A geostationary satellite is taken to a new orbit such that its distance from centre of the earth is doubled. Then find the time period of this satellite in the new orbit.

Question

Aicharya · Accepted Answer

The orbital time period for a satellite is $T = 2 π \sqrt{\frac{r^{3}}{G M_{E}}}$ here, $r$ is the orbital radius, therefore, $T \propto r^{\frac{3}{2}}$

As the satellite is geostationary, therefore, the time period of the satellite is $24 hrs$ or $T_{1} = 24 hrs$

Now, if $T_{2}$ is the time period of the satellite in new orbit $r_{2}$ the ratio of the time period for the satellite in both cases is,

$\Rightarrow \frac{T_{1}}{T_{2}} = {(\frac{r_{1}}{r_{2}})}^{\frac{3}{2}} \Rightarrow T_{2} = T_{1} {(\frac{2 r_{1}}{r_{1}})}^{\frac{3}{2}} \dots \dots (given, r_{2} = 2 r_{1}) \Rightarrow T_{2} = 24 \times {(2)}^{\frac{3}{2}} = 48 \sqrt{2} hrs$

A geostationary satellite is taken to a new orbit such that its distance from centre of the earth is doubled…

Solution