A Problem with a Geostationary Orbit

I’ve been sorting through some old problem sets for my course on Astrophysics and Cosmology, and thought I would post this one in the Cute Problems folder for your amusement. The first part is easy, the second part not so much…

  1. Verify that the radius of a circular geostationary orbit around the Earth is about 42,000 km, i.e. find the radius of a circular  orbit around the Earth which has a period of 24 hours so it is always above the same point on the Earth’s surface . (You will need to look up the mass of the Earth.)
  2. Use the answer to (1)  to estimate what fraction of the Earth’s surface is visible at any  time from a satellite in such an orbit. (You will need to look up the radius of the Earth.)

Answers to (2) through the comments box please – and don’t forget to explain your working!

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10 Responses to “A Problem with a Geostationary Orbit”

  1. Let r be the radius of the Earth and a be the orbital radius. Draw a tangent line to the Earth’s surface passing through the satellite. Draw a triangle whose vertices are the tangent point, the Earth’s center, and the satellite. It’s a right triangle with right angle at the tangent point. Let theta be the angle at the Earth’s center. Then cos theta = r/a. The portion of Earth visible from the satellite is a spherical cap consisting of all points making an angle less than theta with the center-satellite line. Such a cap has area 2 pi r^2 (1-cos theta), so the fraction of Earth that’s visible is (1-cos theta) / 2, or (1-r/a) / 2. I make that about 42%.

    (As I always tell my students, it’s probably worthwhile to check by considering extreme cases. As a tends to r, the answer tends to zero, and as a tends to infinity, it tends to 1/2. Both sound right.)

    • You can also ask what fraction is never visible throughout the satellite’s orbit. I guess that’s two spherical caps with angle (pi/2 – theta), so the answer is 1 – sqrt(1-(r/a)^2), or about 1%.

      Santa Claus can’t watch satellite TV.

    • It was inevitable that the answer would be 42 (%).

  2. … assuming that the Earth is spherical. With the oblate spheroid model I suppose the fraction is probably a bit smaller(?), and real topography is just too difficult for physicists.
    And adding to the quibbling traditional here, the fraction of the _surface_ that is actually _visible_ is much smaller, because (a) cloud makes much of the surface not observable (at visible and NIR wavelengths, at least) and (b) it is night for nearly half the surface. It therefore all depends on what is meant by “visible” (does illumination by the Moon count,…?)

    • I confess that the sign of the oblateness effect isn’t immediately obvious to me, but I think you must be right that it makes the fraction smaller. In the extreme case of maximal oblateness, the Earth becomes a disk, and the satellite orbits in the same plane as the disk, so the fraction tends to zero.

  3. Sorry to keep following up my own post. Just realized that’s a stupid thing to say for a geostationary orbit. I guess I meant “what fraction isn’t visible from any satellite in any geostationary orbit.”

  4. And there was I thinking this was going to be a question about refraction of light… e..g we can see the Sun well after it’s actually set, so presumably the reverse is true, and a satellite in geostationary orbit is ‘visible’ even when below the local horizon…

  5. Jonathan Thornburg Says:

    Isn’t the rotation period that matters here the Earth’s spin rate with respect to an inertial coordinate system? Apart from very small precession-of-equinoxes corrections, that’s the siderial day (about 23 hours 56 minutes) not the solar day (24 hours).

    • telescoper Says:

      Yes, it is the sidereal day that should be used but to the accuracy required this is equivalent to the solar say.

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