Given:
The mass of the mars, M=6.42×10²³ kg
The orbital period of the satellite, T=88640 s
To find:
The radius of the orbit of the satellite.
Explanation:
The orbital period is given by the equation,
[tex]T=2\pi\sqrt[]{\frac{r^3}{GM}}[/tex]
Where G is the gravitational constant.
On rearranging the above equation,
[tex]\begin{gathered} \frac{T^2}{4\pi^2}=\frac{r^3}{GM} \\ \Rightarrow r=\sqrt[3]{\frac{T^2GM}{4\pi^2}} \end{gathered}[/tex]
On substituting the known values,
[tex]\begin{gathered} r=\sqrt[3]{\frac{88640^2\times6.67\times10^{-11}\times6.42\times10^{23}}{4\pi^2}} \\ =20.43\times10^6\text{ m} \end{gathered}[/tex]
Final answer:
The radius of the orbit of the satellite is 20.43×10⁶ m
