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If a projectile is launched from Earth with a speed equal to the escape speed, how high above the Earth's surface is it when its speed is three-fifths the escape speed?

I keep getting the answer 2.56x10^6 m after solving it a few different ways. Help would be great.

  • Physics -

    Its energy at launch is PE at surface of earth.

    1/2 mv^2=GMe*M/Re

    so when is GPE= (2/5)^2*GMeM/Re ?

    or GMeM/(Re+H)=4/25*GMeM/Re

    or 4h=21Re
    h= 21Re/4 or near 5.25Re
    or h= about 6.5*5.2E6 meters which is not your answer. check my work.

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