Answer:
[tex]t=5063s[/tex]
Explanation:
The gravitational force a satellite of mass m orbiting the Earth of mass [tex]M=5.97\times10^{24}Kg[/tex] at a radius equal to the radius of the Earth [tex]R=6371000m[/tex] is:
[tex]F=\frac{GMm}{R^2}[/tex]
Where [tex]G=6.67\times10^{-11}Nm^2/Kg^2[/tex] is the gravitational constant.
Newton's 2nd Law tells us that this satellite of mass m under the mentioned force F will experiment an acceleration given by F=ma.
This acceleration is a centripetal acceleration, so we can use the formula:
[tex]a=\frac{v^2}{R}[/tex]
The velocity will be given by the relation between the circumference of the orbit and the time taken, that is, v=C/t, and we know that the circumference is [tex]C=2\pi R[/tex].
We can put everything together now:
[tex]F=ma=\frac{GMm}{R^2}[/tex]
[tex]a=\frac{GM}{R^2}[/tex]
[tex]\frac{v^2}{R}=\frac{GM}{R^2}[/tex]
[tex]v^2=\frac{GM}{R}[/tex]
[tex](\frac{2 \pi R}{t})^2=\frac{GM}{R}[/tex]
[tex]\frac{4 \pi^2 R^3}{t^2}=GM[/tex]
[tex]t^2=\frac{4 \pi^2}{GM}R^3[/tex]
[tex]t=\sqrt{\frac{4 \pi^2}{GM}R^3}[/tex]
Substituting our values we have:
[tex]t=\sqrt{\frac{4 \pi^2}{(6.67\times10^{-11}Nm^2/Kg^2)(5.97\times10^{24}Kg)}(6371000m)^3}=5063s[/tex]
Which are more than 84 minutes. For reference, the ISS has a period of almost 93 minutes, so we know the result makes sense (and the equation will give us this result considering it orbits 408Km above Earth's surface).
