A skylab of mass `m` kg is first launched from the surface of the earth in a circular orbit of radius `2R` (from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius `3R`. The minimum energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit are

Explanation

(a) The energy of the sky lab on the surface of earth And the total energy of the sky lab in an orbit of radius 2 R is
`E_(1) = -(GMm)/(4R)`
So the energy required to placed the lab from the surface of earth to the orbit of radius 2R is given as
`E_(1) - E_(s) = -(GMm)/(4R) - [-(GMm)/(R )] = (3)/(4)(GM m)/(R )`
or `Delta R = (3)/(4) (m)/(R ) xx gR^(2) = (3)/(4)mgR`
[As `g = (GM)/(R^(2))`]
or `Delta E = (3)/(4)(2 xx 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 of radius 3R the total energy of sky lab is
`E_(1) = -(GMm)/(2(3R)) = -(GMm)/(6R)`
or `E_(2) - E_(1) = -(GMm)/(6R) -(-(GMm)/(4R)) = (1)/(12)(GMm)/(R )`
or `Delta E = (1)/(12) mgR`
`= (1)/(12)(12.8 xx 10^(8))`
`= 1.1 xx 10^(10)J`