<p>Since Mimas is in orbit around Saturn, it is in freefall; just as an astronaut in a space station appears to not experience the Earth's gravity because that gravity is acting equally on the space station and the astronaut, the outside of Mimas will appear to not experience Saturn's gravity, as the center is also experience Saturn's gravity and thus they are moving together. The only effect Saturn will have on the integrity of Mimas is Saturn's tidal force. Also, for the tidal force to rip apart a satellite, it has to overcome not only the satellite's gravity, but also any intermolecular forces. For instance, for a space station to be ripped apart by Earth's gravity, the tidal forces would have to overcome the tensile strength of whatever the station is made out of.</p>

CC BY-SA 4.0
2020-09-15T09:09:33.137Z
For celestial bodies the tensile strength can be considered zero (and it will disintegrate when the orbital radius becomes smaller than the [Roche limit](https://en.wikipedia.org/wiki/Roche_limit)). For space stations it is exactly the opposite: its gravitation can be assumed zero.
Peter - Reinstate Monica
CC BY-SA 4.0
2020-09-15T00:57:12.230Z
2020-09-15T00:57:12.230Z
Acccumulation
<blockquote>
<p>How is it possible that Saturn's gravitational acceleration felt by Mimas is stronger than Mimas' own surface gravity?</p>
</blockquote>
<p>That's just the way it is.  An apple hanging from a tree is more strongly attracted to the Earth than to the tree.  A worm crawling on it is more attracted to the Earth than the apple.  Yet they retain some forces keeping them from falling to the ground.</p>
<p>Since Mimas and any object on it's surface are orbiting Saturn and in <a href="https://en.wikipedia.org/wiki/Free_fall" rel="nofollow noreferrer">free fall</a>, Saturn's force of gravity mainly curves their paths and they don't get tugged straight down to Saturn.  The gravity of Mimas itself is enough to keep things from flying off it's surface.  There are also cohesive forces tending to keep it together.</p>
<blockquote>
<p>How can this be? An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas. Shouldn't Mimas itself get ripped apart or is my math wrong?</p>
</blockquote>
<p>You don't go flying off Mimas because Mimas is being affected by Saturn's gravity as well, and Mimas exerts a large enough pull to keep you in place.  And since you would be travelling in orbit with Mimas you would both be experiencing Saturns's gravity.</p>
<p>What would make Mimas tend to break apart is <a href="https://en.wikipedia.org/wiki/Tidal_force" rel="nofollow noreferrer">tidal forces</a>, except it's dense enough and far enough away from Saturn to avoid that fate.  There's a calculation to tell you if an object in orbit will break apart called the <a href="https://en.wikipedia.org/wiki/Roche_limit" rel="nofollow noreferrer">Roche Limit</a>.  Part of the calculation is the ratio of the density of the primary to the density of the secondary, and Saturn's low density helps keep it small in this case.  Calculating it myself I get 61,826 kilometers for a rigid body.  That fits well with what <a href="https://space.fandom.com/wiki/Roche_limit" rel="nofollow noreferrer">this page</a> says given that the density of Mimas is about 2/3 higher than that of Saturn.  So Mimas orbits about 3 times the Roche limit and will not disintegrate due to Saturn's gravity.  Even for the other extreme of a fluid body, the Roche Limit is just under twice that for a rigid body so Mimas still wouldn't come apart.</p>
<p>Using your calculation for gravity and plugging in an extra 414km for the diameter of Mimas shows that the difference in Saturn's gravity on the near side to Saturn and the far side to Saturn is just 0.005 m/s^2, less than 1/12th the surface gravity of Mimas (0.063 m/s^2)</p>
<hr />
<p>Some thought experiments:</p>
<p>If you were on Mimas and it suddenly disappeared, leaving you in space, you wouldn't be sucked to Saturn.  You would continue in basically the same orbit as Mimas would have.  You are going fast with respect to Saturn's surface, and Saturn's gravity is just enough to curve your path to maintain your orbit so you don't go flying off into space, and so you don't crash into Saturn.</p>
<p>If you could somehow stop Mimas (and you) in it's tracks with respect to Saturn, Mimas and you would still be in free fall, but would both be pulled towards Saturn.  The gravity of Mimas would still tend to pull you towards its center as well so you wouldn't fly off the surface.</p>
<p>If you could somehow stop Mimas, create an adamantite shell around it to keep it's form, and suspend it at a point above Saturn the same distance as it's orbit, you would fly off the surface towards Saturn if you were on the Saturn side.  This is because you are preventing Mimas from falling with you.  You would accelerate at roughly (1.102 - 0.063) m/s^2 because Saturn is pulling you down and Mimas is pulling you up.</p>
<p>If you could somehow stop Mimas and create an adamantite platform under it at the same distance from Saturn, it should collapse and form huge pile of ice on the platform.  The Roche Limit works for orbiting bodies.</p>

CC BY-SA 4.0
2020-09-15T13:33:30.077Z
Sorry but I feel this is too many impossible hypothetical scenarios in prose only, and it invokes purely fictional materials. It also does not even clearly address the question as asked. I usually don't down vote but in this case I must. Firmly.
uhoh
CC BY-SA 4.0
2020-09-15T13:30:12.373Z
2020-09-16T16:50:14.623Z
2020-09-16T16:50:14.623Z
Jason Goemaat
<blockquote>
<p>An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas.</p>
</blockquote>
<p>You are missing that Mimas as a whole accelerates gravitationally toward Saturn. What this means is that a point on the surface of the Mimas will feel the acceleration at that point toward Saturn minus the acceleration of Mimas as a whole toward Saturn. This is the tidal acceleration. It is equal to
<span class="math-container">$$a_\text{tidal} = \left|\frac{GM}{(R\pm r)^2}-\frac{GM}{r^2}\right| \approx 2 \frac{GMr}{R^3} = 2\frac{GM}{R^2}\frac{r}{R}$$</span>
where <span class="math-container">$R$</span> is Mimas' semi major axis length and <span class="math-container">$r$</span> is Mimas' mean radius. The approximation assumes that <span class="math-container">$r\ll R$</span>, which most certainly is the case given that Mimas's radius is about 1/1000 of the semi-major axis length of its orbit about Saturn. The result is rather small, about 0.002355 m/s<sup>2</sup>.</p>

CC BY-SA 4.0
2020-09-15T00:44:06.990Z
"What this means is that a point on the surface of the Mimas will feel the acceleration at that point toward Saturn less the acceleration of Mimas as a whole toward Saturn." I really don't understand this sentence. Are you using "less" to mean "minus"?
Acccumulation
CC BY-SA 4.0
2020-09-15T02:11:58.797Z
That's how I read it. It is a legitimate, though _slightly_ uncommon, use of the word "less".
David Z
CC BY-SA 4.0
2020-09-15T06:06:02.620Z
@DavidZ - I'm not thinking lesser of you for calling that usage "*slightly* uncommon" (pun intended; no slight intended; I couldn't resist). I will change my answer to use *minus* rather than *less* as I have learned that critiques against my writing are almost always valid.
David Hammen
CC BY-SA 4.0
2020-09-15T06:08:45.287Z
@Acccumulation - What I wrote was perfectly valid. "Less" when used as a preposition is a synonym for "minus".  Nonetheless I did edit my answer to replace my use of "less" with "minus".
David Hammen
CC BY-SA 4.0
2020-09-15T16:42:14.960Z
An aside: “Less” meaning minus is generally something I’d expect to hear in conversation, rather than see written in prose, I think. “What does the bill come to?” “£35, less the 20% voucher, plus the tip”, would be the sort of context I expect it in.
Tim
CC BY-SA 4.0
2020-09-15T18:07:11.740Z
Does this mean that an object on Mimas' surface experiences a net acceleration of 0.063 - 0.002355  = 0.060546 m/s^2?
user177107
CC BY-SA 4.0
2020-09-15T19:27:53.590Z
@user177107 -  When Saturn is directly overhead or underfoot, yes. (A better value is 0.06135 m/s^2 based on 0.06370 m/s^2). When Saturn is on the horizon, the tidal acceleration is halved in magnitude and directed toward the center of Mimas, resulting in an acceleration of about 0.06488 m/s^2.
David Hammen
CC BY-SA 4.0
2020-09-16T01:42:06.063Z
Using "less" as "minus" adds confusion to a sentence that is already confusingly worded. You're subtracting acceleration rather than forces, and you're referring to the difference as what is "felt", rather than what's resulting in relative motion.
Acccumulation
CC BY-SA 4.0
2020-09-16T07:29:22.667Z
@Accumulation - One do not feel gravitational acceleration. Nothing does; it's a consequence of the equivalence principle. What one does feel is the normal force pushing one upward so as to keep one on the surface. This is, for example, why an accelerometer at rest on the surface of the Earth registers an acceleration of 1 *g* upward.
David Hammen
CC BY-SA 4.0
2020-09-14T16:10:46.723Z
2020-09-15T06:17:35.353Z
2020-09-15T06:17:35.353Z
David Hammen
<p>The surface gravity on Mimas is <span class="math-container">$≈ 0.063\text{ m}/\text{s}^2$</span> and Saturn's gravitational acceleration at the distance of Mimas' orbit is:</p>
<p><span class="math-container">$$\frac{{GM}}{{r}^2} = \frac{{6.674 \times 10^{-11} \times 568.34 \times 10^{24}}}{{(185.52 \times 10^{6})}^2} ≈  1.102 \text{ m}/\text{s}^2$$</span></p>
<p>How can this be? An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas. Shouldn't Mimas itself get ripped apart or is my math wrong?</p>

CC BY-SA 4.0
2020-09-14T18:14:31.693Z
The earths gravitational acceleration "felt" by astronauts is immensely more than the space station's own "surface gravity".
Michael Hardy
CC BY-SA 4.0
2020-09-14T23:16:40.250Z
related: [If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface?](https://astronomy.stackexchange.com/q/32059/7982)
uhoh
CC BY-SA 4.0
2020-09-15T19:41:06.307Z
@MichaelHardy yes, but that's not really a fair comparison. The ISS isn't bound by gravity but structural trusses, but planetary bodies generally do require gravity. Specifically, if Mimas were within Saturn's [Roche limit](https://en.wikipedia.org/wiki/Roche_limit), like the ISS is withing Earth's, then it _would_ get ripped apart.
leftaroundabout
<gravity><saturn>
How is it possible that Saturn's gravitational acceleration felt by Mimas is stronger than Mimas' own surface gravity?
ChristieToWin
# How is it possible that Saturn's gravitational acceleration felt by Mimas is stronger than Mimas' own surface gravity?

(Asked by: username_0 on 2020-09-14)

The surface gravity on Mimas is $≈ 0.063\text{ m}/\text{s}^2$ and Saturn's gravitational acceleration at the distance of Mimas' orbit is:

$$\frac{{GM}}{{r}^2} = \frac{{6.674 \times 10^{-11} \times 568.34 \times 10^{24}}}{{(185.52 \times 10^{6})}^2} ≈ 1.102 \text{ m}/\text{s}^2$$

How can this be? An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas. Shouldn't Mimas itself get ripped apart or is my math wrong?

**Tags:** gravity, saturn

> **Comments:**
> **username_1**: The earths gravitational acceleration "felt" by astronauts is immensely more than the space station's own "surface gravity".  
> **username_2**: related: [If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface?](https://astronomy.stackexchange.com/q/32059/7982)  
> **username_3**: @username_1 yes, but that's not really a fair comparison. The ISS isn't bound by gravity but structural trusses, but planetary bodies generally do require gravity. Specifically, if Mimas were within Saturn's [Roche limit](https://en.wikipedia.org/wiki/Roche_limit), like the ISS is withing Earth's, then it _would_ get ripped apart.  

---

## 3 answers

### Answer by username_7 on 2020-09-14. Score: 31 | Accepted answer

> An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas.

You are missing that Mimas as a whole accelerates gravitationally toward Saturn. What this means is that a point on the surface of the Mimas will feel the acceleration at that point toward Saturn minus the acceleration of Mimas as a whole toward Saturn. This is the tidal acceleration. It is equal to
$$a\_\text{tidal} = \left|\frac{GM}{(R\pm r)^2}-\frac{GM}{r^2}\right| \approx 2 \frac{GMr}{R^3} = 2\frac{GM}{R^2}\frac{r}{R}$$
where $R$ is Mimas' semi major axis length and $r$ is Mimas' mean radius. The approximation assumes that $r\ll R$, which most certainly is the case given that Mimas's radius is about 1/1000 of the semi-major axis length of its orbit about Saturn. The result is rather small, about 0.002355 m/s2.

> **Comments:**
> **username_4**: "What this means is that a point on the surface of the Mimas will feel the acceleration at that point toward Saturn less the acceleration of Mimas as a whole toward Saturn." I really don't understand this sentence. Are you using "less" to mean "minus"?  
> **username_8**: That's how I read it. It is a legitimate, though _slightly_ uncommon, use of the word "less".  
> **username_7**: @username_8 - I'm not thinking lesser of you for calling that usage "*slightly* uncommon" (pun intended; no slight intended; I couldn't resist). I will change my answer to use *minus* rather than *less* as I have learned that critiques against my writing are almost always valid.  
> **username_7**: @username_4 - What I wrote was perfectly valid. "Less" when used as a preposition is a synonym for "minus".  Nonetheless I did edit my answer to replace my use of "less" with "minus".  
> **username_9**: An aside: “Less” meaning minus is generally something I’d expect to hear in conversation, rather than see written in prose, I think. “What does the bill come to?” “£35, less the 20% voucher, plus the tip”, would be the sort of context I expect it in.  
> **username_10**: Does this mean that an object on Mimas' surface experiences a net acceleration of 0.063 - 0.002355  = 0.060546 m/s^2?  
> **username_7**: @username_10 -  When Saturn is directly overhead or underfoot, yes. (A better value is 0.06135 m/s^2 based on 0.06370 m/s^2). When Saturn is on the horizon, the tidal acceleration is halved in magnitude and directed toward the center of Mimas, resulting in an acceleration of about 0.06488 m/s^2.  
> **username_4**: Using "less" as "minus" adds confusion to a sentence that is already confusingly worded. You're subtracting acceleration rather than forces, and you're referring to the difference as what is "felt", rather than what's resulting in relative motion.  
> **username_7**: @Accumulation - One do not feel gravitational acceleration. Nothing does; it's a consequence of the equivalence principle. What one does feel is the normal force pushing one upward so as to keep one on the surface. This is, for example, why an accelerometer at rest on the surface of the Earth registers an acceleration of 1 *g* upward.  

---

### Answer by username_4 on 2020-09-15. Score: 8

Since Mimas is in orbit around Saturn, it is in freefall; just as an astronaut in a space station appears to not experience the Earth's gravity because that gravity is acting equally on the space station and the astronaut, the outside of Mimas will appear to not experience Saturn's gravity, as the center is also experience Saturn's gravity and thus they are moving together. The only effect Saturn will have on the integrity of Mimas is Saturn's tidal force. Also, for the tidal force to rip apart a satellite, it has to overcome not only the satellite's gravity, but also any intermolecular forces. For instance, for a space station to be ripped apart by Earth's gravity, the tidal forces would have to overcome the tensile strength of whatever the station is made out of.

> **Comments:**
> **username_5**: For celestial bodies the tensile strength can be considered zero (and it will disintegrate when the orbital radius becomes smaller than the [Roche limit](https://en.wikipedia.org/wiki/Roche_limit)). For space stations it is exactly the opposite: its gravitation can be assumed zero.  

---

### Answer by username_6 on 2020-09-15. Score: 4

> How is it possible that Saturn's gravitational acceleration felt by Mimas is stronger than Mimas' own surface gravity?

That's just the way it is. An apple hanging from a tree is more strongly attracted to the Earth than to the tree. A worm crawling on it is more attracted to the Earth than the apple. Yet they retain some forces keeping them from falling to the ground.

Since Mimas and any object on it's surface are orbiting Saturn and in [free fall](https://en.wikipedia.org/wiki/Free_fall), Saturn's force of gravity mainly curves their paths and they don't get tugged straight down to Saturn. The gravity of Mimas itself is enough to keep things from flying off it's surface. There are also cohesive forces tending to keep it together.

> How can this be? An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas. Shouldn't Mimas itself get ripped apart or is my math wrong?

You don't go flying off Mimas because Mimas is being affected by Saturn's gravity as well, and Mimas exerts a large enough pull to keep you in place. And since you would be travelling in orbit with Mimas you would both be experiencing Saturns's gravity.

What would make Mimas tend to break apart is [tidal forces](https://en.wikipedia.org/wiki/Tidal_force), except it's dense enough and far enough away from Saturn to avoid that fate. There's a calculation to tell you if an object in orbit will break apart called the [Roche Limit](https://en.wikipedia.org/wiki/Roche_limit). Part of the calculation is the ratio of the density of the primary to the density of the secondary, and Saturn's low density helps keep it small in this case. Calculating it myself I get 61,826 kilometers for a rigid body. That fits well with what [this page](https://space.fandom.com/wiki/Roche_limit) says given that the density of Mimas is about 2/3 higher than that of Saturn. So Mimas orbits about 3 times the Roche limit and will not disintegrate due to Saturn's gravity. Even for the other extreme of a fluid body, the Roche Limit is just under twice that for a rigid body so Mimas still wouldn't come apart.

Using your calculation for gravity and plugging in an extra 414km for the diameter of Mimas shows that the difference in Saturn's gravity on the near side to Saturn and the far side to Saturn is just 0.005 m/s^2, less than 1/12th the surface gravity of Mimas (0.063 m/s^2)

---

Some thought experiments:

If you were on Mimas and it suddenly disappeared, leaving you in space, you wouldn't be sucked to Saturn. You would continue in basically the same orbit as Mimas would have. You are going fast with respect to Saturn's surface, and Saturn's gravity is just enough to curve your path to maintain your orbit so you don't go flying off into space, and so you don't crash into Saturn.

If you could somehow stop Mimas (and you) in it's tracks with respect to Saturn, Mimas and you would still be in free fall, but would both be pulled towards Saturn. The gravity of Mimas would still tend to pull you towards its center as well so you wouldn't fly off the surface.

If you could somehow stop Mimas, create an adamantite shell around it to keep it's form, and suspend it at a point above Saturn the same distance as it's orbit, you would fly off the surface towards Saturn if you were on the Saturn side. This is because you are preventing Mimas from falling with you. You would accelerate at roughly (1.102 - 0.063) m/s^2 because Saturn is pulling you down and Mimas is pulling you up.

If you could somehow stop Mimas and create an adamantite platform under it at the same distance from Saturn, it should collapse and form huge pile of ice on the platform. The Roche Limit works for orbiting bodies.

> **Comments:**
> **username_2**: Sorry but I feel this is too many impossible hypothetical scenarios in prose only, and it invokes purely fictional materials. It also does not even clearly address the question as asked. I usually don't down vote but in this case I must. Firmly.  

---

