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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.

asked Jul 26 in Physics Problems by JD_PM (490 points)
edited Jul 29 by JD_PM
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

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