The time period of a geostationary satellite is 24 h, at a height 6RE (RE is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 RE from surface will be,

CONCEPT:

According to Kepler's third law we have;

T² ∝ R³

Here we have T as the time period, and R is the radius.

CALCULATION:

The time period of the state is

T2 ∝ R3

For the first geostationary satellite we have, r₁ = R + 6R = 7R

The time period,T1 = 24 h

\(T_1 = r_1^{\frac{3}{2}}\)    ----(1)

For the second geostationary satellite we have, r2 = R + 2.5R = 3.5R

\(T_2 = r_2^{\frac{3}{2}}\)    -----(2)

Now, on dividing the equation (1) by (2) we have;

\(\frac{T_1}{T_2} = \frac{ r_1^{\frac{3}{2}} }{r_2^{\frac{3}{2}}}\)

\(\Rightarrow \frac{T_1}{T_2} = \sqrt \frac{ r_1^{3} }{r_2^{3}}\)

Further solving we have;

\(\Rightarrow \frac{T_2}{24} = \sqrt \frac{ (3.5 R)^{3} }{7^{3}}\)

\(\Rightarrow T_2 = 6 \sqrt2\) h

Hence, option 3) is the correct answer.