Answer:
The period of the orbit is [tex]9.88\times 10^3\ s[/tex].
Explanation:
Given that,
A satellite is in a circular orbit around the Earth at an altitude of, [tex]h=3.58\times 10^6\ m[/tex]
We need to find the period of the orbit. It can be calculated using Kepler's law. It is given by :
[tex]T^2\propto a^3\\\\T^2=\dfrac{4\pi^2}{Gm}a^3[/tex]
a is distance of semi major axis, a = h+ r, r is radius of earth
m is mass of earth
So,
[tex]T^2=\dfrac{4\pi^2}{Gm}(r+h)^3\\\\T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 5.98\times 10^{24}}(6.38\times 10^6+3.58\times 10^6)^3\\\\T=9.88\times 10^3\ s[/tex]
So, the period of the orbit is [tex]9.88\times 10^3\ s[/tex].
