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A sky laboratory of mass 2 xx 10^(3) kg ...

A sky laboratory of mass `2 xx 10^(3) kg` has to be lifted from one circular orbit of radius `2R` into another circular orbit of radius `3R`. Calculate the minimum energy required if the radius of earth is `R= 6.37 xx 10^(6) m` and `g= 9.8 ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
`1.04xx10^(10)J`
Total energy of sky laboratory `E=-(GMm)/(2r)`
`E_(1)=-(GMm)/(2(2R))` and `E_(2) =-(GMm)/(2(3R))`
Therefore the energy required
`/_\E=E_(2)-E_(1)=GMm(1/(4R)-1/(6R))=(GMm)/(12R)`
`((R^(2)g)m)/(12R)=(mgR)/12=1.04xx10^(10)J`
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