A geostationary satellite is orbiting around an arbitary planet
Question:

A geostationary satellite is orbiting around an arbitary planet ‘ $\mathrm{P}$ ‘ at a height of $11 \mathrm{R}$ above the surface of ‘ $P$ ‘, $R$ being the radius of ‘ $P$ ‘. The time period of another satellite in hours at a height

of $2 \mathrm{R}$ from the surface of ‘ $\mathrm{P}$ ‘ is has the time period of 24 hours.

  1. (1) $6 \sqrt{2}$

  2. (2) $\frac{6}{\sqrt{2}}$

  3. (3) 3

  4. (4) 5


Correct Option:

Solution:

(3)

$\mathrm{T} \propto \mathrm{R}^{3 / 2}$

$\frac{24}{\mathrm{~T}}=\left(\frac{12 \mathrm{R}}{3 \mathrm{R}}\right)^{3 / 2} \Rightarrow \mathrm{T}=3 \mathrm{hr}$

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