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_(1)-E_(s)=-(GMm)/(2(2R))-[-(GMm)/(R)]=3/4(GMm)/R`
i.e `DeltaE=3/4m/RxxgR^(2)=3/4mgR" " ["asg" =(GM)/R^2]`
i.e. `DeltaE=3/4(2xx10^(3)xx10xx6.4xx10^6)=`
`3/4(12.8xx10^(10))=9.6xx10^(10)J`
(b) As for II orbit `r = 3R , E_(11)=-(GMm)/(2(3R))=-(GMm)/(6R)`
`:. E_(11)-E_(1)=-(GMm)/(6R)-(-(GMm)/(4R))=1/12 (GMm)/R`
But as `g=(Gm//R^2), ` i.e. `GM=gR^2` or
`DeltaE=1/2mgR=1/12(12.8xx10^(10))=1.1xx10^(10)J`