<p>The speed of a satellite in orbit is directly linked to the <a href="http://en.wikipedia.org/wiki/Orbital_speed" rel="nofollow">radius of that orbit</a>:<br>
$V=\frac{2*\pi*a}{T}$ </p>

<p>V is the orbital speed, a is the radius and T is the orbital period. This means that for any radius, there is only one possible orbital speed and vice versa. So there is only one radius where the period is 24 hours.  </p>

<p>If you really want a satellite to orbit at 70,000 km and still be geostationary, you'll have to use thrusters constantly to provide an extra force towards the Earth. This uses far too much fuel to be economical. </p>

CC BY-SA 3.0
2014-02-17T10:20:51.427Z
2014-02-17T10:20:51.427Z
Hobbes
<p>You have yourself provided us with the answer. </p>

<p>Any satellite revolving over the <strong>Earth's Equator</strong> would be considered as a GeoStationary object. Provided, it should move with the rotation of the Earth and it should be <strong>35,786 kilometres</strong> above of the sea so that the observer, that is on the ground would think that the satellite is <strong>stationary</strong>.</p>

<p>If to say for simple: </p>

<p>A Geostationary object is an object, which is fully synchronized to the movement of the earth. Its position is as it is fixed to the Equator of the earth at zero degree latitude. </p>

<p>Anything that would stay synchronized with the Earth's rotation would be a Geostationary object. </p>

<p>Apart from the Wikipedia article you can read more here: <a href="http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Geostationary_orbit.html" rel="nofollow">http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Geostationary_orbit.html</a> </p>

<p>It has clear and more simple information, and has described that the object must be this_many kms above the sea level and should be at the 0 degree latitude. </p>

CC BY-SA 3.0
2015-06-25T15:53:28.380Z
If an orbit is not on the equatorial plane, it's not geostationary. It may be geosynchronous, though. Similarly, if an orbit isn't circular, the satellite's apparent position in the sky will vary as the angular speed increases, then decreases. It might be close enough for operational purposes, but it's not exactly geostationary.
Hobbes
CC BY-SA 3.0
2014-02-17T10:27:17.327Z
Not correct. Geostationary objects do not have the exact same orbit. They are not on exactly circular orbits, they are not all at the exact same altitude, and they are not all on the Equator plane. Besides, you did not answer the base question.
Envite
CC BY-SA 3.0
2014-02-15T20:01:41.920Z
2014-02-15T20:01:41.920Z
Afzaal Ahmad Zeeshan
<p>@Hobbes put the formula:</p>

<p>$V={2\pi a\over T}$</p>

<p>If you substitute on it $T$ for 1 sidereal day you'll have a one-to-one relation between $V$ and $a$. This seems to imply that you can choose any $a$ and use the corresponding $V$, but this is not true since it does not take into account which force are you using.</p>

<p>The formula above is valid of any object in a circular motion, like a stone on a sling, or a car in a loop. It is also valid for satellites around planets.</p>

<p>But there is another formula stating a relation among $a$ and $T$ when the force is that of Gravity, so you are not free to chose $a$ anymore.</p>

<p>It happens that centripetal acceleration (that of the sling, the Normal force on the loop, or gravity for the satellite) is the Universal Gravity in our case:</p>

<p>$a_c={GM\over r^2}$</p>

<p>but it being the centripetal acceleration means</p>

<p>$a_c=\omega^2 r$</p>

<p>so we have</p>

<p>$\omega^2 r={GM\over r^2}$</p>

<p>$\omega^2 r^3=GM$</p>

<p>$\omega$ is $v\over r$ so</p>

<p>$v^2 r=GM$</p>

<p>which gives you a fixed relation between $v$ and $r$ for any planet with mass $M$.</p>

CC BY-SA 3.0
2014-02-19T08:59:36.010Z
I like your answer, but decide to upvote another because the question asked for a simple explanation.
astabada
CC BY-SA 3.0
2014-02-17T10:41:53.327Z
2014-02-17T10:41:53.327Z
Envite
<p>In a simpler way, if that satellite is above 35,786 kilometres above you country place, for example, Inidia-Dehli, then satellite will remain at the same position throughout its life. That is why it is called as Geostationary satellite. It revolves with the speed of earth, but circumference of earth and circumference of geostationary satellite's orbit is different. Circumference of geostationary satellite is much greater than the earths circumference. So speed of geostationary satellite is greater than speed of earth because it has to stick to you country place above earth so it have to move fast.</p>

<p>Also in this orbit, gravitational force of earth which attracts satellite towards earth and centrifugal force which tries to escape satellite from that orbit (due to the velocity of satellite), works in synchronous way, and it becomes neutral. So that satellite keeps revolving around at fixed position. So, such orbit is known as <strong>Geostationary orbit</strong>.</p>

CC BY-SA 3.0
2014-02-16T07:09:01.580Z
2014-02-17T08:43:16.737Z
2014-02-17T08:43:16.737Z
AmitG
<p>The <a href="http://en.wikipedia.org/wiki/Centrifugal_force" rel="noreferrer">centrifugal force</a> increases with the radius of the orbit for a fixed angular speed. The gravitational pull decreases with the radius of the orbit around Earth. The geostationary orbit is, where the two forces annihilate for the geosynchronous angular speed. The gravitational pull takes the role of the <a href="http://en.wikipedia.org/wiki/Centripetal_force" rel="noreferrer">centripetal force</a> to keep the satellite on the circular orbit.</p>

<p>That's the principle, which keeps satellites, moons, planets on their respective orbit. If there is some additional force, they leave their former orbit.</p>

CC BY-SA 3.0
2014-02-15T20:04:16.267Z
2014-02-15T20:04:16.267Z
Gerald
<p>According to Wikipedia (<a href="http://en.wikipedia.org/wiki/Geostationary_orbit" rel="nofollow noreferrer">link</a>),  a geostationary orbit, geostationary Earth orbit or geosynchronous equatorial orbit (GEO), is a circular orbit 35,786 kilometres (22,236 mi) above the Earth's equator and following the direction of the Earth's rotation. An object in such an orbit has an orbital period equal to the Earth's rotational period (one sidereal day), and thus appears motionless, at a fixed position in the sky, to ground observers.</p>

<p>My question is then couldn't any orbit be a GEO? I mean, what about a satellite in an orbit that's say 70,000km, if the angular speed of the satellite was the same as the earth's rotation such that it always stayed in a fixed position when observed from the Earth, then isn't that a GEO orbit as well?</p>

CC BY-SA 3.0
2014-02-16T03:02:42.093Z
Interesting fact: A geostationary orbit is also called a "Clarke Orbit" after Arthur C. Clarke proposed using satellites in this orbit for global communications in a paper called `"Extra-Terrestrial Relays — Can Rocket Stations Give Worldwide Radio Coverage?"` in 1945. http://lakdiva.org/clarke/1945ww/
Marc
<orbit>
Can someone explain a Geostationary Orbit in a simpler way?
Sean Bernardino
# Can someone explain a Geostationary Orbit in a simpler way?

(Asked by: username_0 on 2014-02-15)

According to Wikipedia ([link](http://en.wikipedia.org/wiki/Geostationary_orbit)), a geostationary orbit, geostationary Earth orbit or geosynchronous equatorial orbit (GEO), is a circular orbit 35,786 kilometres (22,236 mi) above the Earth's equator and following the direction of the Earth's rotation. An object in such an orbit has an orbital period equal to the Earth's rotational period (one sidereal day), and thus appears motionless, at a fixed position in the sky, to ground observers.

My question is then couldn't any orbit be a GEO? I mean, what about a satellite in an orbit that's say 70,000km, if the angular speed of the satellite was the same as the earth's rotation such that it always stayed in a fixed position when observed from the Earth, then isn't that a GEO orbit as well?

**Tags:** orbit

> **Comments:**
> **username_1**: Interesting fact: A geostationary orbit is also called a "Clarke Orbit" after Arthur C. Clarke proposed using satellites in this orbit for global communications in a paper called `"Extra-Terrestrial Relays — Can Rocket Stations Give Worldwide Radio Coverage?"` in 1945. http://lakdiva.org/clarke/1945ww/  

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## 5 answers

### Answer by username_7 on 2014-02-15. Score: 6 | Accepted answer

The [centrifugal force](http://en.wikipedia.org/wiki/Centrifugal_force) increases with the radius of the orbit for a fixed angular speed. The gravitational pull decreases with the radius of the orbit around Earth. The geostationary orbit is, where the two forces annihilate for the geosynchronous angular speed. The gravitational pull takes the role of the [centripetal force](http://en.wikipedia.org/wiki/Centripetal_force) to keep the satellite on the circular orbit.

That's the principle, which keeps satellites, moons, planets on their respective orbit. If there is some additional force, they leave their former orbit.


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### Answer by username_2 on 2014-02-17. Score: 4

The speed of a satellite in orbit is directly linked to the [radius of that orbit](http://en.wikipedia.org/wiki/Orbital_speed):  
$V=\frac{2\*\pi\*a}{T}$

V is the orbital speed, a is the radius and T is the orbital period. This means that for any radius, there is only one possible orbital speed and vice versa. So there is only one radius where the period is 24 hours.

If you really want a satellite to orbit at 70,000 km and still be geostationary, you'll have to use thrusters constantly to provide an extra force towards the Earth. This uses far too much fuel to be economical.


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### Answer by username_3 on 2014-02-15. Score: 1

You have yourself provided us with the answer.

Any satellite revolving over the **Earth's Equator** would be considered as a GeoStationary object. Provided, it should move with the rotation of the Earth and it should be **35,786 kilometres** above of the sea so that the observer, that is on the ground would think that the satellite is **stationary**.

If to say for simple:

A Geostationary object is an object, which is fully synchronized to the movement of the earth. Its position is as it is fixed to the Equator of the earth at zero degree latitude.

Anything that would stay synchronized with the Earth's rotation would be a Geostationary object.

Apart from the Wikipedia article you can read more here: <http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Geostationary_orbit.html>

It has clear and more simple information, and has described that the object must be this\_many kms above the sea level and should be at the 0 degree latitude.

> **Comments:**
> **username_4**: Not correct. Geostationary objects do not have the exact same orbit. They are not on exactly circular orbits, they are not all at the exact same altitude, and they are not all on the Equator plane. Besides, you did not answer the base question.  
> **username_2**: If an orbit is not on the equatorial plane, it's not geostationary. It may be geosynchronous, though. Similarly, if an orbit isn't circular, the satellite's apparent position in the sky will vary as the angular speed increases, then decreases. It might be close enough for operational purposes, but it's not exactly geostationary.  

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### Answer by username_4 on 2014-02-17. Score: 1

@username_2 put the formula:

$V={2\pi a\over T}$

If you substitute on it $T$ for 1 sidereal day you'll have a one-to-one relation between $V$ and $a$. This seems to imply that you can choose any $a$ and use the corresponding $V$, but this is not true since it does not take into account which force are you using.

The formula above is valid of any object in a circular motion, like a stone on a sling, or a car in a loop. It is also valid for satellites around planets.

But there is another formula stating a relation among $a$ and $T$ when the force is that of Gravity, so you are not free to chose $a$ anymore.

It happens that centripetal acceleration (that of the sling, the Normal force on the loop, or gravity for the satellite) is the Universal Gravity in our case:

$a\_c={GM\over r^2}$

but it being the centripetal acceleration means

$a\_c=\omega^2 r$

so we have

$\omega^2 r={GM\over r^2}$

$\omega^2 r^3=GM$

$\omega$ is $v\over r$ so

$v^2 r=GM$

which gives you a fixed relation between $v$ and $r$ for any planet with mass $M$.

> **Comments:**
> **username_5**: I like your answer, but decide to upvote another because the question asked for a simple explanation.  

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### Answer by username_6 on 2014-02-16. Score: 0

In a simpler way, if that satellite is above 35,786 kilometres above you country place, for example, Inidia-Dehli, then satellite will remain at the same position throughout its life. That is why it is called as Geostationary satellite. It revolves with the speed of earth, but circumference of earth and circumference of geostationary satellite's orbit is different. Circumference of geostationary satellite is much greater than the earths circumference. So speed of geostationary satellite is greater than speed of earth because it has to stick to you country place above earth so it have to move fast.

Also in this orbit, gravitational force of earth which attracts satellite towards earth and centrifugal force which tries to escape satellite from that orbit (due to the velocity of satellite), works in synchronous way, and it becomes neutral. So that satellite keeps revolving around at fixed position. So, such orbit is known as **Geostationary orbit**.


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