<p>Of course you would need to specify who the person is - an Olympic athlete? Let us assume so and then you can scale downwards accordingly.</p>

<p>So an Olympic high jumper can jump hard enough to raise their centre of gravity about 2m off the ground. </p>

<p>Let us assume this is a ballistic problem. The athlete actually gives themselves sufficient upward speed to get their centre of gravity from about 1m to 2m in the Earth's gravitational field. Using the usual equations for uniform acceleration, the required initial velocity is <span class="math-container">$v \simeq \sqrt{2gh} = 4.4$</span> m/s.</p>

<p>Now let us assume that the athlete could deliver something similar on an asteroid. This is doubtful, because getting a good run up, while wearing a spacesuit, is probably not going to happen. But if it were possible, then we just equate the jump velocity to the escape speed 
<span class="math-container">$$ v = \sqrt{\frac{2GM}{R}},$$</span>
where <span class="math-container">$M$</span> and <span class="math-container">$R$</span> are the mass and the radius of the (assumed spherical) asteroid. </p>

<p>Thus we do not have separate constraints on the mass and radius of the asteroid, only on their ratio. If
<span class="math-container">$$ \frac{M}{R} &gt; \frac{v^2}{2G},$$</span>
then even an Olympic athlete couldn't jump into space.</p>

<p>To get something more definite we would need to assume a density, <span class="math-container">$\rho$</span>, for the asteroid. This depends on <a href="https://en.wikipedia.org/wiki/Standard_asteroid_physical_characteristics#Density" rel="noreferrer">what type</a> of asteroid you are talking about, but could be between 1500 and 5000 kg/m<span class="math-container">$^3$</span> (<a href="https://arxiv.org/abs/1203.4336" rel="noreferrer">Carry 2012</a>).</p>

<p>If we assume (spherical asteroid again) that <span class="math-container">$M = 4\pi R^3 \rho/3$</span> and substituting this for the mass, we get a constraint on the asteroid radius such that someone is trapped if:
<span class="math-container">$$ R &gt; 4.2 \left(\frac{v}{4.4\ {\rm m/s}}\right) \left(\frac{\rho}{2000\ {\rm kg/m}^3}\right)^{-1/2}\ {\rm km}$$</span></p>

<p>You can mess about with this and assume a different take-off speed (you might want to argue, as MarkP does, that you could work up a larger horizontal speed by just running fast - but I doubt that in a "milligravity environment" - the surface gravity is of order 0.002 m/s<span class="math-container">$^2$</span>) or different asteroid densities to modify the answer (trans-Neptunian objects or comets have densities lower than 1000 kg/m<span class="math-container">$^3$</span>). Or you could come up with an equivalent constraint on the mass by substituting for <span class="math-container">$R$</span> in terms of the mass and density instead.</p>

CC BY-SA 4.0
2019-10-28T08:21:05.583Z
@DonHatch I see it fine, so unless this is only an issue with this answer, then I wouldn't use a beta version.
ProfRob
CC BY-SA 4.0
2019-10-28T14:24:47.937Z
It seems odd that I can more easily jump off an asteroid of fixed density if the radius is *big* but not if it is small.  Perhaps you mean $v > \sqrt{\frac{2 G M}{R}}$, which reverses the inequalities in the last two displays (both of which render fine in Firefox 69.0 on Linux).
Eric Towers
CC BY-SA 4.0
2019-10-28T14:31:36.743Z
@EricTowers a big asteroid has more mass. The mass increases as $R^3$, so the surface gravity increases with $R$. The question asks for a limit to trap someone, so it is if $R$ is bigger than some value.
ProfRob
CC BY-SA 4.0
2019-10-28T16:06:05.897Z
A specific numeric answer would be helpful. Assuming (whatever), it seems no one is likely to jump off an asteroid with radius n km weighing m kg. By the way, using math renderer comon html is a workaround for the Mathml bug (using Brave browser Version 0.70.121 Chromium: 78.0.3904.70 (Official Build) (64-bit)).
Bit Chaser
CC BY-SA 4.0
2019-10-28T16:59:30.793Z
@bitchaser There *is* a specific answer. If $\rho=2000$ kg/m$^3$ and the initial $v=4.4$ m/s, then the answer is $R>4.2$ km. Just plug your own assumptions into the last equation if you want a different answer; those are mine. I might start a meta question about the browser issue - I don't know how to implement your workaround.
ProfRob
CC BY-SA 4.0
2019-10-28T19:25:18.403Z
Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/100388/discussion-on-answer-by-rob-jeffries-gravitational-properties-of-asteroids).
called2voyage
CC BY-SA 4.0
2020-12-22T20:08:28.167Z
Upvoted because of nice form to present the result: a length quantity multiplied by two dimensionless factors.
Incnis Mrsi
CC BY-SA 4.0
2019-10-27T21:10:42.417Z
2019-10-28T14:33:54.153Z
2019-10-28T14:33:54.153Z
ProfRob
<p>Man, maybe it's just late at night, but I can't make out if Rob is giving us a straight answer here, or what it is if he is.</p>

<p>I can maybe help a little though; the density of most smaller asteroids, any time I've seen a measured or reasonably well estimated figure, has been rather less than water ice. They're essentially uncollapsed, porous, icy rubble piles, so their gross density is somewhere in the region of 0.8 to 0.9g/cc (or 800 to 900 kg/cu.m). This both indicates, and causes, their overall gravity to be quite low. I've seen it stated for at least one rock of some kilometres radius that someone bounding around it in a spacesuit would have to be very careful about their speed of travel and force of their jumps, as it would be entirely possible for a human to reach escape velocity under muscle power.</p>

<p>The bigger issue of course, on a loosely compacted pile of dirty slush is getting any kind of purchase for the push-off. You'll just end up foundering, like trying to climb a loose sand dune or scramble up a muddy hill in the rain. So the first job will be to erect some kind of solid platform.</p>

<p>Once you've done that, plug that density (about 850, rather than 2000 kg/m^3) into Rob's equation and see what you get. Also it may be possible to increase the starting velocity some; it's counterproductive to try and jump straight up off the planet, when you could take a running jump at it. We don't launch our rockets straight up, they leave on an expanding spiral trajectory - it works rather better, both for reaching orbit and exceeding it, to accelerate alongside, rather than directly up from the Earth. The initial straight-up phase is mostly to get out of the humongous drag of the lower atmosphere (which is a severe barrier to reaching the required ~17500mph) as rapidly as possible, which won't be a concern on a small asteroid.</p>

<p>(then again, 4.4m/s is about 10mph, which is about what I was going to suggest as a running speed; however, if you then ADD a jump to that, which will be rather more effective than on Earth, you can add to that; if we say there's a forward and an upward vector of 4.4m/s each, the resulting 45 degree vector is more like 6.2m/s)</p>

CC BY-SA 4.0
2019-10-28T08:17:09.327Z
Two things. You can't run when there is only 0.002 m/s$^2$ of gravity. Second, I think you are referring to things like comets and trans-Neptunian objects, rather than asteroids, with those densities. I have added a reference to an authoritative review to my answer.
ProfRob
CC BY-SA 4.0
2019-10-28T08:28:22.310Z
Also the rocket argument isn't really relevant. Rockets have tangential acceleration when they are trying to put something in orbit. If they are just trying to escape (ballistically), then the direction doesn't matter (it's just conservation of energy).
ProfRob
CC BY-SA 4.0
2019-10-29T10:26:13.223Z
To run in small gravity is just a series of long bounces. As long as you can move your leg fast enough and keep your attitude, you'll gain some forward momentum on every bounce. Keeping attitude is the difficult part, but I think practice could overcome that. But still that will only get you up to orbital speed, not the escape speed.
jpa
CC BY-SA 4.0
2019-10-29T11:08:27.433Z
As an aside, 4.4 m/s is a lousy running speed. 4.4 m/s is a lousy speed for a 10K race, let alone a 100 meter dash. Olympic quality athletes have been running sub 10 second 100 meter dashes (> 10 m/s) for over 50 years. We humans evolved to run prey down rather than plucking apples from a tree by jumping straight up.
David Hammen
CC BY-SA 4.0
2019-10-29T23:36:30.133Z
@DavidHammen - wearing a space suit?? I don't think so - watch the Moon landings. 4.4 m/s is a vertical take-off speed in my answer; I still maintain that is probably the only feasible way to try and escape a small asteroid. The suggestion of large bounces made above is just not possible. You would be spending minutes away from the surface between bounces and would be unable to control rotation. - unless you had a jet pack, in which case why don't you use that to escape?
ProfRob
CC BY-SA 4.0
2019-10-29T23:39:48.227Z
Maybe a strategy would be a 2-stage process. Make one lunge in a "forward" direction and then when you come down, aim for a big leap upwards in the style of a triple jumper (providing you hit the surface feet-first).
ProfRob
CC BY-SA 4.0
2019-10-28T05:23:35.740Z
2019-10-28T05:23:35.740Z
MarkP
<p>How large would an asteroid have to be in order to hold a person so that the person could not escape?</p>

CC BY-SA 4.0
2019-10-28T18:00:05.950Z
Larger than the one a certain Prince who loves sheep lives on.
Carl Witthoft
CC BY-SA 4.0
2019-10-29T13:45:13.830Z
@CarlWitthoft Would I be correct in thinking you don't mean [this prince](https://www.mirror.co.uk/news/world-news/prince-charles-visits-sheep-farm-1426329)?
Andrew Morton
CC BY-SA 4.0
2019-10-29T16:22:12.143Z
Well, that would depend on **many** things, especially method of escape... This question should be much more specific.
Mithoron
CC BY-SA 4.0
2019-10-29T19:16:59.497Z
I’m sure there’s an xkcd reference on this prince but I can’t find it.
Max0815
CC BY-SA 4.0
2019-10-31T22:43:33.813Z
@Max0815 https://what-if.xkcd.com/68/
Gerold Broser
CC BY-SA 4.0
2019-11-03T18:16:22.527Z
Is the person an Olympic high-jumper, or just an average Joe?
John Canon
<gravity><astrophysics><asteroids>
Gravitational Properties of Asteroids
RUDY NUNEZ
# Gravitational Properties of Asteroids

(Asked by: username_0 on 2019-10-27)

How large would an asteroid have to be in order to hold a person so that the person could not escape?

**Tags:** gravity, astrophysics, asteroids

> **Comments:**
> **username_1**: Larger than the one a certain Prince who loves sheep lives on.  
> **username_2**: @CarlWitthoft Would I be correct in thinking you don't mean [this prince](https://www.mirror.co.uk/news/world-news/prince-charles-visits-sheep-farm-1426329)?  
> **username_3**: Well, that would depend on **many** things, especially method of escape... This question should be much more specific.  
> **username_4**: I’m sure there’s an xkcd reference on this prince but I can’t find it.  
> **username_5**: @username_4 https://what-if.xkcd.com/68/  
> **username_6**: Is the person an Olympic high-jumper, or just an average Joe?  

---

## 2 answers

### Answer by username_7 on 2019-10-27. Score: 20

Of course you would need to specify who the person is - an Olympic athlete? Let us assume so and then you can scale downwards accordingly.

So an Olympic high jumper can jump hard enough to raise their centre of gravity about 2m off the ground.

Let us assume this is a ballistic problem. The athlete actually gives themselves sufficient upward speed to get their centre of gravity from about 1m to 2m in the Earth's gravitational field. Using the usual equations for uniform acceleration, the required initial velocity is $v \simeq \sqrt{2gh} = 4.4$ m/s.

Now let us assume that the athlete could deliver something similar on an asteroid. This is doubtful, because getting a good run up, while wearing a spacesuit, is probably not going to happen. But if it were possible, then we just equate the jump velocity to the escape speed
$$ v = \sqrt{\frac{2GM}{R}},$$
where $M$ and $R$ are the mass and the radius of the (assumed spherical) asteroid.

Thus we do not have separate constraints on the mass and radius of the asteroid, only on their ratio. If
$$ \frac{M}{R} > \frac{v^2}{2G},$$
then even an Olympic athlete couldn't jump into space.

To get something more definite we would need to assume a density, $\rho$, for the asteroid. This depends on [what type](https://en.wikipedia.org/wiki/Standard_asteroid_physical_characteristics#Density) of asteroid you are talking about, but could be between 1500 and 5000 kg/m$^3$ ([Carry 2012](https://arxiv.org/abs/1203.4336)).

If we assume (spherical asteroid again) that $M = 4\pi R^3 \rho/3$ and substituting this for the mass, we get a constraint on the asteroid radius such that someone is trapped if:
$$ R > 4.2 \left(\frac{v}{4.4\ {\rm m/s}}\right) \left(\frac{\rho}{2000\ {\rm kg/m}^3}\right)^{-1/2}\ {\rm km}$$

You can mess about with this and assume a different take-off speed (you might want to argue, as username_12 does, that you could work up a larger horizontal speed by just running fast - but I doubt that in a "milligravity environment" - the surface gravity is of order 0.002 m/s$^2$) or different asteroid densities to modify the answer (trans-Neptunian objects or comets have densities lower than 1000 kg/m$^3$). Or you could come up with an equivalent constraint on the mass by substituting for $R$ in terms of the mass and density instead.

> **Comments:**
> **username_7**: @DonHatch I see it fine, so unless this is only an issue with this answer, then I wouldn't use a beta version.  
> **username_8**: It seems odd that I can more easily jump off an asteroid of fixed density if the radius is *big* but not if it is small.  Perhaps you mean $v > \sqrt{\frac{2 G M}{R}}$, which reverses the inequalities in the last two displays (both of which render fine in Firefox 69.0 on Linux).  
> **username_7**: @EricTowers a big asteroid has more mass. The mass increases as $R^3$, so the surface gravity increases with $R$. The question asks for a limit to trap someone, so it is if $R$ is bigger than some value.  
> **username_9**: A specific numeric answer would be helpful. Assuming (whatever), it seems no one is likely to jump off an asteroid with radius n km weighing m kg. By the way, using math renderer comon html is a workaround for the Mathml bug (using Brave browser Version 0.70.121 Chromium: 78.0.3904.70 (Official Build) (64-bit)).  
> **username_7**: @bitchaser There *is* a specific answer. If $\rho=2000$ kg/m$^3$ and the initial $v=4.4$ m/s, then the answer is $R>4.2$ km. Just plug your own assumptions into the last equation if you want a different answer; those are mine. I might start a meta question about the browser issue - I don't know how to implement your workaround.  
> **username_10**: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/100388/discussion-on-answer-by-rob-jeffries-gravitational-properties-of-asteroids).  
> **username_11**: Upvoted because of nice form to present the result: a length quantity multiplied by two dimensionless factors.  

---

### Answer by username_12 on 2019-10-28. Score: 3

Man, maybe it's just late at night, but I can't make out if Rob is giving us a straight answer here, or what it is if he is.

I can maybe help a little though; the density of most smaller asteroids, any time I've seen a measured or reasonably well estimated figure, has been rather less than water ice. They're essentially uncollapsed, porous, icy rubble piles, so their gross density is somewhere in the region of 0.8 to 0.9g/cc (or 800 to 900 kg/cu.m). This both indicates, and causes, their overall gravity to be quite low. I've seen it stated for at least one rock of some kilometres radius that someone bounding around it in a spacesuit would have to be very careful about their speed of travel and force of their jumps, as it would be entirely possible for a human to reach escape velocity under muscle power.

The bigger issue of course, on a loosely compacted pile of dirty slush is getting any kind of purchase for the push-off. You'll just end up foundering, like trying to climb a loose sand dune or scramble up a muddy hill in the rain. So the first job will be to erect some kind of solid platform.

Once you've done that, plug that density (about 850, rather than 2000 kg/m^3) into Rob's equation and see what you get. Also it may be possible to increase the starting velocity some; it's counterproductive to try and jump straight up off the planet, when you could take a running jump at it. We don't launch our rockets straight up, they leave on an expanding spiral trajectory - it works rather better, both for reaching orbit and exceeding it, to accelerate alongside, rather than directly up from the Earth. The initial straight-up phase is mostly to get out of the humongous drag of the lower atmosphere (which is a severe barrier to reaching the required ~17500mph) as rapidly as possible, which won't be a concern on a small asteroid.

(then again, 4.4m/s is about 10mph, which is about what I was going to suggest as a running speed; however, if you then ADD a jump to that, which will be rather more effective than on Earth, you can add to that; if we say there's a forward and an upward vector of 4.4m/s each, the resulting 45 degree vector is more like 6.2m/s)

> **Comments:**
> **username_7**: Two things. You can't run when there is only 0.002 m/s$^2$ of gravity. Second, I think you are referring to things like comets and trans-Neptunian objects, rather than asteroids, with those densities. I have added a reference to an authoritative review to my answer.  
> **username_7**: Also the rocket argument isn't really relevant. Rockets have tangential acceleration when they are trying to put something in orbit. If they are just trying to escape (ballistically), then the direction doesn't matter (it's just conservation of energy).  
> **username_13**: To run in small gravity is just a series of long bounces. As long as you can move your leg fast enough and keep your attitude, you'll gain some forward momentum on every bounce. Keeping attitude is the difficult part, but I think practice could overcome that. But still that will only get you up to orbital speed, not the escape speed.  
> **username_14**: As an aside, 4.4 m/s is a lousy running speed. 4.4 m/s is a lousy speed for a 10K race, let alone a 100 meter dash. Olympic quality athletes have been running sub 10 second 100 meter dashes (> 10 m/s) for over 50 years. We humans evolved to run prey down rather than plucking apples from a tree by jumping straight up.  
> **username_7**: @DavidHammen - wearing a space suit?? I don't think so - watch the Moon landings. 4.4 m/s is a vertical take-off speed in my answer; I still maintain that is probably the only feasible way to try and escape a small asteroid. The suggestion of large bounces made above is just not possible. You would be spending minutes away from the surface between bounces and would be unable to control rotation. - unless you had a jet pack, in which case why don't you use that to escape?  
> **username_7**: Maybe a strategy would be a 2-stage process. Make one lunge in a "forward" direction and then when you come down, aim for a big leap upwards in the style of a triple jumper (providing you hit the surface feet-first).  

---

