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)?

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

Explanation:

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.

pe = weightxheight = 2000x1000=2,000,000joules

scientific laws are the result of testing theories and proving them to be facts.