\(E_{i}-\frac{G M_{E} m}{6 R_{E}}\)
Final total energy of the satellite is
\(E_{f}=-\frac{G M_{E} m}{18 R_{E}}\)
The change in the total energy is
\(\Delta E=E_{f}-E_{i}\)
\(\Delta E=-\frac{G M_{E} m}{18 R_{E}}-\left(-\frac{G M_{E} m}{6 R_{E}}\right)\)
\(=-\frac{G M_{E} m}{18 R_{E}}+\frac{G M_{E} m}{6 R_{E}}=\frac{G M_{E} m}{9 R_{E}}\)
Thus, the energy required to transfer the satellite to the desired orbit \(=\frac{G M_{E} m}{9 R_{E}}\)
($\mathrm{g}=10\; \mathrm{ms}^{-2}$, ઉત્તર ધ્રુવ પાસે પૃથ્વીની ત્રિજ્યા $=6400\; \mathrm{km}$ )
($R_{\text {earth }}=6400\;km$ $, r =2000\;km$ $, M _{\text {earth }}=6 \times 10^{24}\;kg$ આપેલ છે $)$