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Equation for Final velocity: Projectile Fired from Earth

  1. Apr 14, 2012 #1
    Suppose a projectile is fired upward from the surface of the Earth. If the initial speed of the projectile is v = 12.0 km/s (more than the escape velocity): how fast will it be moving it is so far from Earth that the force of gravity is approximately zero?

    The equation I was given to solve is:

    1/2 v(initial)^2 - G(Mass of Earth) / radius of Earth = 1/2v(final)^2

    I do NOT think this equation is correct. Can someone please help/advise? Many thanks.
  2. jcsd
  3. Apr 14, 2012 #2


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    Homework Helper

    "so far from Earth that the force of gravity is approximately zero" means basically at infinite separation.

    At all times, the sum of Kinetic and Potential energy is constant.

    Kinetic energy is given by 1/2 m v2

    Potential energy is given by -GMEm/R

    when r = ∞ , Potential Energy = zero

    thus comparing "at the surface" - the initial position - to infinite separation - the final position - we have

    1/2 m vi2 + -GMEm/RE = 1/2 m vf2

    Once you cancel the common term - the mass of the satellite - you get the equation you were given.

    As an aside:
    If you wanted to find out how fast it was travelling 4 Earth radii above the surface [which means a centre - centre separation of 5RE] you would use:

    1/2 m vi2 + -GMEm/RE = 1/2 m vf2+ -GMEm/5RE
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