# Why can large objects at greater distance be treated as a point particle?

Why can large objects at greater distance be treated as a point particle?

"The bodies of our solar system are so far apart compared with their diameters that they can be treated as particles to an excellent approximation."

was the statement written in a textbook.

It got me thinking whether apart from mere visualization of these giant bodies as point particles due to the great distance between them, is there any explanation about why they can be treated as point particles (to an excellent approximation).

How does the large distance between the planets influence the way the planets or any other objects are treated?

• Maybe I'll write a full answer at some point (or someone else can), but the key is something called the multipole expansion. You can write the gravitational field of a complicated body as a sum of terms, each of which decreases with a higher power of distance. So at large distances only the lowest order term matters (since it decreases the slowest) and that term is just the gravitational field of a point mass. – octonion Jan 1 '18 at 2:42
• Could you explain it in detail? – suiz Jan 1 '18 at 2:44

## 2 Answers

It's a little hard for me to answer this question because I see the answer as obvious - if I lose you somewhere please say so.

These "giant bodies" can be treated as point particles because from a great distance away, the bodies can be treated as points without losing much accuracy. Let's take the gravitational attraction between you and a person as an example. If you're standing next to the person, the gravitational attraction you feel is very complex since the mass of the person isn't uniformly distributed. If I told you you're allowed to approximate the person as a uniform sphere (the easiest body to model, since by the shell theorem it can be treated as though it's a point mass in the center of the sphere), then suddenly the calculation is very easy. But if you aren't allowed to do that, it's very complex - you'd have to calculate the attraction from the person's hands, legs, etc, and neglecting any of these can lead to substantial errors - say, of 10%.

Now let's assume the person is on the moon. In that case, doing the very complex calculation is effectively pointless, since even if you approximate the person as a sphere, you'll get almost exactly the same answer. You can try it with some numbers: let's say the person is only made of two hands of mass $m$ (a simplifying assumption) and that one hand is 3 meters away while the other is 2 meters. Then:

$F_{far}=GMm/9; F_{close}=GMm/4; F_{total} = F_{far} + F_{close}$

where $M$ is your mass. Now if the person is on the moon standing at the same angle, then:

$F_{far}=GMm/(384,400,001)^2; F_{close}=GMm/(384,400,000)^2$

Comparatively, you could also approximate the person as a single object of mass $2m$ located at an in-between distance. Then:

$F_{point}=2GMm/6.25$ if you're close to each other, and $F_{point}=2GMm/(380,400,000.5)^2$ if the person's on the moon.

You can put that into a calculator and see how good the approximation is. You should find that on the scales of the moon, it's fine for almost all purposes to just treat the person as a point particle. Treating the person as being made of two hands simply makes the calculation more complex for the same result! Note I also completely neglected the fact that you might be on the opposite side of the Earth, in which case you have to add the diameter of the Earth (6400 km) to the distance calculation. But on the scales of the Earth-Moon system, even that is negligible.

The same thing applies on planetary scales. For example, to calculate the gravitational force between the Sun and the Earth, we don't have to care whether Mt. Everest is pointing in the direction of the Sun or not. The impact on the result is negligible.

• So did you mean that the great distance causes the force to reduce to such a great extent due to its inverse squared nature that the small error in approximation of mass doesn't have great impact on the gravitational force between them? – suiz Jan 1 '18 at 2:40
• No, it's about the size of the error. For example if you're solving a question that needs the acceleration due to Earth's gravity, using g = 10 isn't that different from g = 9.8. It depends on how precise an answer you want. If you don't need a really precise answer, g = 10 makes computations a lot easier. If you're after real precision, even g = 9.8 is not enough - you need the local value of g, since g varies by location (in turn because the Earth is not exactly a sphere). – Allure Jan 1 '18 at 2:45
• Similarly if you treat planets as extended objects (as opposed to point particles) then you make your calculations a lot harder. However because the distances between planets is so large, it's not likely to affect the precision of your results until like the 14th (or some other large number) decimal place. If you care about that level of precision then you can't use point particles, but for most applications you can. – Allure Jan 1 '18 at 2:46
• Now it seems to sink in. – suiz Jan 1 '18 at 3:02
• Glad to hear it. I'll edit the answer to improve it, since this explanation looks better. – Allure Jan 1 '18 at 8:22

You can expand the gravitational field of an arbitrarily shaped body systematically in something called the multipole expansion (this is also done in electromagnetism). The lowest order term will be the slowest to decrease with distance and that lowest order term is just that of a point particle.

For instance say you have a two point masses of mass m at the points (0,0,0) and (0,0,dz)

The gravitational potential is $$U=-Gm\left(\frac{1}{\sqrt{x^2+y^2+z^2}}+\frac{1}{\sqrt{x^2+y^2+(z-dz)^2}}\right)$$ Taylor expanding about $dz=0$, and using $r\equiv\sqrt{x^2+y^2+z^2}$ $$U=-Gm\left(\frac{2}{r}+\frac{z\,dz}{r^3}+\mathcal{O}(dz^2)\right)$$ The first term is just the gravitational potential of a point mass of 2m, and this is the term that is most important when $dz/r\ll 0$.

The second term is the dipole moment term $$\frac{\vec{p}\cdot \hat{r}}{r^2}$$ where $\vec{p}=m\vec{dz}$. And higher order terms are 'multipole moment' terms that have to do with higher order spherical harmonics.

This example I'm using is very similar to user3727079's answer, but the idea is we can do this expansion for an arbitrarily shaped body and it is always the case that the most important term far away is that of a point mass (and the next most important is the dipole moment).

• Yours is the more technically accurate answer [but harder to understand :)]. – Allure Jan 1 '18 at 8:46
• Another, possibly more intuitive, way to understand why the dipole moment becomes less important at larger distances is to note that it is proportional to $z/r$, i.e. the ratio of the size of the body to the distance away from it. Which is to say that as the distance from a body becomes much greater than its linear size, it becomes better and better approximated as a point particle. This ends up being true for e.g charge distributions in electromagnetism. Close up, distribution matters. Far away it doesn't. – PhillS Jan 1 '18 at 9:15