Question

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?

Answer 1

Time period of the satellite is given as
\(T = \frac{2π(R+h)^{3/2}}{\sqrt{GM}}\)

When the height of the satellite is h₁, it takes T time to revolve around the Earth. Thus, at height h₁
∴ \(T = \frac{2π(R+h_1)^{3/2}}{\sqrt{GM}}\) .... (i)

When the satellite takes \(2\sqrt{2}T\) time to revolve around the Earth, let it be at height h₂. 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+2h₁ = R+h₂

∴ h₂=R+2h₁