Question

A geostationary satellite is orbiting the earth at a height 6R above the surface of earth, where R is radius of earth. What will be the time period of another satellite at a height 2.5 R from the
surface of earth?​

Answer 1

$T_0\degree =6\sqrt{2}$

Explanation:

According to Kepler's Law,the square of the time period of a satellite is a proportional to the cube  of the distance frrom the center.

Hence$,T^2 = R^3$

For the geo stationary satellite,T  =24 hours

Here,$T^2$∝$(R+6R)^3= (7R)^3$

Similarly,for other satellite

$T_0{\degree}^{2}$∝$(R+2.5R)^3$ =$(3.5R)^3$

∴,$\frac{T_0{\degree}^{2}}{T^2} = \frac{(3.5R)^3}{(7R)^3}$

                               = $(\frac{3.5R}{7R})^3$

                              =$(\frac{1}{2})^3$

                             = $\frac{1}{8}$

∴.$T_0{\degree}^{2} = \frac{T^2}{8}$

            =$\frac{24\times 24}{8}=72$

∴ $6\sqrt{2} = T_0\degree$

Answer 2

Answer:

it has full solution

hope this helps you