Question

The time period of a satellite very close to earth is T. The time period of geosynchronous satellite will be:
A. $2\sqrt 2 (T)$
B. $6\sqrt 6 (T)$
C. $7\sqrt 7 (T)$
D. $\dfrac{1}{{7\sqrt 7 }}(T)$

Answer 1

Hint Geosynchronous satellite is a satellite which revolves in the geosynchronous orbit. For a geosynchronous satellite the orbital period is equal to the earth's rotation period. After each sidereal day, this satellite returns to the same position in the sky. Geosynchronous satellites are very near to the earth 's surface.

Complete Step by step solution
As we all know that geosynchronous satellite orbits at the height of 6R from the surface of earth.
According to Kepler's third law,
The third law of Kepler states that the square of the time period of a planet is proportional to the cube of the semimajor axis. For a circular orbit the semimajor axis is the same as the radius.
Mathematically,
  ${T^2}\infty {r^3}$
 ${(\dfrac{{{T_1}}}{{{T_2}}})^2} = {(\dfrac{{{r_1}}}{{{r_2}}})^3}$
Now we have already said that geosynchronous satellite orbits at a distance 6R from the earth surface.
So,
$
   \Rightarrow {(\dfrac{{{T_1}}}{{{T_2}}})^2} = {(\dfrac{R}{{R + 6R}})^3} \\
   \Rightarrow {(\dfrac{{{T_1}}}{{{T_2}}})^2} = {(\dfrac{R}{{7R}})^3} \\
    \\
$
$
   \Rightarrow {(\dfrac{{{T_1}}}{{{T_2}}})^2} = {(\dfrac{1}{7})^3} \\
   \Rightarrow (\dfrac{{{T_1}}}{{{T_2}}}) = \dfrac{1}{{\sqrt {{7^3}} }} \\
   \Rightarrow {T_2} = {T_1}\sqrt {{7^3}} = 7\sqrt 7 \\
$
Hence, time period for geosynchronous satellite will be $7\sqrt 7 $

Therefore, Correct answer will be option number C

Note
Satellites are launched from the earth so as to move round it. A number of rockets are fired from the satellite at proper time to establish the satellite in the desired orbit. Once the satellite is placed in the desired orbit with the correct speed for that orbit , it will continue to move in that orbit under gravitational attraction of the earth.