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A 75 kg man weighs 735 N on the Earth's surface. How far above the surface of the Earth would he have to go to "lose" 15% of his body weight?

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  1. m g = G M m /r^2
    so m does not matter if we only want ration

    G M is constant
    r = earth radius (any units) but be consistent. You might use 6.38*10^6 kg

    R = r + d

    .85 G M/r^2 = G M/(r+d)^2

    r^2 = .85 (r+d)^2
    r = .922 (r+d)
    r - .922 r = .922 d
    d = r (1-.922)/.922 = .0846 r

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  2. The value of acceleration due to gravity on surface of the earth is
    g=weight of the person/mass of the
    person=735/75=9.8ms^-2
    The man loses 15% of weight,
    g'=g85/100
    Required height =g'=g*R^2/(R+H)^2

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  3. only 6 years late

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  4. Thanks

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