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physics

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A 70 kg man weighs 686 N on the Earth's surface. How far above the surface of the Earth would he have to go to "lose" 14% of his body weight?

i don't even know where to start with this question... i need help!

  • physics - ,

    .86 = Rearth^2/r^2

    because
    W = G m M/R^2
    and G m M does not change
    remember that they really ask for
    r-Rearth
    the distance above ground, not the bigger radius from earth center

  • physics - ,

    could try and explain more in depth? like wat does W stand for? his weight? and i don't really get how your solving it..

  • physics - ,

    F=(G*m1*m2)/r^2
    F- force
    G- gravitational constant
    m1- mass on earth
    m2- mass of object
    r- radius on earth plus given distance over it

    .86*686=589.96 to find final force
    plug in everything..
    589.96=(6.67*10^-11*70*5.98*10^24)/ (h+6,370,000)^2
    Solve for h!

    h=509411
    =]

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