<p>The Roche limit is where the tidal forces exerted on an orbiting object are sufficient to overcome the <em>self-gravity</em> of that object.</p>

<p>The "self-gravity" of an astronaut is tiny. We can estimate it as something like <span class="math-container">$$F_{\rm grav} \sim \frac{Gm^2}{(h/2)^2},$$</span>
where <span class="math-container">$m$</span> is the mass of the astronaut (+ equipment) and <span class="math-container">$h$</span> is their size (height). Assuming <span class="math-container">$m=100$</span> kg and <span class="math-container">$h=2$</span> m, then the self-gravity  force is <span class="math-container">$2.7\times 10^{-6}$</span> N. This is a force that is too small to feel.</p>

<p>The tidal force on the astronaut a distance <span class="math-container">$R$</span> from a body of mass <span class="math-container">$M$</span> is approximately
<span class="math-container">$$ F_{\rm tidal} \simeq 2 \frac{GMm}{R^3} h,$$</span>
assuming they are pointing feet-first towards the Earth.</p>

<p>The Roche limit is where <span class="math-container">$F_{\rm grav} &lt; F_{\rm tidal}$</span>, so where
<span class="math-container">$$ \frac{Gm^2}{(h/2)^2}&lt; 2 \frac{GMm}{R^3} h$$</span>
<span class="math-container">$$ R &lt; h\left(\frac{M}{2m}\right)^{1/3}$$</span>
For example if <span class="math-container">$M=M_{\rm Earth}$</span> and for the astronaut above, then <span class="math-container">$R &lt; 60,000$</span> km for tidal break up. Which seems odd, because astronauts work quite happily in low-Earth orbit where the tidal forces are much stronger.</p>

<p>The problem with this calculation is that astronauts are not held together by self-gravity and a tidal field at the Roche limit has a negligible effect on a small body that is actually held together by atomic forces.</p>

<p>In order to experience a tidal field that can be felt on astronaut scales, let's say larger than 10 N (imagine hanging a 1 kg weight from your ankle on Earth), you would have to get much closer to the source of gravity. </p>

<p>Assuming a fixed mass equal to that of the Earth, we can work out you would need to get to within 500 km <em>of the centre of mass</em> in order to feel the tidal force. For the Earth (and also for other solar system bodies you could do the same calculation for) this would put you well <em>inside</em> the Earth, which isn't possible and in any case we could not assume that <span class="math-container">$M$</span> was fixed in that case, because it is the mass interior to <span class="math-container">$R$</span> that counts.</p>

<p>The only way that an astronaut could "feel" a tidal force would be to approach a compact star - a high density neutron star, white dwarf or black hole. There you can generate a very strong tidal field and, because they are compact, an astronaut could get close enough to feel it.</p>

CC BY-SA 4.0
2018-07-14T08:23:10.280Z
2020-04-27T18:05:54.453Z
2020-04-27T18:05:54.453Z
ProfRob
<p>Expanding on Peterh's answer, we could try to find how should be an astronomical object for the tidal forces be felt by an astronaut orbiting it.</p>

<p>I don't have any reliable data on how strong need the tidal force to be felt. However, with a big simplification, we can very roughly model the upper and lower body of an astronaut as two masses placed about 1 meter apart. For a 70 kg astronaut under a tidal force of <span class="math-container">$0.1·g$</span> per meter (where <span class="math-container">$g$</span> is the acceleration of gravity), the difference of pull between those two 35 kg masses would be <span class="math-container">$0.1·35 kg = 3.5 kg$</span>. Those force would stretch the astronaut's waist and would be clearly noticeably (maybe a force ten times weaker would be noticeable, too, but I'll will stick to <span class="math-container">$0.1m^{-1}·g$</span>).</p>

<p>From Peterh's formulas:</p>

<p><span class="math-container">$$r=\sqrt[3]{\frac{2·G·M}{0.1m^{-1}·g}}$$</span></p>

<p>For a 1 solar mass object:</p>

<p><span class="math-container">$$r=\sqrt[3]{\frac{2·6.67·10^{-11}·2·10^{30}}{0.1·g}}=6481168 m = 6481 km$$</span></p>

<p>Then, an astronaut orbiting a sun-sized mass at a distance similar to Earth's radius would clearly feel tide forces when their head or feet point to the object. Of course the object would need to be a black hole or a neutron star to fit inside of the orbit.</p>

<p>With a more massive object the orbit could be larger, but given that mass is inside a cubic root, radius would grow very slowly.</p>

CC BY-SA 4.0
2018-07-15T09:35:37.460Z
You don't need a black hole for this.  A neutron star is quite enough (typical mass: one solar mass, typical radius: 10 km).
Martin Bonner supports Monica
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2018-07-15T09:44:49.797Z
@MartinBonner Thank you. Neutron stars added.
Pere
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2018-07-15T15:52:26.803Z
Might also refer to Niven's *Neutron Star*
DJohnM
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2018-07-16T17:55:37.897Z
@DJohnM ack you ninja'd me.   Sorry for posting my comment against the OP
Carl Witthoft
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2018-07-17T04:14:45.233Z
@DJohnM  I don't understand the reference to the Niven's Neutron Star?
Muze
CC BY-SA 4.0
2018-12-28T01:00:32.513Z
@DJohnM Stack Exchange has no [spoiler alert](https://meta.stackexchange.com/q/1191/303080) function for comments, perhaps you can link to Wikipedia as [CarlWitthoft did](https://astronomy.stackexchange.com/questions/26955/is-there-any-orbit-at-which-the-roche-limit-can-be-felt#comment46889_26955) instead of posting the surprise ending of the story here?
uhoh
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2018-07-14T20:57:46.863Z
2020-04-27T17:25:50.313Z
2020-04-27T17:25:50.313Z
Pere
<p>Roche limit happens where the gravity of the object, trying to pull the object together, becomes smaller than the tidal force (trying to pull the object apart).</p>
<p>But the astronaut is bound by not gravity, rather by the electromagnetic interaction between his/her atoms. The own gravity of the astronaut is negligible, compared to the electromagnetic interaction.</p>
<p>However, the tidal force affecting an astronaut, should require a little calculation. We can derive the formula of the gravitational acceleration around a point-like body (<span class="math-container">$F=\frac{GM}{r^2}$</span>), we get</p>
<p><span class="math-container">$$\frac{dF}{dr}=\frac{2GM}{r^3}$$</span></p>
<p>(We can ignore the sign on obvious reasons.)</p>
<p>Here <span class="math-container">$G$</span> is the gravitational constant, <span class="math-container">$M$</span> is the mass of the body, and <span class="math-container">$r$</span> is the distance.</p>
<p>Substituting the values of the Sun, we get <span class="math-container">$\frac{2\cdot 6.67 \cdot 10^{-11} \cdot 2 \cdot 10^{30}}{(7\cdot 10^8)^3} \approx 7.78\cdot 10^{-7} \frac{\mathrm{m/s^2}}{\mathrm{m}} \approx \underline{\underline{8 \cdot 10^{-8} \frac{g}{\mathrm{m}}}}$</span>.</p>
<p>More clearly, if we are orbiting the Sun just above its surface, a roughly 2m long astronaut feel that his head and foot are pulled apart by around <span class="math-container">$1.6\cdot 10^{-7}g$</span> weight. In the case of a <span class="math-container">$70\:\mathrm{kg}$</span> astronaut, it is around the weight of <span class="math-container">$0.0112$</span> gram on the Earth.</p>
<p>The astronaut wouldn't feel it, but not very sensitive sensors could already measure it.</p>
<hr />
<p><sub>This calculation sometimes used <span class="math-container">$\mathrm{g}$</span> for &quot;gram&quot;, as unit of mass, and <span class="math-container">$g$</span> as the (non-standard) unit of acceleration.</sub></p>

CC BY-SA 4.0
2018-07-14T17:01:08.030Z
Ignoring the obvious fact that any astronaut or instrument that close to the sun would be instantly vaporized of course...
Darrel Hoffman
CC BY-SA 4.0
2018-07-15T14:00:21.313Z
@DarrelHoffman The Sun gives $\approx$ 6000K thermal radiation, which is hard but not chanceless to protect against it. I think some strong defense, for example well-polished tungsten mirrors, maybe combined with some cooling from behind, could handle it. The Parker Solar Probe will near the Sun to 8 solar radii.
peterh
CC BY-SA 4.0
2018-07-14T08:22:59.133Z
2021-02-04T21:05:10.693Z
2021-02-04T21:05:10.693Z
peterh
<p>Do any of the planets have a Roche limit that is strong enough to be felt by an astronaut whilst in orbit?  </p>

CC BY-SA 4.0
2018-07-16T17:54:56.007Z
https://en.wikipedia.org/wiki/Neutron_Star_(short_story)
Carl Witthoft
CC BY-SA 4.0
2018-12-28T00:57:52.777Z
https://scifi.stackexchange.com/q/128375/51174
uhoh
<gravity><jupiter><roche-limit>
Is there any orbit at which the Roche limit can be "felt"?
Muze
# Is there any orbit at which the Roche limit can be "felt"?

(Asked by: username_0 on 2018-07-14)

Do any of the planets have a Roche limit that is strong enough to be felt by an astronaut whilst in orbit?

**Tags:** gravity, jupiter, roche-limit
> **Comments:**
> **username_1**: https://en.wikipedia.org/wiki/Neutron_Star_(short_story)  
> **username_2**: https://scifi.stackexchange.com/q/128375/51174  

---

## 3 answers

### Answer by username_7 on 2018-07-14. Score: 24 | Accepted answer

Roche limit happens where the gravity of the object, trying to pull the object together, becomes smaller than the tidal force (trying to pull the object apart).

But the astronaut is bound by not gravity, rather by the electromagnetic interaction between his/her atoms. The own gravity of the astronaut is negligible, compared to the electromagnetic interaction.

However, the tidal force affecting an astronaut, should require a little calculation. We can derive the formula of the gravitational acceleration around a point-like body ($F=\frac{GM}{r^2}$), we get

$$\frac{dF}{dr}=\frac{2GM}{r^3}$$

(We can ignore the sign on obvious reasons.)

Here $G$ is the gravitational constant, $M$ is the mass of the body, and $r$ is the distance.

Substituting the values of the Sun, we get $\frac{2\cdot 6.67 \cdot 10^{-11} \cdot 2 \cdot 10^{30}}{(7\cdot 10^8)^3} \approx 7.78\cdot 10^{-7} \frac{\mathrm{m/s^2}}{\mathrm{m}} \approx \underline{\underline{8 \cdot 10^{-8} \frac{g}{\mathrm{m}}}}$.

More clearly, if we are orbiting the Sun just above its surface, a roughly 2m long astronaut feel that his head and foot are pulled apart by around $1.6\cdot 10^{-7}g$ weight. In the case of a $70\:\mathrm{kg}$ astronaut, it is around the weight of $0.0112$ gram on the Earth.

The astronaut wouldn't feel it, but not very sensitive sensors could already measure it.

---

This calculation sometimes used $\mathrm{g}$ for "gram", as unit of mass, and $g$ as the (non-standard) unit of acceleration.

> **Comments:**
> **username_8**: Ignoring the obvious fact that any astronaut or instrument that close to the sun would be instantly vaporized of course...  
> **username_7**: @DarrelHoffman The Sun gives $\approx$ 6000K thermal radiation, which is hard but not chanceless to protect against it. I think some strong defense, for example well-polished tungsten mirrors, maybe combined with some cooling from behind, could handle it. The Parker Solar Probe will near the Sun to 8 solar radii.  

---

### Answer by username_3 on 2018-07-14. Score: 14

The Roche limit is where the tidal forces exerted on an orbiting object are sufficient to overcome the *self-gravity* of that object.

The "self-gravity" of an astronaut is tiny. We can estimate it as something like $$F\_{\rm grav} \sim \frac{Gm^2}{(h/2)^2},$$
where $m$ is the mass of the astronaut (+ equipment) and $h$ is their size (height). Assuming $m=100$ kg and $h=2$ m, then the self-gravity force is $2.7\times 10^{-6}$ N. This is a force that is too small to feel.

The tidal force on the astronaut a distance $R$ from a body of mass $M$ is approximately
$$ F\_{\rm tidal} \simeq 2 \frac{GMm}{R^3} h,$$
assuming they are pointing feet-first towards the Earth.

The Roche limit is where $F\_{\rm grav} < F\_{\rm tidal}$, so where
$$ \frac{Gm^2}{(h/2)^2}< 2 \frac{GMm}{R^3} h$$
$$ R < h\left(\frac{M}{2m}\right)^{1/3}$$
For example if $M=M\_{\rm Earth}$ and for the astronaut above, then $R < 60,000$ km for tidal break up. Which seems odd, because astronauts work quite happily in low-Earth orbit where the tidal forces are much stronger.

The problem with this calculation is that astronauts are not held together by self-gravity and a tidal field at the Roche limit has a negligible effect on a small body that is actually held together by atomic forces.

In order to experience a tidal field that can be felt on astronaut scales, let's say larger than 10 N (imagine hanging a 1 kg weight from your ankle on Earth), you would have to get much closer to the source of gravity.

Assuming a fixed mass equal to that of the Earth, we can work out you would need to get to within 500 km *of the centre of mass* in order to feel the tidal force. For the Earth (and also for other solar system bodies you could do the same calculation for) this would put you well *inside* the Earth, which isn't possible and in any case we could not assume that $M$ was fixed in that case, because it is the mass interior to $R$ that counts.

The only way that an astronaut could "feel" a tidal force would be to approach a compact star - a high density neutron star, white dwarf or black hole. There you can generate a very strong tidal field and, because they are compact, an astronaut could get close enough to feel it.


---

### Answer by username_4 on 2018-07-14. Score: 5

Expanding on Peterh's answer, we could try to find how should be an astronomical object for the tidal forces be felt by an astronaut orbiting it.

I don't have any reliable data on how strong need the tidal force to be felt. However, with a big simplification, we can very roughly model the upper and lower body of an astronaut as two masses placed about 1 meter apart. For a 70 kg astronaut under a tidal force of $0.1·g$ per meter (where $g$ is the acceleration of gravity), the difference of pull between those two 35 kg masses would be $0.1·35 kg = 3.5 kg$. Those force would stretch the astronaut's waist and would be clearly noticeably (maybe a force ten times weaker would be noticeable, too, but I'll will stick to $0.1m^{-1}·g$).

From Peterh's formulas:

$$r=\sqrt[3]{\frac{2·G·M}{0.1m^{-1}·g}}$$

For a 1 solar mass object:

$$r=\sqrt[3]{\frac{2·6.67·10^{-11}·2·10^{30}}{0.1·g}}=6481168 m = 6481 km$$

Then, an astronaut orbiting a sun-sized mass at a distance similar to Earth's radius would clearly feel tide forces when their head or feet point to the object. Of course the object would need to be a black hole or a neutron star to fit inside of the orbit.

With a more massive object the orbit could be larger, but given that mass is inside a cubic root, radius would grow very slowly.

> **Comments:**
> **username_5**: You don't need a black hole for this.  A neutron star is quite enough (typical mass: one solar mass, typical radius: 10 km).  
> **username_4**: @MartinBonner Thank you. Neutron stars added.  
> **username_6**: Might also refer to Niven's *Neutron Star*  
> **username_1**: @DJohnM ack you ninja'd me.   Sorry for posting my comment against the OP  
> **username_0**: @DJohnM  I don't understand the reference to the Niven's Neutron Star?  
> **username_2**: @DJohnM Stack Exchange has no [spoiler alert](https://meta.stackexchange.com/q/1191/303080) function for comments, perhaps you can link to Wikipedia as [CarlWitthoft did](https://astronomy.stackexchange.com/questions/26955/is-there-any-orbit-at-which-the-roche-limit-can-be-felt#comment46889_26955) instead of posting the surprise ending of the story here?  

---

