# Satellite on circular orbit around Earth

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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.

edited Jul 29, 2018
a) Your method should give you the correct solution. I don't know what you mean by "Earth has always an initial angular displacement of$2\pi$." What is the rest of your calculation?

b) You need to include gravitational potential energy.
a) With the statement "Earth has always an initial angular displacement of 2π" I meant that at the moment both align I considered Earth had already made a turn, having an initial angular displacement of $2\pi$. I did this because I saw it in an exercise but not sure neither it is suitable to this one nor the reasoning is correct. Then, $T=\frac{2\pi}{\omega_{s}-\omega_{e}}$ but I got 6.29s so I must have left something.

b) How would it be? I thought about $mgh$ and how it would apply to this case. I used $\frac{m_{s}Gm_{s}m_{e}}{r_{e}+h}$ but did not get the provided solution.
a) Sorry I do not understand your explanation about "Earth has always an initial angular displacement of $2\pi$." It is inadvisable to copy methods you have seen in books unless you understand them. But it does not make any difference to the solution. The point is that while the Earth turns through angle $\theta$ the satellite turns through angle $\theta+2\pi$ - a whole extra revolution - to get to the same point. $$\omega_e T=2\pi+\omega_s T$$ $$\frac{1}{T_e}=\frac{1}{T}+\frac{1}{T_s}$$ Calculate the period of the satellite's orbit $T_s$, you know $T_e$.

b) See https://deutsch.physics.ucsc.edu/6A/book/gravity/node15.html.
a) Why do you have to add a whole extra revolution to get to the same point? Is it because satellite is also rotating?
Try drawing a diagram of a fixed point P on the Earth and the satellite S both rotating about the centre of the Earth C. Points CPS are aligned at $t=0$. How far has each moved when they are aligned again? S starts ahead of P and must make an extra lap to catch up with P again. This is similar to asking when two hands of a clock will be aligned again.

If the satellite S did not rotate (ie it remains in a fixed position in space, like the minute hand remaining fixed while the second hand moves) then the Earth point P would still have to make 1 complete revolution to be aligned with the satellite again.
Now I see it with clarity! Thanks. I will update my answers.
It is updated. I have difficulty getting a) result.
BTW your attempt says "Supposing the satellite is **geostationary**..." but this means that it will always be above the same position on the Earth, then the question in part a) is meaningless.
My bad, I supposed it wrongly. The exercise does not state it.
a) Your calculation of $\omega_s$ is way off. A value of $1 rad/s$ is enormous. Compare with $\omega_s$. There are $2\pi$ rad in a full circle, so this means the satellite takes 6.28s to make one revolution. The Earth makes 1 rev per day, the ISS makes 1 rev per 93 mins. This satellite takes only 6.28s!?

Another quick way to get $T_s$ is from Kepler's 3rd Law ($T^2 \propto R^3$), if you know the period and radius of another Earth satellite, eg the Moon or the ISS.

b) Yes I get -2.71x10^9 J also.
a) Yes the answer does not make sense, that is why I knew $\omega_{s}$ was miscalculated. But why? I mean, it should be satellite's velocity divided by $7370\times 10^{3}$ but that division outputs 1rad/s if I am not mistaken
Your value of $v$ is also incorrect. It is a factor of 1000 too large. Perhaps you got the right numerical value for $v$ as 7356.64 but interpreted this as km/s instead of m/s?
It is fixed. Thanks