# Number of Geostationary Orbits

It is stated that there is only one geostationary orbit whose height can be calculated using:-

$$H = [\frac{GM_ET^2}{4π^2}]^{\frac{1}{3}} - R$$

But there can be more than one geostationary orbits if I use a different trick.

$$ω = \frac{2π}{T}$$

$$ω = \frac{v_c}{R+H}$$

$$\therefore H = \frac{v_cT}{2π} - R$$

$$H$$ represents height of geostationary orbit;

$$G$$ represents Gravitational constant;

$$M_E$$ represents Mass of Earth;

$$T$$ represents time period of rotation of Earth;

$$v_c$$ represents centripetal velocity;

$$R$$ represents Radius of Earth;

$$\omega$$ represents angular frequency;

My first equation states that there can be only one geostationary orbit as everything on Right Hand Side is constant.

My second equation states that there can be more than one geostationary orbit depending on the centripetal velocity.

Question:-

Can there be more than one geostationary orbits depending on the centripetal velocity?

• I'm 70% sure the answer is that these are equivalent equations, expressing the same height in terms of different variables. But more detail would be really helpful. Considering that these equations are not in terms of the same quantities, can you more clearly explain why you think these equations imply that there is more than one geostationary orbit? Also, saying "the symbols have their usual meanings" does not suffice. I'm fairly sure I figured out the meanings of each symbol, but it would have been a lot easier to read if you just listed all the definitions. Commented Jun 26, 2023 at 15:37
• "there can be more than one geostationary orbit depending on the centripetal velocity." How? The velocity of a circular orbit at a given height is a function of the gravitational acceleration at that height. Commented Jun 26, 2023 at 16:37
• @AXensen The second equation is just a definition of angular velocity, and simply defines how fast you need to go to complete a circular path of a particular size in a certain amount of time. It has no particular relationship to the concept of a freefall orbit, however, as it entirely ignores the notion of gravity. It's not the same as the first equation - the second one can be solved for arbitrary H, but that doesn't mean it can actually be achieved under the force of gravity alone. Commented Jun 26, 2023 at 18:32