Question

An earth satellite moves in a circular orbit at a speed of $5500m/s$. What is its orbital period?

Short answer

The orbital period is$4.18hours.$

Step-by-step solution

Given information

Speed of orbiting =$4.18hours.$

Calculation

The centripetal acceleration required for circular motion of satellite is provided by the gravitational force between the satellite and the earth, therefore : $G{M}_e{m}_s/r_s^2=m_s{v}_s^2/r_s$

Here, ${\mathrm{M}}_{\mathrm{e}}$is the mass of earth, ${\mathrm{m}}_{\mathrm{s}}$is the mass of satellite, G is the gravitational constant, r is the radius of the orbit.

Now solving we get,

$$\begin{gathered} G{M}_e{m}_s/r_s^2=m_s{v}_s^2/r_s \\ {r}_s=G{M}_e/v_s^2 \\ {r}_s=\frac{(6.67\times {10}^{-11}N{m}^2/k{g}^2)(5.98\times {10}^{24}kg)}{{(5500m/s)}^2} \\ r=13.18\times {10}^6\mathrm{m} \end{gathered}$$

Orbital period T is given by according to Kepler's Law :

$$\begin{gathered} T=2\pi \sqrt{\frac{r_s^3}{G{M}_e}} \\ T=2\pi \sqrt{\frac{{\left(13.18\times {10}^{6}m\right)}^3}{(6.67\times {10}^{-11}N{m}^2/k{g}^2)(5.98\times {10}^{24}kg)}} \\ T=2\pi \sqrt{5740096.7s^2} \\ T=2\pi \times 2395.84s \\ T=15053\sec \\ T=4.18\text{hours} \end{gathered}$$

Therefore, time period = $4.18hours.$

Final answer

The orbital period is$4.18hours.$