How does one calculate the gravitational field of an irregularly shaped object? I was thinking specifically of the gravitational field due to a human, and tried to see if modeling a human as an infinite wire would work, and of course it's a horrible approximation*.

Is there anyway to do it without relying on a computer, or is numerical approximation the only option? And even with a computer, what process does it actually do to get an answer? I would assume that it uses superposition with $\vec{F}= \frac{GMm}{ \vec{r}^{2}} \hat{r}$, which would work for an irregularly body acting on a point particle, but how would you calculator the field of two irregularly shaped bodies acting on each other?

  • I used Gauss's law for Gravitation, and ended up with a distance of 289 Å, which would make the acceleration equal to $g$, which sounds really wrong, considering that the distance between atoms is only on the order of a few Å.
  • $\begingroup$ "..tried to see if modeling a human as an infinite wire would work, and of course it's a horrible approximation" -- I would suggest modeling a human using the 'spherical cow approximation'... Kidding aside, surely you can model a human as a short stack of spherical masses? Renders the gravitational computation trivial. $\endgroup$ – Johannes Jun 29 '13 at 12:18
  • $\begingroup$ I don't see where you specify the distance from the object at which you would like to know the force. After all, far enough away we are all monopoles. Nor the accuracy needed. Without these 2 facts there is no answer. $\endgroup$ – user27777 Aug 15 '13 at 23:17

Firstly, I tried to reproduce your thin wire calculation : if you assume a wire of mass 60kg with a length of 2 metres, then I reckon the magnitude radially symmetric field near the middle of the wire from your Gauss law technique at $\frac{2\,G\,M}{\ell\,r}$ with $m = 60\mathrm{kg}$ and $\ell = 2\mathrm{m}$, which works out for me to give $r = 4\times10^{-10}\mathrm{m}$ to get the same attraction as at the Earth's surface with $G = 6.7\times10^{-11}\mathrm{m}^3\mathrm{kg}^{-1}\mathrm{s}^{-2}$. Gravitation is $very$ weak in comparison with other everyday forces - think of the classic trick of picking a nail up with a magnet: the currents in the tiny magnet are overwhelming the gravitational attraction of the $whole\,Earth$. Your wire model will be more accurate than you might think (see a few paragraphs below).

Really, for irregular shapes, numerical methods are the only way. And, to calculate the gravitational field at any point, they work pretty much as you would think, either summing up the fields from finite element masses calculated individually with the $\frac{G\,\Delta M}{r} \hat{\mathbf{r}}$ formula, or by solving the Poisson equation $\nabla^2 \phi = 4\,\pi\,G\,\rho$, where $\rho$ is the mass density and $\phi$ the gravitational potential. There are highly developed numerical algorithms for the Poisson equation, and then one just works out $\nabla \phi$ to get the field.

To get the force between bodies is a little less straightforward, as you seem to realise in your question, because the field is the force on a small test particle ($i.e.$ one too small to disturb the outside field). What you need to do is sum up the forces on all the point masses making up one body exerted by the point masses exerted by the other body. You do $\mathbf{not}$ include the force exerted by points in the same body on the other points in the same body. Thin of it this way. Suppose we have a human (human 1)begetting a gravitational field: then we bring in the first piont mass making up the other human (human 2): we calculate the force on the test mass. Then we bring in the second point mass making up human 2. The total force on this one is of course that owing to the human 1 $plus$ that from the first test mass in human 2. However, this latter force is balanced somehow by an internal pressure force within human 2. Otherwise human 2 would gravitationally collapse! So the nett force on the second point mass considered as part of human 2 is just the sum of the forces arising from human 1.

I would approximate two humans as long cylinders, not wires. Your Gauss technique will work just as well for cylinders in isolation (inside a long cylinder the gravitational field rises linearly with distance from the centre, inversely with radius outside). Then calcalate the forces by an integration keeping in mind the last paragraph. You won't be far wrong.

Another, beautiful way to get a good working idea of solutions to Laplace's equation (which is what you are doing outside the body) if you don't want to use a computer is the graphical technique of curvillinear squares. See http://www.ie.itcr.ac.cr/acotoc/Maestria_en_Computacion/Sistemas_de_Comunicacion_II/Material/Biblio2/chapt06.pdf.

  • $\begingroup$ I tried to do it using cylinders, but I ended up with a strange expression. $$ \oint \vec{g} \cdot d \vec{A} = \int _{V} \nabla \cdot \vec{g} dV$$ $$2g \pi r L = -4 \pi G (\rho \pi R^{2} L) \rightarrow g = \frac{-2G \rho R^{2}}{r}$$ When I did this with an infinite line, I had done gotten an answer of $ \frac{2G \lambda}{r}$, how can you cancel out the infinity, if you don't consider $M=\lambda L$ and $M= \rho A$? $\endgroup$ – Astrum Jun 29 '13 at 6:48
  • $\begingroup$ This is correct: what you have is the expression outside the cylinder. Now do it again inside the cylinder: inside the cylinder your big $R$ is the same as little $r$, so we get $g = -2 G \rho r$ and your infinity is gone! Also, you keep the infinite line expression for the finite cylinder - this assumes you're calculating fields at radiusses which are much less than the cylinder's length - so it's only an approximation. If you have two rods side by side, the force on the central parts of the rods will be good, but the approximation coarsens near the ends. $\endgroup$ – WetSavannaAnimal Jun 29 '13 at 8:20
  • $\begingroup$ But if you take r to equal R, that would mean the the radius is at the surface of the solid cylinder that we're trying to find the gravitational field of. I thought that the idea was to find the radius r, at which the gravitational field would be the same as earths. And this also means that the gravitational field is not dependent on its mass, but rather its mass density? If I understand what you're saying, we can estimate the field at a point r, by using this: $g= \frac{-2GM}{rL}$, and using r as the distance from the cylinder (human) and L as the height of the cylinder (human)? $\endgroup$ – Astrum Jun 30 '13 at 1:50
  • $\begingroup$ What I mean by $r$ equal to $R$ is that you're putting your cylindrical integration volume inside the cylinder to work out the field inside the cylinder - given the axisymmetry, the field depends only on the mass inside the cylinder and is uninfluenced by that outside. So at a radius $r < R$, the field varies linearly as $g = -2 G \rho r$ from nought (at the centre) to its maximum $g = -2 G \rho R$ at the cylinder's surface. Then outside the cylinder, the inverse relationship holds: $g = \frac{- 2 G \rho R^2}{r}$ note that the two expressions are equal at the surface. $\endgroup$ – WetSavannaAnimal Jun 30 '13 at 12:30
  • $\begingroup$ But this comment was just to show that there are no problems with infinities inside the cylinder. What you'll need to do is use the formula outside the cylinder to work out the force on the other cylinder. You can't get the cylinders anywhere near enough to one another to reach an attraction of $g$. $\endgroup$ – WetSavannaAnimal Jun 30 '13 at 13:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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