JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Gravitation

A geostationary satellite is orbiting around an arbitrary planet '

P

' at a height of

11 R

above the surface of '

P

R

being the radius of '

P

'. The time period of another satellite in hours at a height of 2R from the surface of '

P

' is ____ '

P

' has the time period of 24 hours.

(A) $62$
(B)3✓ Correct
(C)5
(D) $\frac{6}{2}$

Answer: (B)

Solution

For circular orbits

T^{2} \propto r^{3}

, where

r

is measured from the centre of the planet. The geostationary satellite sits at

r_{1} = R + 11 R = 12 R

with

T_{1} = 24

h, and the second satellite at

r_{2} = R + 2 R = 3 R

. Hence

\frac{T _{2}}{T _{1}} = (\frac{r _{2}}{r _{1}})^{3/2} = (\frac{3 R}{12 R})^{3/2} = (\frac{1}{4})^{3/2} = \frac{1}{8}

T_{2} = \frac{24}{8} = 3

hours.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option

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