Answer :
To solve the problem we need to know the Satellite Period, given by,
[tex]T= 2\pi sqrt{\frac{r^3}{GM_E}}[/tex]
It is also necessary to identify the satellite speed, which is given,
[tex]v=\sqrt{\frac{GM_E}{r}}[/tex]
Finally, it is required to know the satellite acceleration, given by,
[tex]a=\frac{v^2}{r}[/tex]
By definition, the radius of the earth is 6.371 * 10 ^ 6m measured from the center of the earth and that the mass is [tex]M_E=5.972*10^{24}Kg[/tex] and that the gravitational constant is [tex]6.67*10^{-11}kg^{-1}s{-2}[/tex]
On the other hand, the distance from the earth's surface to the satellite point of [tex]3.76 * 10 ^ 6m.[/tex]
Then the net distance, from the center of the earth to the point of the Satellite would be given by,
[tex]r=6.371*10^6m+3.76*10^6m\\r=10.131*10^6m[/tex]
With the previous values obtained it is possible to obtain the period, as well,
[tex]T= 2\pi sqrt{\frac{(10.131*10^6)^3}{(6.67*10^{-11})(5.972*10^{24})}}[/tex]
[tex]T= 10150.12s = 2h45m[/tex]
In turn with the previously found data and using the velocity formula we can calculate it,
[tex]v=\sqrt{\frac{GM_E}{r}}[/tex]
[tex]v=\sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{10.131*10^6}}[/tex]
[tex]v=6.27*10^3m/s[/tex]
Once the speed is obtained we can proceed to calculate the acceleration, for this we apply in the acceleration equation,
[tex]a= \frac{(6.27*10^3)^2}{10.131*10^6}[/tex]
[tex]a=3.88m/s^2[/tex]
So we have,
Part a= 2h45m
Par b = [tex]6.27*10^3m/s[/tex]
Part c= [tex]3.88m/s^2[/tex]