<p>The formula for orbital velocity is <span class="math-container">$\sqrt{GM/r}$</span> and for a &quot;low&quot; orbit you would mean orbit at, or close to the surface, ie with a radius equal to the radius of the planet.  This makes calculating the velocity possible (SI units metres and seconds):</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">Body</th>
<th style="text-align: center;">GM</th>
<th style="text-align: center;">r</th>
<th style="text-align: center;">v</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">Sun</td>
<td style="text-align: center;">1.33E+20</td>
<td style="text-align: center;">696340000</td>
<td style="text-align: center;">436561</td>
</tr>
<tr>
<td style="text-align: center;">Mercury</td>
<td style="text-align: center;">2.20E+13</td>
<td style="text-align: center;">2439500</td>
<td style="text-align: center;">3005</td>
</tr>
<tr>
<td style="text-align: center;">Venus</td>
<td style="text-align: center;">3.25E+14</td>
<td style="text-align: center;">6052000</td>
<td style="text-align: center;">7327</td>
</tr>
<tr>
<td style="text-align: center;">Earth</td>
<td style="text-align: center;">3.99E+14</td>
<td style="text-align: center;">6378000</td>
<td style="text-align: center;">7905</td>
</tr>
<tr>
<td style="text-align: center;">Moon</td>
<td style="text-align: center;">4.90E+12</td>
<td style="text-align: center;">1737500</td>
<td style="text-align: center;">1680</td>
</tr>
<tr>
<td style="text-align: center;">Mars</td>
<td style="text-align: center;">4.28E+13</td>
<td style="text-align: center;">3396000</td>
<td style="text-align: center;">3551</td>
</tr>
<tr>
<td style="text-align: center;">Ceres</td>
<td style="text-align: center;">6.26E+10</td>
<td style="text-align: center;">473000</td>
<td style="text-align: center;">364</td>
</tr>
<tr>
<td style="text-align: center;">Jupiter</td>
<td style="text-align: center;">1.27E+17</td>
<td style="text-align: center;">71492000</td>
<td style="text-align: center;">42096</td>
</tr>
<tr>
<td style="text-align: center;">Saturn</td>
<td style="text-align: center;">3.79E+16</td>
<td style="text-align: center;">60268000</td>
<td style="text-align: center;">25087</td>
</tr>
<tr>
<td style="text-align: center;">Uranus</td>
<td style="text-align: center;">5.79E+15</td>
<td style="text-align: center;">25559000</td>
<td style="text-align: center;">15056</td>
</tr>
<tr>
<td style="text-align: center;">Neptune</td>
<td style="text-align: center;">6.84E+15</td>
<td style="text-align: center;">24764000</td>
<td style="text-align: center;">16615</td>
</tr>
<tr>
<td style="text-align: center;">Pluto</td>
<td style="text-align: center;">8.71E+11</td>
<td style="text-align: center;">1188000</td>
<td style="text-align: center;">856</td>
</tr>
<tr>
<td style="text-align: center;">Eris</td>
<td style="text-align: center;">1.11E+12</td>
<td style="text-align: center;">1163000</td>
<td style="text-align: center;">976</td>
</tr>
</tbody>
</table>
</div>
<p>For planets with an atmosphere, a practical low orbit will have a slightly larger radius, and so lower velocity.  There may be practical issues with a &quot;low solar orbit&quot;!</p>
<p>If you multiply these orbital velocities by <span class="math-container">$\sqrt2$</span>, you will get the escape velocity from the surface.</p>
<p>There would be a &quot;Jovistationary&quot; orbit, but note that different parts of Jupiter rotate at different speeds. On the other hand, Venus rotates so slowly that a &quot;veneralstationary&quot; orbit would be so far from the planet that you would no longer be able to orbit.  </p>

CC BY-SA 4.0
2022-08-10T13:45:07.433Z
For anyone wondering: A "Veneralstationary" orbit would have to be at about 100 000 AU  or about 1.5 ly away from Venus.
SirHawrk
CC BY-SA 4.0
2022-08-10T18:37:30.073Z
' Junostationary' would be stationary about 3 Juno. For Jupiter, the prefix is usually *jovi-*, or sometime *zeno-*; so, jovistationary, a term which does have some usage. For Venus, as you point out that orbit does not exist, but the term would probably be "venerostationary"; one would remove the ending "-al" from the prefix.
Mark Beadles
CC BY-SA 4.0
2022-08-10T19:58:34.620Z
yes, but "veneralstationary" is funnier.....and yes I am 14.
James K
CC BY-SA 4.0
2022-08-11T02:56:40.040Z
An object 1.5ly from Venus would have to travel over 9ly to complete an orbit - and to do so in 243 days would need to travel at 14 times the speed of light. 
This answer https://www.quora.com/Is-geostationary-orbit-possible-on-Venus suggests that the aphroditosynchronous (!) orbit is at 7.132×10⁵km.
Barrington
CC BY-SA 4.0
2022-08-10T06:48:26.480Z
2022-08-11T05:12:33.773Z
2022-08-11T05:12:33.773Z
James K
<p>&quot;Low-Earth-Orbit&quot; is kind of arbitrarily defined, and I don't believe there's a widely accepted general definition of a low orbit that can be applied to other planets.</p>
<p>If you know what the radius of your orbit is, your satellite is of negligible mass compared to the body it orbits, and the orbit is roughly circular, you can approximate the orbital velocity with</p>
<p><span class="math-container">\begin{equation}
\ v = \sqrt\frac{GM}{r}
\end{equation}</span></p>
<p>If we assume the altitude of a Low-Jupiter-Orbit is about the same as the minimum altitude of the Juno spacecraft's orbit, we'd get a velocity of around 41 km/s for a circular orbit.</p>
<p>The simple equation for a stationary orbit's radius would be
<span class="math-container">\begin{equation}
\ r = \sqrt[3]\frac{GMT^2}{4\pi^2}
\end{equation}</span></p>
<p>(r being the orbit's radius, G being the gravitational constant, M being the mass of the body you're orbiting, and T being that body's rotational period for both equations.)</p>
<p>Note that because Jupiter isn't a solid body, different parts of its surface rotate at different speeds, so I'm not sure you could really call a Jupiter orbit &quot;stationary&quot;.</p>

CC BY-SA 4.0
2022-08-10T06:22:58.490Z
2022-08-10T06:22:58.490Z
toxicPsychotic
<p>I was pondering this question recently, but most sites I can find only mention the speeds/velocities of the planets around the Sun when I look for 'orbital velocities' of the planets.</p>
<p>I tried examining the velocities of moons around other planets, or NASA satellites, but these are all over the proverbial map, and often in very eccentric orbits.</p>
<p>Is there a 'Low-Jupiter-Orbit' like there is for Earth (LEO)?  What is its altitude and speed?</p>
<p>What about a Juno-synchronous or Jupiter-stationary high(er) orbit, as there is for Earth?</p>
<p>P.S.:  Are there (simple) equations for these orbits?</p>

CC BY-SA 4.0
2022-08-10T05:46:12.310Z
Yes, there's a simple equation for circular orbit speed. I'm surprised you didn't find it in your research. ;) However, for low orbits you need to account for atmosphere, which is a bit trickier, especially for the giant planets, which don't have a well-defined surface. Here's some info about atmospheric pressure and altitude: https://en.wikipedia.org/wiki/Scale_height
PM 2Ring
CC BY-SA 4.0
2022-08-10T09:41:44.413Z
Using the state vector output, JPL Horizons allows you to set a planet as the coordinate centre and then find the velocity vectors of that planet's moons.
Peter
CC BY-SA 4.0
2022-08-13T06:11:51.670Z
Similar question asked in Space.SE: https://space.stackexchange.com/questions/54937/is-there-a-standard-low-orbit-for-other-solar-system-bodies
Nilay Ghosh
<orbit><orbital-mechanics><orbital-elements><planetary-science><eccentric-orbit>
What are the orbital velocities of the other planets? For objects in a 'Low-Earth-Orbit' around planets other than Earth, e.g.?
Kurt Hikes
# What are the orbital velocities of the other planets? For objects in a 'Low-Earth-Orbit' around planets other than Earth, e.g.?

(Asked by: username_0 on 2022-08-10)

I was pondering this question recently, but most sites I can find only mention the speeds/velocities of the planets around the Sun when I look for 'orbital velocities' of the planets.

I tried examining the velocities of moons around other planets, or NASA satellites, but these are all over the proverbial map, and often in very eccentric orbits.

Is there a 'Low-Jupiter-Orbit' like there is for Earth (LEO)? What is its altitude and speed?

What about a Juno-synchronous or Jupiter-stationary high(er) orbit, as there is for Earth?

P.S.: Are there (simple) equations for these orbits?

**Tags:** orbit, orbital-mechanics, orbital-elements, planetary-science, eccentric-orbit
> **Comments:**
> **username_1**: Yes, there's a simple equation for circular orbit speed. I'm surprised you didn't find it in your research. ;) However, for low orbits you need to account for atmosphere, which is a bit trickier, especially for the giant planets, which don't have a well-defined surface. Here's some info about atmospheric pressure and altitude: https://en.wikipedia.org/wiki/Scale_height  
> **username_2**: Using the state vector output, JPL Horizons allows you to set a planet as the coordinate centre and then find the velocity vectors of that planet's moons.  
> **username_3**: Similar question asked in Space.SE: https://space.stackexchange.com/questions/54937/is-there-a-standard-low-orbit-for-other-solar-system-bodies  

---

## 2 answers

### Answer by username_4 on 2022-08-10. Score: 20

The formula for orbital velocity is $\sqrt{GM/r}$ and for a "low" orbit you would mean orbit at, or close to the surface, ie with a radius equal to the radius of the planet. This makes calculating the velocity possible (SI units metres and seconds):

| Body | GM | r | v |
| --- | --- | --- | --- |
| Sun | 1.33E+20 | 696340000 | 436561 |
| Mercury | 2.20E+13 | 2439500 | 3005 |
| Venus | 3.25E+14 | 6052000 | 7327 |
| Earth | 3.99E+14 | 6378000 | 7905 |
| Moon | 4.90E+12 | 1737500 | 1680 |
| Mars | 4.28E+13 | 3396000 | 3551 |
| Ceres | 6.26E+10 | 473000 | 364 |
| Jupiter | 1.27E+17 | 71492000 | 42096 |
| Saturn | 3.79E+16 | 60268000 | 25087 |
| Uranus | 5.79E+15 | 25559000 | 15056 |
| Neptune | 6.84E+15 | 24764000 | 16615 |
| Pluto | 8.71E+11 | 1188000 | 856 |
| Eris | 1.11E+12 | 1163000 | 976 |

For planets with an atmosphere, a practical low orbit will have a slightly larger radius, and so lower velocity. There may be practical issues with a "low solar orbit"!

If you multiply these orbital velocities by $\sqrt2$, you will get the escape velocity from the surface.

There would be a "Jovistationary" orbit, but note that different parts of Jupiter rotate at different speeds. On the other hand, Venus rotates so slowly that a "veneralstationary" orbit would be so far from the planet that you would no longer be able to orbit.

> **Comments:**
> **username_5**: For anyone wondering: A "Veneralstationary" orbit would have to be at about 100 000 AU  or about 1.5 ly away from Venus.  
> **username_6**: ' Junostationary' would be stationary about 3 Juno. For Jupiter, the prefix is usually *jovi-*, or sometime *zeno-*; so, jovistationary, a term which does have some usage. For Venus, as you point out that orbit does not exist, but the term would probably be "venerostationary"; one would remove the ending "-al" from the prefix.  
> **username_4**: yes, but "veneralstationary" is funnier.....and yes I am 14.  
> **username_7**: An object 1.5ly from Venus would have to travel over 9ly to complete an orbit - and to do so in 243 days would need to travel at 14 times the speed of light.  This answer https://www.quora.com/Is-geostationary-orbit-possible-on-Venus suggests that the aphroditosynchronous (!) orbit is at 7.132×10⁵km.  

---

### Answer by username_8 on 2022-08-10. Score: 7

"Low-Earth-Orbit" is kind of arbitrarily defined, and I don't believe there's a widely accepted general definition of a low orbit that can be applied to other planets.

If you know what the radius of your orbit is, your satellite is of negligible mass compared to the body it orbits, and the orbit is roughly circular, you can approximate the orbital velocity with

\begin{equation}
\ v = \sqrt\frac{GM}{r}
\end{equation}

If we assume the altitude of a Low-Jupiter-Orbit is about the same as the minimum altitude of the Juno spacecraft's orbit, we'd get a velocity of around 41 km/s for a circular orbit.

The simple equation for a stationary orbit's radius would be
\begin{equation}
\ r = \sqrt[3]\frac{GMT^2}{4\pi^2}
\end{equation}

(r being the orbit's radius, G being the gravitational constant, M being the mass of the body you're orbiting, and T being that body's rotational period for both equations.)

Note that because Jupiter isn't a solid body, different parts of its surface rotate at different speeds, so I'm not sure you could really call a Jupiter orbit "stationary".


---

