Answer :
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
Answer:
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.