Question

(a) Determine the amount of work (in joules) that must be done on a 103 kg payload to elevate it to a height of 1007 km above the Earth's surface.
  MJ
(b) Determine the amount of additional work that is required to put the payload into circular orbit at this elevation.
J

Answer 1

F = G * m * M / r^2

where G = 6.67384 * 10^-11
m = 103 kg
M = 5.9736 * 10^24 kg (the mass of the earth)
r = radius from the centre of the earth in metres

Mean radius of the earth = 6371 km
Payload height = 6371 + 1007 = 7378 km

a) Potential Energy gained by payload in lifting it through 1007 km
= integral (6371 * 10^3,7377 * 10^3) (G * m * M / r^2) dr
= (6371 * 10^3,7377 * 10^3) (-G * 102 * 5.9736 * 10^24 / r)
= 6.67384 * 102 * 5.9736 * 10^7 * (1/6.371 - 1/7.378)
= 854 MJ

With this amount of energy, the payload will just reach the height of 1007 km and will be stationary for a moment before falling back to earth. This is the amount of work necessary to lift the payload to 1007 km above the earth.

b)
To stay in orbit, it will need to be travelling fast enough so that the centrifugal force upwards caused by the payload's orbit exactly cancels the attraction downwards due to gravity.

F = G * m * M / r^2 (attraction due to gravity)
and
F = ma (acceleration downwards)

where a = the acceleration of the payload necessary to hold it in a circular orbit
a = v^2 / r (acceleration upwards due to centrifugal force)

where v = the velocity in m/s

The Kinetic Energy of the payload is m * v^2 / 2 J due to its motion

So
G * m * M / r^2 = m * v^2 / r
or
G * m * M / (2 * r) = m * v^2 / 2 = KE

KE = 6.67384 * 102 * 5.9735 * 10^7 / (2 * 7.378) J
= 4066.419 * 10^7 * 0.067778
= 275.615 * 10^7

= 2.7562 * 10^9 J
or
2756.2 MJ