Question

Two satellites $A$ and$B$ are revolving around the earth in circular orbits of radius $r_1$ and $r_2$ respectively with $r_1<r_2$. Plane of motion of the two are same. At position $1$, $A$ is given an impulse in the direction of velocity by firing a rocket so that it follows an elliptical path to meet $B$ at position $2$ as shown. Focal length of the elliptical path are $r_1$ and $r_2$ respectively. At position $2$, $A$ is given another impulse so that velocities of $A$ and $B$ at $2$ become equal and the two move together.
For any elliptical path of the satellite time period of revolution is given by Kepler's planetary law as $T^2\propto r^3$ where $a$ is semi major axis of the ellipse which is $\frac{r_1+r_2}{2}$ in this case. Also angular momentum of any satellite revolving around the Earth will remain a constant about Earth's centre as force of gravity on the satellite which keeps it in elliptical path is along its position vector relative to the earth centre.

If the two bodies have same mass then,

• A. A would have more potential energy than $B$ while on their initial circular paths
• B. A would have more kinetic energy than $B$ while on their initial circular paths
• C. Relative to Earth's centre, angular momentum of $A$ when it is in elliptical path would be less than angular momentum of $B$
• D. During the whole process angular momentum of $B$ would be more than angular momentum of $A$

Answer 1

Answer: (B), (C)

Relative to Earth's centre, angular momentum of $A$ when it is in elliptical path would be less than angular momentum of $B$
A would have more kinetic energy than $B$ while on their initial circular paths

We know the expressions for kinetic energy and potential energy os a satellite in orbit are,

$KE=\frac{1}{2}m{\left(\sqrt{\frac{GM}{R}}\right)}^2=\frac{GMm}{2R}$,

$PE=-\frac{GMm}{R}$.

From these expressions, although magnitude of potential energy of A will be higher, but due to negative sign, potential energy of B will be higher than that of A.

Angular momentum is given by $L=mvR=m\sqrt{\frac{GM}{R}}R=m\sqrt{GMR}$. HEnce angular momentum of

A relative to earth will be lesser than that of B as $R_A<R_B$.