Question

A satellite is moving in a low, nearly circular orbit around the earth. Its radius is roughly equal to that of the earth’s radius $R_e$. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it becomes \[\sqrt{\frac{3}{2}}\] times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is R. Value of R is:
1. $2{R_e}$
2. $3{R_e}$
3. $4{R_e}$
4. $2.5{R_e}$

Answer 1

Hint: Most of the problems in physics can be solved with the help of conservation laws. Here conservation of angular momentum and conservation of energy is used to solve the value of orbital velocity near the earth’s surface and for finding the value of farthest distance from the centre of earth. There are so many satellites revolving around the planet. They move in the same orbit due to the gravitational force of attraction between earth and satellites.

Formula used:
Using conservation of angular momentum,
Initial angular momentum = final angular momentum
$m{{v}_{o}}{{R}_{e}}=mvR$

Complete answer:
Conservation of angular momentum means initial angular momentum is same as final and conservation of energy means total energy is constant.

Total energy at earth’s surface = Potential energy at earth’s surface +Kinetic energy at earth’s surface Similarly total energy at farthest distance is the sum of kinetic and potential energy at farthest distance. Let ‘m’ be the mass of a satellite and ‘M’ be the mass of earth.

Using conservation of angular momentum,
Initial angular momentum = final angular momentum
$m{{v}_{o}}{{R}_{e}}=mvR$
Given
${{v}_{o}}=\sqrt{\frac{GM}{{{R}_{e}}}}\times \sqrt{\frac{3}{2}}$

Therefore, from conservation law velocity at farthest distance is
$v=\sqrt{\frac{GM}{{{R}_{e}}}}\times \sqrt{\frac{3}{2}}{{R}_{e}}\times \frac{1}{R}$

Using conservation of energy,
Total energy at earth’s surface = Total energy at farthest distance
\[\frac{-GMm}{{{R}_{e}}}+\frac{1}{2}m{{v}_{o}}^{2}=\frac{-GMm}{R}+\frac{1}{2}m{{v}^{2}}\]

On substituting values of vo and v obtained from applying conservation of angular momentum,
\[\frac{-1}{4{{R}_{e}}}=\frac{-1}{R}+\frac{3{{R}_{e}}^{2}}{4{{R}^{2}}}\]
\[\therefore R=3{{R}_{e}}\]

Therefore, the Answer is (b).

Note: So many satellites are revolving around the earth and many rockets are fired from it at the correct time to achieve desired orbits. All the equations derived above for planets can be used for satellites also. But here M will be the mass of the earth and m will be the mass of the satellite.