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 2.5R from the surface of earth?​

Answer 1

Answer:

please see the attachment  

and for geostationary satellite T = 24 hr

so T2 = 24/2√2 = 6√2 hr

Explanation:

thank you

Answer 2

Answer:-

According to Kepler's third law.

$\boxed{\sf T^2\propto R^3}$

$\\ \rm\longmapsto \dfrac{T_1^2}{T_2^2}=\dfrac{R_1^3}{R_2^3}$

$\\ \rm\longmapsto \left(\dfrac{T_2}{T_1}\right)^2=\left(\dfrac{R_2}{R_1}\right)^3$

$\\ \rm\longmapsto T_2^2=T_1^2\left[\dfrac{3.5R}{7R}\right]^3$

$\\ \rm\longmapsto T_2^2=\dfrac{T_1^2}{8}$

$\\ \rm\longmapsto T_2=\dfrac{T_1}{2\sqrt{2}}$

$\\ \rm\longmapsto T_2=\dfrac{24}{2\sqrt{2}}$

$\\ \rm\longmapsto T_1=6\sqrt{2}hours$

$\\ \underline{\boxed{\bf{\therefore T_2=6\sqrt{2}hours}}}$