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7 January, 05:53

How much work would be needed to raise the payload from the surface of the moon (i. e., x = r) to an altitude of 4r miles above the surface of the moon (i. e., x = 5r) ?

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  1. 7 January, 08:02
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    The gravitational attraction between the moon and the payload is given as

    F (x) = r²P/x²

    Where P is the load weight.

    The total amount of work for raising the load from x = r to x = (4r+r) I. e from x = r to x = 5r is given as

    W = ∫ F (x) dx. From x = r to x = 5r

    W = ∫ r²P/x² dx

    W = r²P ∫ x^-2 dx

    W = r²P [ x^ (-2+1) / (-2+1) ]

    W = r²P [ x^-1 / - 1]

    W = - r²P•x^-1 ... From x = r to x = 5r

    W = - rP²• (1/x) ... From x = r to x = 5r

    W = - r²P• (1/5r - 1/r)

    W = - r²P * (-4/5r)

    W = 4r²P / 5r

    W = 4rP / 5 milepounds

    So, the work need to raise the payload from x=r to x=4r is 4rP / 5

    Where P is the weight of the load.
  2. 7 January, 09:07
    0
    Gravitational potential at surface:

    V1 = - k / r where k = G M m

    V2 = - k / (5 r)

    V2 - V1 = k (1 / r - 1 / (5 r)) = k / r (1 - 1/5) = 4 k / (5r)

    Which is the work required to raise a payload of mass m to 5 r
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