A satellite of 100 kg is on equatorial circular orbit around the Earth, spinning in the same Earth direction, at a height of 1000 km over the blue planet's surface.

a) How much time takes the satellite to go through the same point of the Earth’s vertical (both have rotation movement).

b) What’s the total energy of the satellite on the orbit?

Use: G= $6.67\times 10^{-11} \frac{Nm^2}{kg^2}$ ; Earth’s radius: $6370 km$; Earth’s mass= $5.98\times 10^{24} kg$

What I thought:

a) I interpreted this question is asking for the time both satellite and Earth take to align. I came up with the idea I had to equal the two angular displacement $\theta_{1}$ and $\theta_{2}$ (satellite's one and Earth's one respectively). We have $\theta_{1} = \omega_{s}T= \theta_{2} = 2\pi + \omega_{e}T$ I also supposed Earth has always an initial angular displacement of $2\pi$. Then I just had to isolate T factor. However I did not get the given outcome $6.79\times 10^{3}s$

The velocity of satellite:

$v= \sqrt{\frac{GM}{R_{e}+h}} = 7.36\times 10^{3}m/s$

$\omega_{s} = \frac{v}{R_{e}+h} = 9.98\times 10^{-4} rad/s$

$\omega_{e} = \frac{2\pi}{T} = 7.27\times 10^{-5} rad/s$

$T=\frac{2\pi}{\omega_s-\omega_e} = 6.79\times 10^{3} s$

b) Thanks to the provided link I got the following=

$E = K + U = -\frac{GMm}{2(R_{e}+h)} = -2.71\times 10^{9} J$

I do think the provided answer $-2.71\times 10^{13}$ is wrong . It is just a factor issue.

b) You need to include gravitational potential energy.