Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass me, radius re) and place it in a circular low each orbit, that is, an orbit whose altitude above the earth's surface is much less than re. (as an example, the international space station is in low earth orbit at an altitude of about 400 km, much less than re=6380 km.) you can ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation. express your answer in terms of the variables m, me, re and appropriate constants. b) calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. you can ignore the gravitational effects of the sun, the moon, and the other planets. express your answer in terms of the variables m, me, re, and appropriate constants.

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)

this could be newton#'s law of cooling. the cooling curve is exponential