Question

A satellite is in a circular orbit around an unknown planet. The satellite has a speed of $1.70\times {10}^4{\text{ms}}^{-1}$, and radius of the orbit is $5.25\times {10}^6\text{m}$. A second satellite also has a circular orbit around the same planet. The orbital radius of second satellite is $8.6\times {10}^6\text{m}$. Determine the orbital speed of the second satellite.

Answer 1

Orbital speed equation:

Consider a satellite with mass ‘m’ orbiting around a central body of mass M.

Let R be the radius of orbit for a satellite, then the velocity of a satellite moving around a central body, $v=\sqrt{\frac{GM}{R}}$, and G is the gravitational constant.=$6.67\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$.

So, $v\alpha \frac{1}{\sqrt{R}}\to \left(1\right)$

Given data and calculation:

For the first satellite, orbital speed is $v_1=1.70\times {10}^4{\text{ms}}^{-1}$ and the radius of the orbit is $R_1=5.25\times {10}^6\text{m}$

For the second satellite, the orbital radius is $R_2=8.6\times {10}^6\text{m}$

Let $v_2$ be the orbital speed of the second satellite.

Using (1) we can compare the orbital velocities of satellites as

$$\begin{gathered} \frac{v_1}{v_2}=\sqrt{\frac{R_2}{R_1}} \\ {v}_2=v_1\times \sqrt{\frac{R_2}{R_1}} \end{gathered}$$

Substituting the given values,

$\begin{array}{rcl}{v}_2& =& 1.7\times {10}^4\times \sqrt{\frac{5.25\times {10}^6}{8.6\times {10}^6}}\\ & =& 1.33\times {10}^4{\text{ms}}^{-1}\end{array}$

Thus, the orbital speed of the second satellite is $1.33\times {10}^4{\text{ms}}^{-1}$