# How much mass does an object in space need to keep a human on its surface?

Assume there were a roughly spherical object in space like a meteorite or a comet.

If I weighed 80kg on Earth, how much mass would be required for an object in space so that I could stay on its surface without "flying away?" It's not necessary have the same gravity as Earth, but I am wondering what minimal mass the object would require for it to have meaningful gravity for someone to remain on the surface.

• This depends highly upon what the people are expected to be doing. To be clear, Earth is not 100% effective at keeping people from flying off of it. – Mitch Goshorn Apr 3 '15 at 16:07

Any object with mass (even you) has gravity. The mutual attractive force between two objects is given by the formula $$F = G \frac{M_1 M_2}{R_{12}^2},$$ where the two mass are $M_1$ and $M_2$ and $R_{12}$ is the separation of their centres of mass.

So to answer your question we need to define some sort of parameter that specifies what you mean by "meaningful gravity".

One way of doing this would be to demand that the forces due to tides from the Sun are greater than the gravitational force holding you to the body in question. Even here though we need to make an assumption about how far away the object is from the Sun - let it be $r$. Now let your mass be $m$, the mass of the body be $M$, the radius of the body be $R$, and the mass of the Sun be $M_{\odot}$.

To stay attached to the object when you are at the "subsolar point" - i.e. the body rotates so that you are closest to the Sun, then we require $$G \frac{M m}{R^2} > G \frac{m M_{\odot}}{(r-R)^2} - G\frac{m M_{\odot}}{r^2}$$

We can assume that $r \gg R$, so cancelling out $Gm$ and performing a binomial expansion on the middle term, keeping only the first two terms of the expansion: $$\frac{M}{R^{2}} > \frac{M_{\odot}}{r^{2}}(1 + \frac{2R}{r}) - \frac{M_{\odot}}{r^{2}}$$ $$\frac{M}{R^{2}} > 2M_{\odot}\frac{R}{r^3}$$ Thus this sets a limit on the density of the object in question $$\frac{3M}{4\pi R^3} = \rho > \frac{3}{2\pi} \frac{M_{\odot}}{r^3}$$

At $r=1\ au$, this means the density only needs to exceed $3 \times 10^{-4}$ kg/m$^3$, which will be satisfied by any solid body. NB: This limit would be a limit where you would actually be pulled from the surface by the Sun's tides.

A more stringent requirement might be to ensure that you can't jump off the surface. On Earth, an average person might be able to jump vertically about 50cm. Using the equations of uniform acceleration (SUVAT), we know $u^2 = 2g s$, where $g$ is the gravitational acceleration, $s$ is the height jumped and $u$ is the initial upward velocity. This tells us that you can jump upwards at about 3m/s. Assuming this is the same on any other body (its hard to say how well you could jump in much lower gravity) you could equate this to the escape velocity of the object, given by $v_{esc} = (2GM/R)^{1/2}$. Thus this gives a constraint of $M/R > u^2/2G$.

If we set a realistic density of $\rho = 5000$ kg/m$^{3}$ for an asteroid, we can replace $M$ with $4\pi R^{3} \rho/3$, to say that you would be able to jump off an asteroid if it were smaller than: $$R < u \left(\frac{3}{8\pi G \rho}\right)^{1/2}$$ For the numbers discussed this means $R< 1.8$ km and $M<1.2\times 10^{14}$ kg.

Incidentally, as you will have noticed, the result doesn't depend on your mass.

• quote ", this means the density only needs to exceed 3×10−4 m/s^2" you are using m/s^2 as a unit for Density ? I thought meters per seconds per seconds was acceleration, am I missing something here? – fahadash Feb 3 '15 at 0:50
• @fahadash Just a typo. – Rob Jeffries Feb 3 '15 at 0:52