a *gravitationally bound* system of N particles interacting solely under gravity, |KE| = 1/2|PE|, where KE is the total kinetic energy of the system and PE is its total gravitational potential energy. In our situation, so long as |KE| > 1/2|PE|, the car will not be able to remain in a bound orbit of radius R, and will try to move into a higher orbit. This is precisely the condition you derived, and the factor of 2 discrepancy has now been explained.

    Note that the higher orbit might not necessarily be circular. If it's elliptical, and it partly intersects the lower circular orbit that the car used to be in, then what will actually happen is that the car will go flying into the air on some curved trajectory and eventually come crashing back down into the ground, just like any other projectile that is launched from the surface.
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