Question

A satellite is in a circular equatorial orbit of radius $7000\text{km}$ around the earth. If it is transferred to a circular orbit of double the radius.

Column I	Column -II
(A) angular momentum	(P) increases
(B) area of earth covered by satellite signal	(Q) decreases
(C) potential energy	(R) becomes double
(D) kinetic energy	(S) becomes half

Which of the following option has the correct combination considering column-I and column-II.

• A. $D\to R,S$
• B. $A\to P$
• C. $B\to Q,S$
• D. $C\to S$

Answer 1

The correct option is B $A\to P$

(A) Angular momentum, $L=mvr=m\left(\sqrt{\frac{GM}{r}}\right)r$
$\therefore L\propto \sqrt{r}$
So angular momentum increases and becomes $\sqrt{2}$ times
$A\to P$

(B)

Area of the earth covered by satellite signal is given as, $A=4\pi R^2-2\times 2\pi R^2(1-\cos \theta )=4\pi R^2\cos \theta$ where $R$ is radius of the earth.
As satellite goes away from the earth, $\theta$ decrease, hence $\cos \theta$ increase. So area covered by satellite incraeses.
$B\to P$

(C) Potential energy, $U=\frac{-GMm}{r}$
So, when $r$ is doubled $PE$ becomes half and increases.
$C\to P,S$

(D) Kinetic energy, $KE=\frac{GMm}{2r}$
So, when $r$ is doubled, $KE$ decreases and becomes half.
$D\to Q,S$