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 (from the centre of earth) and then it is shifted from this circular orbit to another circular orbit of raidus 3R. Calculate the minimum energy required (a) to place the lab in the first orbit (b) to shift the lab from first orbit to the second orbit. GIven , R = 6400 Km and g = 10 m//s^(2).

A sky lab of mass 2 xx 10^(3) kg is first launched from the surface of

1.	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 (from the centre of earth) and then it is shifted from this circular orbit to another circular orbit of raidus 3R. Calculate the minimum energy required (a) to place the lab in the first orbit (b) to shift the lab from first orbit to the second orbit. GIven , R = 6400 Km and g = 10 m//s^(2).
Answer» Solution :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`