Question

If the length of the day is $ T $ , the height of that TV satellite above the earth’s surface which always appears stationary from earth, will be
(A) $ h = {\left[ {\dfrac{{4{\pi ^2}GM}}{{{T^2}}}} \right]^{1/3}} $
(B) $ h = {\left[ {\dfrac{{4{\pi ^2}GM}}{{{T^2}}}} \right]^{1/2}} $
(C) $ h = {\left[ {\dfrac{{{T^2}GM}}{{4{\pi ^2}}}} \right]^{1/3}} $
(D) $ h = {\left[ {\dfrac{{4{\pi ^2}GM}}{{4{\pi ^2}}}} \right]^{1/3}} $

Answer 1

Hint : To solve this question we have to find out the angular velocity of the earth from the given time. Then, as the satellite is geostationary, this will also be the angular velocity of the satellite. Finally we have to equate the centripetal force and the gravitational force acting on the satellite to get the final answer.

Formula used: The formulae used for solving this question are given by
 $ {F_C} = \dfrac{{m{v^2}}}{r} $ , here $ {F_C} $ is the centripetal force acting on a body of mass $ m $ which is performing a uniform circular motion in a circle of radius $ r $ with a speed of $ v $ .
 $ \omega = \dfrac{{2\pi }}{T} $ , here $ \omega $ is the angular velocity, and $ T $ is the time period.
 $ {F_G} = \dfrac{{GMm}}{{{r^2}}} $ , here $ {F_G} $ is the gravitational force acting between two objects of masses $ M $ and $ m $ , which are separated by a distance of $ r $ .
 $ v = r\omega $ , here $ v $ is the speed of a particle which is revolving in a circle of radius $ r $ with an angular velocity of $ \omega $ .

Complete step by step answer:
We know that the satellite which revolves around the earth and appears stationary from earth is known as a geostationary satellite. For being stationary with respect to the earth, it must revolve around the earth with the same angular velocity as that of the earth.
According to the question, the length of the day is given to be $ T $ . Now in one day, the earth completes its single rotation. So the time period of the rotation of the earth is equal to $ T $ . So the angular velocity of the earth becomes
 $ \omega = \dfrac{{2\pi }}{T} $
As the TV satellite has the same angular velocity as that of the earth, so the angular velocity of the satellite is also equal to $ \dfrac{{2\pi }}{T} $ . That is,
 $ {\omega _s} = \dfrac{{2\pi }}{T} $ (1)
Now, we know that the satellite rotates under the effect of the centripetal force which is provided by the gravitational attraction of the earth. The centripetal force is given by
 $ {F_C} = \dfrac{{m{v^2}}}{r} $
If the speed of the satellite is $ {v_s} $ , then
 $ {F_C} = \dfrac{{m{v_s}^2}}{r} $ (2)
We know that
 $ v = r\omega $
Therefore for the satellite
 $ {v_s} = \omega r $
Substituting (1)
 $ {v_s} = \dfrac{{2\pi }}{T}r $
Substituting this in (2) we get
 $ {F_C} = \dfrac{m}{r}\dfrac{{4{\pi ^2}{r^2}}}{{{T^2}}} $
 $ \Rightarrow {F_C} = \dfrac{{4{\pi ^2}mr}}{{{T^2}}} $ (3)
The satellite is rotating at a height of $ h $ above the earth’s surface. Let the radius of the earth be $ R $ . So the total radius of the circle in which the satellite rotates becomes $ \left( {h + R} \right) $ . This means that
 $ r = \left( {h + R} \right) $ (4)
Substituting (4) in (3) we get
 $ {F_C} = \dfrac{{4{\pi ^2}m\left( {h + R} \right)}}{{{T^2}}} $ (5)
Now, we know that the gravitational force is given by
 $ {F_G} = \dfrac{{GMm}}{{{r^2}}} $
Substituting (4)
 $ {F_G} = \dfrac{{GMm}}{{{{\left( {h + R} \right)}^2}}} $ (6)
Equating (5) and (6) we get
 $ \dfrac{{4{\pi ^2}m\left( {h + R} \right)}}{{{T^2}}} = \dfrac{{GMm}}{{{{\left( {h + R} \right)}^2}}} $
 $ \Rightarrow {\left( {h + R} \right)^3} = \dfrac{{{T^2}GMm}}{{4{\pi ^2}m}} $
On simplifying we get
 $ {\left( {h + R} \right)^3} = \dfrac{{{T^2}GM}}{{4{\pi ^2}}} $
If $ h > > R $ , then $ h + R \approx h $ . Therefore we get
 $ {h^3} = \dfrac{{{T^2}GM}}{{4{\pi ^2}}} $
Taking cube root both the sides, we finally get
 $ h = {\left[ {\dfrac{{{T^2}GM}}{{4{\pi ^2}}}} \right]^{1/3}} $
Hence, the correct answer is option C.

Note:
The height of a geostationary satellite is about $ 35800km $ , while the earth’s radius is equal to about $ 6400km $ . As we can see that the radius of the earth is very small compared to the height of the satellite. So we were able to assume that $ h > > R $ so that $ h + R \approx h $ .