Question

A $400kg$ satellite is in a circular orbit of radius $2R_e$ 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 $4R_e$.	1	$-6.26\times 10^{9}J$
B	Change in $KE,\Delta K$$(KE\to$ Kinetic Energy)	2	$3.13\times 10^{9}J$
C	Change in potential energy, $△PE$	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 $E_i=-\frac{G{M}_{e}m}{4R_e}$
Finally, energy of satellite, $E_f=\frac{-G{M}_{e}m}{8R_e}$
Change in total energy, $\Delta E=E_f-E_i$
$=\left(-\frac{G{M}_{e}m}{8R_e}\right)-\left(-\frac{G{M}_{e}m}{4R_e}\right)$ $=\frac{G{M}_{e}m}{8R_e}$ or $\frac{g{R}_{e}m}{8}\left[\because g\frac{GM}{R_e^2}\right]$ $=3.13\times 10^{9}J$
Thus, $\Delta E=\frac{g{R}_{e}m}{8}=\frac{9.8\times 400\times 6.37\times 10^6}{8}$
$\left[\because g=9.8m{s}^2,m=400kg,R=6.37\times 10^{6}m\right]$
$=3.13\times 10^{9}J$
B. The kinetic energy is reduced and change in $KE$ is just negative of $\Delta E$.
$\to \Delta K=K_f-K_i=-3.13\times 10^{9}J$
C. The change in potential energy is twice the change in the total energy namely,
$\Delta PE=P{E}_f-P{E}_i=-6.26\times 10^{9}J$
Hence, $A\to 2,B\to 3$ and $C\to 1$.