A satellite orbiting the earth in a circular orbit of radius R completes one revolution in 3h. If orbital radius of geostationary satellite is 36000 km, orbital radius of earth is

6000 km

9000 km

12000 km

15000 km

Solution

9000 km

Time period of satellite

T² = $\frac{4 π^{2}}{GM} r^{3}$

T² = $\frac{4 π^{2}}{g R^{2}} r^{3}$ [ since GM = gR² ]

⇒ T = $\frac{2 π}{R} \sqrt{\frac{r^{3}}{g}}$

⇒ T ∝ r^3/2

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

Given T₁ = 3h , T₂ = 24 h

(geostationary satellite) r₁ = R, r₂ = 36000 km

⇒ $\frac{3}{24} = {(\frac{R}{36000})}^{\frac{3}{2}}$

⇒ R = ${(\frac{1}{8})}^{\frac{2}{3}} \times 36000$

⇒ R = $\frac{1}{4} \times 36000$

⇒ R = 9000 km

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