Question

A satellite orbits the sun with a period of $1day.$What is the radius of its orbit?

Short answer

The radius of the orbit of the satellite is$2.9\times {10}^{9}m.$

Step-by-step solution

Given information

Time period of the satellite =$1day=8.64\times {10}^{4}s.$

Calculation

According to the Kepler's third law, the time period is given by : $T^2=\frac{4{\pi }^2{R}^3}{GM}.$

Where, T is time period, R is the required radius of the orbit, G is the universal gravitational constant and M is mass.

We know, mass of the sun = $1.989\times {10}^{30}kg.$

Solving we get,

$$\begin{gathered} T^2=\frac{4{\pi }^2{R}^3}{GM} \\ {R}^3=T^2\times \frac{GM}{4{\pi }^2} \\ R={\left(T^2\times \frac{GM}{4{\pi }^2}\right)}^{1/3} \\ R={\left({\left(8.64\times {10}^{4}s\right)}^2\times \frac{(6.67\times {10}^{-11}N{m}^2/k{g}^2)(1.989\times {10}^{30}kg)}{4{\pi }^2}\right)}^{1/3} \\ R=2.9\times {10}^9\mathrm{m} \end{gathered}$$

Final answer

The radius of the orbit of the satellite is$2.9\times {10}^9\mathrm{m}.$