Here, `m = 400`,
`R_(E) = 6.37 xx 10^(6) m`
Initial total energy, `E_(i) = PE + KE`
`= (-GM m)/(r ) + (1)/(2) m upsilon_(0)^(2) = (-GM m)/(r ) + (1)/(2) m ((GM)/(r ))`
`= (-GM m)/(2r) = (-GM m)/(2 xx (2 R_(E))) = (-GM m)/(4R_(E))`
Final total energy, `E_(f) = (-GM m)/(2 xx (4 R_(E))) = (-GM m)/(8R_(E))`
The change in total energy is, `Delta E = E_(f) - E_(i)`
`(-GM m)/(8R_(E)) - ((-GM m)/(4R_(E))) = (GM m)/(8R_(E))`
`= ((GM)/(R_(E)^(2))) xx (mR_(E))/(8) = (g m R_(E))/(8)`
`= ((9.81) xx (400) xx (6.37 xx 10^(6)))/(8)`
`= 3.13 xx 10^(9) J` Thus, teh energy required to transfer the satllite to the desired orbit `= 3.13 xx 10^(9) J`
the decrease in `K.E. = K_(i) - K_(f)`
`= (1)/(2) m ((GM)/(2R_(E))) - (1)/(2) m ((GM)/(4R_(E)))`
`= (1)/(8) (GM m)/(R_(E)) = (g mR_(E))/(8) = 3.13 xx 10^(9) J`
Gain in `P.E. = U_(f) - U_(i)`
`= (-GMm)/(4R) - ((-GMm)/(2R)) = (GM m)/(4R)`
`= (g mR_(E))/(4) = 2 xx 3.13 xx 10^(9) J = 6.26 xx 10^(9) J`
Thus, the again in `P.E.` is twice the loss in `K.E.`