If the height of a satellite completing one revolution around the earth in T seconds is h1 meter, then what would be the height of a satellite taking \(2\sqrt{2}T\) seconds for one revolution?

Time period of the satellite is given as

\(T = \frac{2π(R+h)^{3/2}}{\sqrt{GM}}\)

When the height of the satellite is *h*_{1}, it takes *T* time to revolve around the Earth. Thus, at height *h*_{1}

∴ \(T = \frac{2π(R+h_1)^{3/2}}{\sqrt{GM}}\) .... (i)

When the satellite takes ** \(2\sqrt{2}T\)** time to revolve around the Earth, let it be at height *h*_{2}. Thus,

\(2\sqrt{2}T =\frac{2π(R+h_2)^{3/2}}{\sqrt{GM}}\) .....(ii)

Dividing (ii) by (i), we get

\(2\sqrt{2} =\frac{(R+h_2)^{3/2}}{(R+h_1)^{3/2}}\)

\(2^{3/2} =\frac{(R+h_2)^{3/2}}{(R+h_1)^{3/2}}\)

\(2 =\frac{(R+h_2)}{(R+h_1)}\)

2R+2h_{1} = R+h_{2}

∴ h_{2}=R+2h_{1}