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.


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

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    $\begingroup$ 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. $\endgroup$
    – AXensen
    Commented Jun 26, 2023 at 15:37
  • $\begingroup$ "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. $\endgroup$
    – PM 2Ring
    Commented Jun 26, 2023 at 16:37
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    $\begingroup$ @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. $\endgroup$ Commented Jun 26, 2023 at 18:32

1 Answer 1


Orbit is not like a circular racetrack where you can choose however fast you'd like to move along a fixed path - in orbit, your speed and trajectory are inextricably linked. Your error is in assuming that you can define an orbital path at whatever height you want, and then choose the speed/acceleration required to complete it in 24 hours.

That's not how it works - orbit is freefall, meaning the only acceleration comes from gravity alone. If you are too far away, gravity is too weak and your orbit will take more than 24 hours, and if you are too close, gravity is too strong and your orbit will take less than 24 hours. For a circular orbit, the centripetal acceleration, and therefore period, is determined solely by the height. Geosynchronous orbit can only occur at one particular height.

At an intuitive level, you're suggesting that something in low earth orbit like the ISS (orbital altitude 400km, period 90 minutes) could be in a synchronous orbit instead just by slowing down so that the same trajectory takes 24 hours instead of 90 minutes. But that's not possible, the ISS would just crash to the ground if it reduced its speed without changing its altitude. Similarly, you couldn't put the moon into synchronous obit just by speeding it up, as it would need to go so fast it would escape the planet entirely. Orbital velocities and trajectories are not independent.


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