A sky lab of mass `2 xx 10^(3)kg` is first launched from the surface of earth in a circular orbit of radius `2R` and them it is shifted from this circular orbit to another circular orbit of radius `3R`. Calculate the energy required
(a) to place the lab in the first orbit,
(b) to shift the lab from first orbit to the second orbit. `(R = 6400 km`, `g = 10 m//s^(2))`

Explanation

The energy of the sky lab on the surface of earth
`E_(s) = KE + PE = 0 + (-(GMm)/(R)) = -(GMm)/(R)`
And the energy of the sky lab in an orbit of radius r
`E = (1)/(2)mv_(0)^(2) + [-(GMm)/(r)] = (-GMm)/(2r) ["as " v_(0) = sqrt((GM)/(r))]`
(a) So the energy required to palce tha lab from the surface of earth to the orbit of radius 2R,
`E_(1) - E_(s) = -(GMm)/(2(2R)) -[-(GMm)/(R)] = (3)/(4)(GMm)/(R)`
i.e., `Delta E = (3)/(4)(m)/(R) xx g R^(2) = (3)/(4) mgR " "["as g" = (GM)/(R^(2))]`
i.e., `Delta E = (3)/(4)(2 x 10^(3) xx 10 xx 6.4 xx 10^(6)) = (3)/(4)(12.8 xx 10^(10)) = 9.6 xx 10^(10)J`
(b) As for II orbit r = 3R,
`E_(11) = -(GMm)/(2(3R)) = -(GMm)/(6R)`
`therefore E_(11) - E_(1) = -(GMm)/(6R) -(-(GMm)/(4R)) = (1)/(12) (GMm)/(R)`
But as `g = (GM//R^(2))`, i.e., `GM = gR^(2)` or `Delta E = (1)/(12)mgR = (1)/(12)(12.8 xx 10^(10)) = 1.1 xx 10^(10)J`