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=1/2mv_0^2 + [-(GMm)/r]=(-GMm)/(2r)[as v_0=sqrt((GM)/r)]`
(a) So the energy required to place the lab from the surface of earth to the orbit of radius 2R,
`E_l-E_S=-(GMm)/(2(2R)) -[-(GMm)/R]=3/4 (GMm)/R`
i.e., `DeltaE=3/4 m/R xx gR^2 =3/4 mgR [ as g=(GM)/R^2]`
i.e., `DeltaE=3/4 (2xx10^3xx10xx6.4xx10^6)=3/4(12.8xx10^10)=9.6xx10^10` J
(b) As for II orbit r=3R , `E_(||)=-(GMm)/(2(3R)) =-(GMm)/(6R)`
`therefore E_(||)-E_| =-(GMm)/(6R) - (-(GMm)/(4R))=1/12 (GMm)/R`
But as `g=(GM//R^2)`, i.e., GM=`gR^2` or
`DeltaE=1/12 mgR = 1/12 (12.8xx10^10)=1.1xx10^10 J`