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(B) The total energy of the satellite on the surface of the earth is the sum of its potential energy and kinetic energy. Since it is at rest on the su

Vedclass · Accepted Answer

$\frac{13GMm}{14R}$

Vedclass · Answer

(B) The total energy of the satellite on the surface of the earth is the sum of its potential energy and kinetic energy. Since it is at rest on the surface,its kinetic energy is $0$. Thus,$E_1 = -\frac{GMm}{R}$.When the satellite is in a circular orbit of radius $r = 7R$,its total energy is given by $E_2 = -\frac{GMm}{2r}$.Substituting $r = 7R$,we get $E_2 = -\frac{GMm}{2(7R)} = -\frac{GMm}{14R}$.The minimum energy required to be spent by the launching vehicle is the difference between the final energy and the initial energy: $\Delta E = E_2 - E_1$.$\Delta E = -\frac{GMm}{14R} - (-\frac{GMm}{R}) = -\frac{GMm}{14R} + \frac{GMm}{R}$.$\Delta E = \frac{-GMm + 14GMm}{14R} = \frac{13GMm}{14R}$.

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