When calculating the gravitational field of a continuous mass, like a rod, at some point outside it, why can't we just use the centre of mass of the rod and then directly plug in the distance between the point and the centre of mass in the gravitational field equation. Why do we need to integrate over the whole rod?
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The basic idea is that for generic objects, one cannot simply use the distance to the center of mass because there is not enough symmetry. When the object is in different orientations, the point "sees" a different mass distribution which leads to a generically different value for the gravitational field.
I think this is best illustrated by a concrete example. Let's consider a rod as you suggest. For simplicity, let the rod be thin and of length $\ell$ and constant linear mass density $\lambda$. Let the center of mass of the rod be at the origin, and for simplicity, let the rod lie in the $x$-$y$ plane. Consider computing the gravitational field at the point $\mathbf x_0 = (0, 2\ell)$ in the following two cases:
The rod lies along the $x$-axis.
The rod lies along the $y$-axis.
In both cases, the gravitational field vector at $\mathbf x_0$ points in the $y$-direction by symmetry, but in case 1, its magnitude is \begin{align} g_1 = \int_{-\ell}^{\ell} dx \frac{G\lambda}{x^2+(2\ell)^2} = \cot^{-1}(2)\frac{G\lambda}{\ell} \end{align} while in case 2, its magnitude is \begin{align} g_2 = \int_{-\ell}^{\ell} dy \frac{G\lambda}{(2\ell-y)^2} = \frac{2}{3}\frac{G\lambda}{\ell} \end{align} and a numerical approximation shows that \begin{align} \frac{g_2}{g_1} = 1.44. \end{align} The strength of the gravitational field is greater in case 2. What has happened is that in the second orientation, some of the mass is closer to $\mathbf x_0$ than in the first orientation, and some of the mass is further away, but the net effect of changing the mass distribution is that the strength of the gravitational field at point $\mathbf x_0$ increases. Mathematically, you can see why this happens by examining the two integrands. In case 2, the denominator can be rewritten as \begin{align} y^2 + (2\ell)^2 - 4\ell y. \end{align} This is the same as the denominator in the case 1 integrand except for the term $-4\ell y$ which makes it smaller for every value of the integration variable, so the entire integrand is always larger, and thus one obtains larger value for the strength of the gravitational field.
Contrast this with an object that possess spherical symmetry. In that case, once you have chosen a point, the mass distribution looks the same if you are viewing it from that point no matter how you change its orientation. The distribution of the mass doesn't matter and you can in fact simply treat all the mass as if it's concentrated at the center of mass.
joshphysics
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You're in the milky way galaxy, whose center of mass is nowhere near Earth - so why are you feeling a force towards Earth? (note that the Earth and you would accelerate towards the center of the galaxy at the same rate, so "the earth is accelerating faster" would not pan out.)
If that doesn't satisfy you, note that the fundamental law is $F=G\frac{m M}{r^2}$ for point masses. First you apply this fundamental law using calculus arguments, that is, $dF=G \frac{m dM}{r^2}$, where $dM=\sigma dx \cdot dy\cdot dz$ where $dM$ is a portion of mass, $\sigma$ is the density of the body, and $dx \cdot dy\cdot dz$ is a volume element. When you do that you find that it is not the general case that a body always feels a force towards the center of mass of a matter distribution. This makes sense as intuition (using my argument in the first paragraph), but even if you're still confused by it you have to admit that it follows from the fundamental inverse square law!
So, your intuition was wrong.