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A few days ago I obervered a satellite with my telescope. But when I later looked at Stellarium (after adding all available satellite databases), there was no satellite passing the spot that I observed at the given time (one was close but way too fast and should not have been visible). Since I knew how big my FOV was and had roughly counted how many seconds it took to cross it, I thought it should be possible to determine the height of the satellite's orbit (assuming it's circular). I had a few different approaches, one was to set the formula for orbital velocity $v = \sqrt{\frac{GM}{R+h}}$ equal to the formula for apparent size (per second to get velocity) $v = 2h \cdot tan(a/2)$ [M=mass of earth, R=radius of earth, h=orbital height, a=angle the satellite moves in one second measured in degrees] Using WolframAlpha, I got a very long formula as a result with cot and stuff but in the end it turned out it wasn't right, as I checked with other satellites, it gave wrong heights, and for LEO objects it gave an error since the result would be imaginary. So I tried a different approach: $$\sqrt{\frac{GM}{R}} = \frac{(2\pi R)}{(2\pi/\omega)}$$ [R=total radius of satellite orbit, w=angular velocity in rad/s] (Basically orbital velocity using gravity = orbital length depending on radius / time it takes for a $2\pi$ (i.e. one complete) orbit)

In the end you should get something like $$R = \sqrt[3]{GM/\omega^2}$$ And to get the height just subtract the earth's radius It seems to work out better, as it can return solutions in a LEO, but it still gives the wrong solutions.

Is there a well-known equation for determining orbital radius from angular velocity? Did I just make some calculation error? Please help!

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  • $\begingroup$ I think at least one of the equations is wrong because it would assume the observer as the center of rotation. $\endgroup$
    – 299792458
    Commented Jan 4, 2022 at 14:49

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The transit time across your field of view would depend on your position on the surface of the earth as well as the speed and altitude of the satellite. That gets really complicated. I'm thinking that to estimate the altitude, you should observe it passing overhead several times and use your best estimate of the period of revolution to calculate the altitude. Don't forget to allow for your rotation with the earth.

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  • $\begingroup$ Ok, thanks. Makes sense! $\endgroup$
    – 299792458
    Commented Jan 4, 2022 at 18:23
  • $\begingroup$ Only problem is I'll never see that satellite again, and even if do, I won't be able to identify it since it wasn't in any to me available database (which means it's probably a secret spy-sat XD) $\endgroup$
    – 299792458
    Commented Jan 4, 2022 at 18:36
  • $\begingroup$ If the satellite passes nearly overhead, then it defines an angle within your field of view: θ = vt/h. But for a circular orbit: $v^2$ = GM/R = GM/($R_o$ + h) where $R_o$ is the radius of the earth and h is the altitude of the satellite. Combine to eliminate v and solve for h. (This assumes you know the angle for the field of view in radians.) $\endgroup$
    – R.W. Bird
    Commented Jan 5, 2022 at 15:20

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