fetsBedscurce4why1
2022-05-10
Answered

Why do we integrate along the whole length while finding the gravitational force between a object of mass (m) and a rod of length L and mass M? Can't we simply use $F=\frac{GmM}{{r}^{2}}\phantom{\rule{thinmathspace}{0ex}}?$?

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empatteMattmkezo

Answered 2022-05-11
Author has **22** answers

We have to consider the contribution of each little piece of mass $\mathrm{d}M=\rho \mathrm{d}V$ and how far it is from the other mass, m resulting in a small force $\mathrm{d}F$. If you have uniform distribution of mass, $\rho =M/(LA)$, where A is the cross sectional area of the rod, and we probably don't integrate over that because it's a small diameter rod. So $\mathrm{d}V=A\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$. Then we have a $1/{r}^{2}$ force behavior where r is the distance from each $\mathrm{d}M$

$\mathrm{d}F=\frac{-Gm}{{r}^{2}}\mathrm{d}M=\frac{-Gm}{{r}^{2}}\frac{M}{L}\mathrm{d}x$

To find r, you have to give a specific geometry, and let x be the variable of integration that runs along the length and integrate from 0 to $L$. $r$ will be a function of x. The geometry dictates what that function is. Because of the nature of $1/{r}^{2}$, the force does not change linearly, but the distance from rod to mass m might (again, the geometry is important).

In your comment, you mentioned the sphere situation. That means that for a sphere of mass $M$ and radius ${R}_{0}$

$\mathrm{d}M=\rho \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}V=\frac{M}{{V}_{sp}}{R}^{2}\mathrm{sin}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}R\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\varphi $

$0<R<{R}_{0}$

where R is a spherical variable of integration, along with $\theta $ and $\varphi $. The distance r will be a function of R, $\theta $, and $\varphi $. Without going into the details, an examination of the form of $\mathrm{d}M$ shows that the mass is not linearly distributed, so there's a chance, at least, that a $1/{r}^{2}$ force will reduce to something simple.

$\mathrm{d}F=\frac{-Gm}{{r}^{2}}\mathrm{d}M=\frac{-Gm}{{r}^{2}}\frac{M}{L}\mathrm{d}x$

To find r, you have to give a specific geometry, and let x be the variable of integration that runs along the length and integrate from 0 to $L$. $r$ will be a function of x. The geometry dictates what that function is. Because of the nature of $1/{r}^{2}$, the force does not change linearly, but the distance from rod to mass m might (again, the geometry is important).

In your comment, you mentioned the sphere situation. That means that for a sphere of mass $M$ and radius ${R}_{0}$

$\mathrm{d}M=\rho \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}V=\frac{M}{{V}_{sp}}{R}^{2}\mathrm{sin}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}R\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\varphi $

$0<R<{R}_{0}$

where R is a spherical variable of integration, along with $\theta $ and $\varphi $. The distance r will be a function of R, $\theta $, and $\varphi $. Without going into the details, an examination of the form of $\mathrm{d}M$ shows that the mass is not linearly distributed, so there's a chance, at least, that a $1/{r}^{2}$ force will reduce to something simple.

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I have this expression for the gravitational force between 2 masses when the gr term is added:

$F=-\frac{GMm}{{r}^{2}}+\frac{4{G}^{2}mM(M+m)}{{r}^{4}{c}^{2}}$

which I got from the internet since I haven't bee able to find a similar expression in any book. But i'm not sure if it's right or needs fixing.

Is the second term in the force correct?

$F=-\frac{GMm}{{r}^{2}}+\frac{4{G}^{2}mM(M+m)}{{r}^{4}{c}^{2}}$

which I got from the internet since I haven't bee able to find a similar expression in any book. But i'm not sure if it's right or needs fixing.

Is the second term in the force correct?

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(a) Calculate the gravitational force exerted on a 5.00 kg baby by a 90 kg father 0.200 m away at birth (assisting so he is close).

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