A sky lab of mass 2xx10^3 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 radius 3R . Calculate the minimum energy required (a) to place the lab in the first orbit (b) to shift the lab form first orbit to the second orbit . Given , R = 6400 Km and g = 10m//s^2

A sky lab of mass 2xx10^3 kg is first launched from the surface of ear

1.	A sky lab of mass 2xx10^3 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 radius 3R . Calculate the minimum energy required (a) to place the lab in the first orbit (b) to shift the lab form first orbit to the second orbit . Given , R = 6400 Km and g = 10m//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=1/2mv_(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`