Question

A $400\, kg$ satellite is in a circular orbit of radius $2 R_{e}$ about the earth, where $R_{e}=$ radius of the earth. With reference to the above situation, match the Column I (quantities) with Column II (calculated values) and select the correct answer from the codes given below.

Column I		Column II
A	Energy required to transfer it to a circular orbit of radius $4 R_{e}$.	1	$-6.26 \times 10^{9} J$
B	Change in $KE , \Delta K$$( KE \rightarrow$ Kinetic Energy)	2	$3.13 \times 10^{9} J$
C	Change in potential energy, $\triangle PE$	3	$- 3.13 \times 10^9\,J$

• A. $A -(2), B - (3), C -(1)$
• B. $A -(1), B - (2), C -(3)$
• C. $A -(3), B - (2), C -(1)$
• D. $A -(2), B - (1), C -(3)$

Answer 1

Answer: (A)

A. Initially, energy of satellite $E_{i}=-\frac{G M_{e} m}{4 R_{e}}$
Finally, energy of satellite, $E_{f}=\frac{-G M_{e} m}{8 R_{e}}$
Change in total energy, $\Delta E=E_{f}-E_{i}$
$=\left(-\frac{G M_{e} m}{8 R_{e}}\right)-\left(-\frac{G M_{e} m}{4 R_{e}}\right)$ $=\frac{G M_{e} m}{8 R_{e}}$ or $\frac{g R_{e} m}{8}\left[\because g \frac{G M}{R_{e}^{2}}\right]$ $=3.13 \times 10^{9} J$
Thus, $\Delta E=\frac{g R_{e} m}{8}=\frac{9.8 \times 400 \times 6.37 \times 10^{6}}{8}$
$\left[\because g=9.8\, ms ^{2}, m=400 \,kg , R=6.37 \times 10^{6}m\right]$
$=3.13 \times 10^{9} J$
B. The kinetic energy is reduced and change in $KE$ is just negative of $\Delta E$.
$\rightarrow \Delta K=K_{f}-K_{i}=-3.13 \times 10^{9} J$
C. The change in potential energy is twice the change in the total energy namely,
$\Delta PE = PE _{f}- PE _{i}=-6.26 \times 10^{9} J$
Hence, $A \rightarrow 2, B \rightarrow 3$ and $C \rightarrow 1$.