Question

A satellite circles the earth in an orbit whose radius is twice the earth’s radius. The mass of earth is $6\times {10}^{24}kg$ and its radius is $6.4\times {10}^{6}m$. The time period of the satellite is approximately

• A. $14346s$
• B. $2yr$
• C. $84h$
• D. $1h$

Answer 1

The correct option is A

$14346s$

Step 1: Given data
$$\begin{gathered} \mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{earth}\left({\mathrm{M}}_{\mathrm{E}}\right)=6\times {10}^{24}\mathrm{kg}. \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{earth}\mathrm{is}\left({\mathrm{r}}_{\mathrm{E}}\right)=6.4\times {10}^6\mathrm{m}. \end{gathered}$$

Formula used:
Time period of the satellite around the planet is,

$$\begin{gathered} T=\frac{2\pi r}{\nu } \\ \mathrm{Velocity}\mathrm{of}\mathrm{a}\mathrm{satellite}\mathrm{is}, \\ v=\sqrt{\frac{G{M}_E}{r}} \end{gathered}$$

Step 2 : Calculating time period
A satellite circles the earth in an orbit whose radius is twice the earth’s radius. So, $(r=2r_E)$.
Substitute velocity of the satellite in Time period ‘T’,
$$\begin{gathered} T=\frac{2\pi r}{\sqrt{{\displaystyle \frac{G{M}_E}{r}}}} \\ \left(r=2r_E\right) \\ T=\frac{2\pi .2r_E}{\sqrt{{\displaystyle \frac{G{M}_E}{2r_E}}}} \\ T=\frac{2\pi .2\left(6.4\times {10}^6\right)}{\sqrt{{\displaystyle \frac{(6.674\times {10}^{-11})\left(6\times {10}^{24}\right)}{\left(2\times 6.4\times {10}^6\right)}}}} \\ T=\frac{(80424771.932)}{\sqrt{31284375}} \\ T=\frac{80424771.932}{5593.2437} \\ T=14,378s. \end{gathered}$$

Hence, the answer is option (a)