# Can a satellite stay in one place but not above equator? [closed]

I know that there are geostationary satellites that are always seen in one place in the sky and that they all are located in the equatorial area. I wonder if it's possible for a satellite to "hang" still, for example, above Moscow (55.7558° N, 37.6173° E)?

• I’m voting to close this question because it's more suitable for the space exploration stackexchange. Sep 11 at 15:24
• Thanks, I didn't know there is one Sep 11 at 15:49

In order to keep at least one satellite in the sky over Russia (or any other high N or S latitude area) at all times, the Molniya orbit was invented.

It is not stationary or even synchronous, but with a period of exactly 1/2 sidereal day it is repeat-ground track. It will appear to linger for about 8 of it's ~12 hour orbit well above the horizon for one region centered around 63.4° degrees north or south.

Twelve hours later it will do the same for an area 180 degrees offset in longitude, so you need three of them to make a Molniya constellation.

Because they are not synchronous, they slowly but continuously move across the sky, so a dish antenna communicating with the constellation would have to be equipped with motors slowly but constantly moving to track one of them at a time, then jump back to the next about every 8 hours (one third of a sidereal day).

The inclination of 63.4° comes from $$4 - 5 \sin^2 \theta = 0$$, a term which is in the numerator of the expression for the precession of the orbit's argument of perigee due to Earth's oblateness as expressed by $$J_2$$. At other inclinations the perigee would slowly precess such that the repeat ground track would drift east or west on the timescale of months.

No.

A geostationary satellite is in an orbit around the earth with a 24 hour period - in the same sense of rotation as the Earth rotates. That makes sure that its orientation with respect to the surface of Earth does not change.

If you put a satellite into another orbit which is not around the equator, it changes the latitude, thus cannot ever be geo-stationary.

A satellite moving around another, but constant latitude than the equator, is not in orbit as the center of its path is not the center of the Earth, and thus would need constant propulsion.

• A (so far theoretical) device called a statite is able to hover synchronously over any point, even a pole, but it is not in orbit. It uses a solar sail to supply thrust at all times. Sep 12 at 3:28

Depending on your definition of satellite; maybe yes.

Only over the equator can you have a satellite in a Keplerian geostationary orbit. This idea is covered in depth in other answers. I would note that even these satellites are not totally without means of propulsion. They need to be able to generate thrust occasionally to make corrections to their orbit.

However, if we allow for satellites that are in non-Keplerian orbits, there are pole-sitters. A pole sitter remains stationary above the pole of a planet. This is achieved by a combination of thrust, and often being much higher up than ordinary satellites. Being higher up means having less gravity to counter, although eventually there are other issues with not drifting out of place. These is also often a compromise between height, and being able to make good observations - pole-sitters are normally used to observe the polar caps.

There are some cool suggestions to use solar sails to keep pole-sitters in place; https://arc.aiaa.org/doi/abs/10.2514/1.G003952

A satellite has a centrifugal force* radially away from the center of its curvature. If the orbit is in a plane that does not go through the center of mass of the Earth, then the centrifugal force will be at an angle to the ray going from the satellite and the center of mass of the Earth. Since gravity acts only through this ray, gravity can't completely cancel out the centrifugal force, and there will be a force perpendicular to this ray. So there's no way to orbit a non-zero latitude without a force pushing towards another latitude.

*Yes, the centrifugal force is a pseudoforce, but that doesn't affect this argument.