JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Gravitation

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 become

\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:

Answer: (C)

Solution

In the low circular orbit the speed is

v_{0} = \frac{G M _{e}}{R _{e}}

. After the burn the speed becomes

v = \frac{3}{2} v_{0}

at the same radius

R_{e}

, which is now the perigee of an ellipse. Conserving angular momentum between perigee

R_{e}

and apogee

R

v R_{e} = v_{a} R

, and conserving energy,

\frac{1}{2} v^{2} - \frac{G M _{e}}{R _{e}} = \frac{1}{2} v_{a}^{2} - \frac{G M _{e}}{R}

. Eliminating

v_{a}

and using

v^{2} = \frac{3}{2} \frac{G M _{e}}{R _{e}}

leads to

R = 2 R_{e}

, so the farthest distance reached is

2 R_{e}

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option