On the Earth you can walk - of course we all know it - in the southern hemisphere without falling out in the open space, and you feel no difference between being at the North Pole or on the Equator.

Now suppose a planet with the same density of the Earth.

What would be the minimum size of this planet which would allow a human being to walk around without falling out in the open space?

If we could change the planet size and make it a little bigger or smaller than this hypothetical size threshold, what would be the feeling to walk around the Equator in such a planet?

One meter wider would mean to feel head-up / feet down? A small jump would be enough to fall in the open space?

One meter smaller, the opposite?

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    $\begingroup$ A reasonable interpretation of this question would be "what is the minimum size planet where you couldn't accidentally reach escape velocity while walking." But, with that level of gravity, actually walking would be difficult. $\endgroup$ Nov 3, 2015 at 3:14

3 Answers 3


Lets just start with the cosmic velocity assuming you dont want to escape fully.

$v \le \sqrt{\frac{2GM}{r}}$, where M is the mass of the planet and r is the distance to the center of Mass. We assume a spherical planet. We know the average density $\rho$ of our planet earth and since you will probably want to live on the new planet we will assume it has the same density. We can then replace M in our previous formula by $\rho \frac{4}{3}\pi r^3$ Plug in and solve for r to get

$r \ge \sqrt{\frac{3v^2}{8 G \rho}}$/ Plug in the numbers. $\rho=5515kg/m^3$ is the average density of our planet earth and the gravitational constant. Lets assume you play Basketball like I do, to dunk I would need to jump off with a velocity of $4m/s$ according to my dunk calculator. This is the maximum velocity you will jump with when you play basketball on the new planet.

Thus we get

$ r \ge 6446.4m$.

This was fun just doing random assumptions and calculations :D I didnt get to specific, but still this should help you with your homework.

For the walking part I would just assume your average walking speed should be smaller than the first cosmic velocity or you will just fall around the planet. Start there and backtrack the way I did above.

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    $\begingroup$ Thank you for your explanation. It's not a homework, it's my pure curiosity. :) I admit I made some basic mistakes talking about "falling into open space". $\endgroup$ Nov 3, 2015 at 18:26
  • $\begingroup$ So did I answer your question? If I did, please accept it; if not, please let us know what other information you require. $\endgroup$
    – hdhondt
    Nov 4, 2015 at 2:54

The problem is in your assumption that "we fall into open space" unless the planet is large enough.

Even if there were no planet at all, we would not fall. In open space you just stay where you are - unless you are affected by some star or planet. In other words you will always drift slowly towards something or other.

Now the earth's gravity is so large that we need to move at $11km/s$ (or $40,000km/h$) in order to escape from its gravity. To escape from the sun, your velocity would need to be $618 km/s$. As the body gets smaller, the escape velocity gets smaller too - but it never gets to $0$. That means that, if you are standing on a small asteroid, you may be able to escape it by jumping. On an even smaller one you might be able to push yourself away with your little pinkie - but you always need some velocity to escape. Using the equations qacwnfq q provided you can work out the size of the asteroid needed to stop you from escaping at walking speed (about $5km/h$ or $1.4m/s$). The radius turns out to about $1.4km$.

But you NEVER just "fall off".

  • $\begingroup$ I think you have a typo in your units. Shouldn't km/s be km/hr or kph? $\endgroup$ Nov 4, 2015 at 14:11
  • $\begingroup$ You missed the second one, i.e., "about $5km/s$ or $1.4m/s$" should be "about $5km/h$ or $1.4m/s$". $\endgroup$ Nov 6, 2015 at 12:14
  • $\begingroup$ @hdhondt So on a small asteroid with almost no gravity, if no other planet is influencing with its gravity, you don't feel any difference between being with your head up or upside down? This feeling is in some way linked with the gravity? $\endgroup$ Nov 7, 2015 at 1:25
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    $\begingroup$ On a tiny asteroid, gravity is extremely small, and hence you may not feel any different standing on your head. On earth, gravity causes you to weigh, say, 80kg. On a minuscule asteroid, you might weigh less than a gram, and you would simply not feel it. Your mass would still be 80kg, but your weight is caused by gravity and hence changes depending on where you are. On the moon an 80kg person would only weigh about 13kg. $\endgroup$
    – hdhondt
    Nov 7, 2015 at 9:54
  • $\begingroup$ Fixed that one too, honeste. Just shows I'd make a lousy proofreader $\endgroup$
    – hdhondt
    Nov 8, 2015 at 6:51

The Earth has an average density of about 5500 kg/m^3. For a small planet, the density would be pretty similar throughout. Therefore the force of gravity you would experience on a planet with Earth density is: Your Mass * 5500 * 4/3pi * r^3 / r^2 * 6.6723 * 10^-11

This is equal to about 1.456 * 10^-6 * Your Mass * Radius of Small Planet. (I know, if you check the multiplication, it doesn't work out, that is because the simplified equation I wrote down is more accurate)

This is also assuming all of your mass is concentrated on your feet, this is not true, but the overall equations remain the same. We could try to integrate the force on every cross-section of your body combined with the cross-sections density. This statement shows how complicated it would be to accurately find the force. Instead we could just assume your mass is concentrated halfway up your body.

So our new equation is: 1.456 * 10^-6 * Your Mass * Radius of Mini Planet Cubed / (Radius of Mini Planet + Half of Your Height) Squared

An excellent article on what it would feel like to walk around, stand on, and jump on a mini planet is found at the following link:


When the radius was increased, you would feel stronger gravity, but less tidal forces. When the radius was decreased, you would feel weaker gravity, but more tidal forces. Although the tidal forces involved may be small, it would feel as though you were laying on a ball, or laying with your head near the center of a merry-go-round.

Escape Velocity for a planet is found with the formula sqrt(2GM/r). An average human can jump at a velocity of .153 m/s upward. So to jump off of a mini planet with Earth density, the planet would have to be smaller than 87.33 m in radius. An interesting point noted in the article is that an object will escape the pull of a planet if the object is moving at escape velocity in any direction, provided that it is not moving at the planet. An average human can run at 5.6 m/s. So a mini-planet would have to be smaller than 3193.81 m in radius to run off of it. To just walk off a mini planet, it would have to be smaller than 798.45 m in radius.

  • $\begingroup$ I think I messed up the average humans jump velocity in my calculations. I'm also sorry I don't know how to format equations. $\endgroup$ Nov 3, 2015 at 14:46
  • $\begingroup$ I gave you +1 for the very interesting link. Thank you! $\endgroup$ Nov 3, 2015 at 18:33

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