Answer:
a) W = - (G mM/ Re²) r (1 - r/Re)⁻¹
Explanation:
Work is defined as the product force by distance
dW = F. dr
Where the blacks indicate vectors, the force in this exercise the gravitational attraction force and is a radial force, if we assume that the launch is vertical to orbit the Angle between this force and the displacement vector is zero degrees, so the product scalar is reduced to an ordinary product
dW = ∫ F dr
Let's replace in the gravitational force equation and integrate
∫dW = ∫ (G m M / r²) dr
W = G m M ∫ dr / r²
W-Wo = G m M (- 1 / r)
Let us evaluate this integral from the lower limit the surface of the Earth r = Re where the work is zero (Wo = 0) to the upper limit r = R ’dorniel work vouchers W
W = G m M (-1 / R’+ 1 / Re)
Let's simplify this expression
R ’= Re + r r << Re
W = G m M [1 / Re - 1 / (Re -r)]
W = G m M [(Re-r) - Re] / ((Re-r) Re]
W = G mM (-r) / [Re² (1- r/Re)]
W = - (G mM/ Re²) r (1 - r/Re)⁻¹
This exact exposed scale, if we expand the quantity (1-x)⁻¹ = 1 + x + ...
The expression is
W = - (G m M / Re²) r (1+ r/Re)
b) It is requested to calculate work from this orbit to a very high orbit, we can repeat the integral evaluating the lower orbit from the lower limit to the upper limit the very high orbit
R ’= Re + r r << Re
R₂ = Re + R R >> Re
W = G m M (-1 / R₂ + 1 / R ’)
W = G m M [-1 / (Re + R) + 1 / (Re + r)]
Let's simplify
W = (G mM) [(R-r) / (Re + r) (Re + R)]
Let's develop the parentheses
(Re + r) (Re + R)] = Re² + Re R + Re r + r R
(Re + r) (Re + R)] = Re R (Re / R + 1 + r / R + r / Re)
W = (G mM) [R (1-r/R) / Re R (Re / R + 1 + r / R + r / Re)]
W = (G mM /Re) (1-r / Re) [(1 + Re / R + r / R + r / Re)]⁻¹
This solution is accurate.
If an approximate solution is desired, some terms can be neglected and a binomial expansion carried out.
Remembering
r << Re Re << R ⇒ r <<< R
W = (G mM / Re) (1-r / Re) (1 + Re / R + r / Re)
