According to this answer the surface gravity of Ceres is estimated to be only $0.27 m/s^2$. With a rotation period of 9 hours. The gravity seems light enough to overcome by leg muscle alone, and if you add in centrifugal force at the equator, it would seem even easier.

Could an average human in a NASA space suit jump off of Ceres? If not, could they achieve orbit by Jumping?

  • $\begingroup$ I was going to add the new reduced-gravity-sports tag, but interestingly all of your tags, even crewed-spaceflight actually apply nicely :) $\endgroup$
    – uhoh
    Oct 22, 2016 at 23:27
  • $\begingroup$ It would take a moderately competent rocket to get off Ceres. For example, the backpack maneuvering units (MMU) used for some EVAs in the Shuttle era has more than enough thrust, but not enough total Delta-v to get away from Ceres, and the smaller SAFER unit used nowadays has neither the thrust nor delta-v to get anywhere. $\endgroup$ Jan 6, 2022 at 16:37
  • $\begingroup$ This abstract is like...yeah but it suggests typical standing-jump speeds are 2.7 m/s. So for a C, S, and M class asteroid, that would get you off of non-rotating, spherical asteroids of diameter (respectively) 6.2 km, 4.4 km, and 3.1 km. $\endgroup$
    – Nick T
    Aug 11, 2022 at 19:05

7 Answers 7


Wikipedia gives $0.51 {km \over s}$ or $510 {m \over s}$ escape velocity, so, no, no leaving Ceres by jumping.

Following my earlier calculations, an asteroid of the radius of Ceres would have orbital speed at near-surface orbit of about $336 {m\over s}$, which is way beyond jump strength of anyone as well.

Gravitational acceleration on the Moon is $1.6249 {m\over s^2}$, barely 6 times more.

With 9h4m day long (32640s) and 3061km equator length, the rotation adds only about $94{m\over s}$ to orbital speed of whatever is standing there - $242 {m\over s}$ remain for orbital speed, so still quite a bit out of human's body reach.


FYI, the relevant equation is to set the kinetic energy equal to the gravitational potential energy. This is written on a per-mass basis, because both kinetic energy and gravitational potential energy is proportional to mass.

$$ \frac{ G M m }{ r } = \frac{ 1 }{ 2} m v_{\infty}^2 $$

For Ceres:

$$ v_{\infty} = \sqrt{ \frac{ 2 G M }{r} } = \sqrt{ \frac{ 2 G ( 9.43 \times 10^{20} ) }{4.87 \times 10^5 m} } = 508.2 \frac{m}{s} $$

So obviously this is too high for someone to accomplish with human capabilities. It's also interesting to note that this follows a different mathematical form than the surface gravity. You could jump nearly 37 times as high on Ceres than on Earth, assuming the initial mechanics are fairly similar. This still isn't sufficient because Ceres' gravitational sphere of influence extends much further than this.

Also, the orbital velocity is lower by a factor of $\sqrt{2}$. So to obtain orbit you only need to reach $360 m/s$. While this is still not obtainable, you have a better shot at indefinitely departing from the ground if you exert force to the side, as opposed to straight up.

Also, note that Ceres has a relatively fast rotation. I calculate the equatorial velocity to be $93.7 m/s$, which would help you a good deal. If you get on the equator and jump in the direction of rotation, then you're down to $265 m/s$ to obtain orbit. Again, this is still unobtainable but it's the best shot you've got.

  • $\begingroup$ Jumping into orbit does not mean indefinitely departing from the ground. If you manage to jump to orbit you will be in an orbit with the periapsis underground, probably far underground. $\endgroup$ Jan 7, 2022 at 2:47

You'd have to go a lot smaller to achieve escape velocity with musclepower alone. According to xkcd, you can escape from Deimos using a bike and a ramp...

  • 1
    $\begingroup$ Could you get enough trackion to get up to speed on Deimos? $\endgroup$ Nov 12, 2013 at 20:19
  • 3
    $\begingroup$ Oh I got it! Use one of the circles that they use in the circus, the faster you go the more force holds to the track and then you angle off to the jump. $\endgroup$ Nov 12, 2013 at 20:21
  • 1
    $\begingroup$ Related Can you ride a bicycle on Deimos? $\endgroup$ Jun 12, 2016 at 10:35

If you could power a "massless" propulsion method (like a bicycle that climbed out of the gravity field), you could easily escape Ceres, but this highlights the need for huge power thrust at the early stage, because propulsion needs mass that it can "throw backwards", and if you accelerate slowly, that means you have to carry all that mass with you for longer.


If Ceres had an earth like atmosphere, you should be able to fly pretty easily with a pair of wings taped to your arms and you just might gain enough velocity with hard flapping to fly your way off the planet. Probably not, but maybe.

I think manual powered flight would be one of the really fun aspects of low gravity living.

  • 2
    $\begingroup$ This would definitely not work. Low gravity => low atmospheric pressure, so even if there were vast amounts of breathable gas on Ceres, you could never use if for lifting. $\endgroup$ Sep 9, 2018 at 3:43
  • $\begingroup$ @FelixDombek You're probably right. It would be interesting to calculate how much gas it would take to give Ceres 1 atm. There's no way that it works in reality. The atmosphere would just fly off the planet. $\endgroup$
    – userLTK
    Sep 9, 2018 at 15:02
  • $\begingroup$ Do it indoors then $\endgroup$
    – Innovine
    Jan 6, 2022 at 9:25
  • $\begingroup$ @userLTK Yup--the escape velocity is similar to the average particle speed in Earth's atmosphere. It departs rapidly. $\endgroup$ Jan 7, 2022 at 2:49
  • $\begingroup$ @FelixDombek No. It's possible to have low gravity and high atmospheric pressure. You need a very fluffy central mass and then just pile up enough atmosphere. Consider a Dyson Sphere the size of Earth's orbit. The surface gravity is milligees, but it's quite capable of holding an Earthlike atmosphere. Strap on your wings and enjoy. $\endgroup$ Jan 7, 2022 at 2:52

Surface gravity gives no information about escape velocity. Saturn for example has a surface gravity comparable to Earth, but an escape velocity three times higher. Similarly, if the Moon had a 10% smaller radius than its current one but was made entirely of iridium, it would have exactly the same surface gravity as Earth but half of Earth's escape velocity.

You can easily calculate the escape velocity of a body knowing its radius and its density, with

$$V_{e}=\sqrt{2G\frac{4}{3}\pi\rho r^{2}}$$

where $\rho$ is the density and is $r$ the radius.

NBA players can jump up to $1\,\mathrm{m}$ high. Knowing that jumping as high as $h$ corresponds to leaving the ground with a velocity equal to

$$V_j=\sqrt{2 h g}$$

we obtain an initial speed of about $4.3\,\mathrm{m/s}$ for the best jumpers.

For having such an escape velocity an asteroid made of stone would need to be no larger than $4\,\mathrm{km}$. That is about less than $1\%$ of Ceres' radius.


Folks here are assuming that a human cannot run 265 m/s. But it seems to me that such a speed would be attainable on Ceres, especially if you allow me to build a track and ride a bicycle. On Earth humans can run 10 m/s, but they must contend with much stronger gravity and air resistance. The fastest cyclists reach 37 m/s. Just how fast could you go on Ceres, if you minimized frictional losses?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.