Gravitational Force Exerted by a Rod

 
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Gravitational Force Exerted by a Rod

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In this lesson, we'll derive a formula which will allow us to calculate the gravitational force exerted by a rod of length \(L\) on a particle a horizontal distance \(x\) away from the rod as illustrated in Figure 1. We'll assume that the width and depth of the rod are negligible and approximate all of the mass comprising the rod as being distributed along only one dimension. We'll model the rod as being composed of an infinite number of particles of mass \(dm\). The mass of the rod is given by the infinite sum of all the mass elements (or particles) comprising the rod:

$$M_{rod}=\int{dm}.\tag{1}$$

 Figure 1: A rod of mass \(M\) and a particle of mass \(m\) are separated from each other by a distance of \(d\) along the \(x\)-axis. Each mass element \(dm\) comprising the rod is located at some position \(x\) along the \(x\)-axis and is separated from the particle \(m\) by some amount \(r\). By summing all the gravitational forces \(d\vec{F}_g\) exerted by each mass element \(dm\) comprising the rod, we can find the total gravitational force exerted on \(m\) by the entire rod.

Figure 1: A rod of mass \(M\) and a particle of mass \(m\) are separated from each other by a distance of \(d\) along the \(x\)-axis. Each mass element \(dm\) comprising the rod is located at some position \(x\) along the \(x\)-axis and is separated from the particle \(m\) by some amount \(r\). By summing all the gravitational forces \(d\vec{F}_g\) exerted by each mass element \(dm\) comprising the rod, we can find the total gravitational force exerted on \(m\) by the entire rod.

We're interested in finding the gravitational force exerted by the rod on a particle of some mass \(m\). Now, of source, the notion of a particle is something that is very abstract—an object of zero size with all of its mass concentrated at a single point (more precisely, a geometrical point which is another very abstract notion) in space. No object is actually a particle (except a black hole), but if the object is very small compared to the size of the rod then it is reasonable to ignore the size of the dimensions of that object and to approximate it as a point mass.

Newton's law of gravity is defined as

$$\vec{F}_{m_1,m_2}=G\frac{m_1m_2}{r^2}\hat{r}_{1,2}.\tag{2}$$

where \(\vec{F}_{m_1,m_2}\) is the gravitational force exerted by a particle of mass \(m_1\) on another particle of mass \(m_2\), \(r\) is their separation distance, and \(\hat{r}_{1,2}\) is a unit vector pointing from \(m_1\) to \(m_2\). For the moment, we'll just be interested in the magnitude of the gravitational force exerted on \(m_2\) which is given by

$$F_{m_1,m_2}=G\frac{m_1m_2}{r^2}.\tag{3}$$

When I was telling you the definition of Newton's law of gravity, notice how I was very specific about how Equation (2) (and thus Equation (3) as well) gives the gravitational force exerted by one particle (or point-mass) on another particle. The famous equation representing Newton's law of gravity only deals with particles and for this reason we cannot use Equation (2) or (3) to compute the gravitational force exerted by a rod on a particle—the rod isn't a particle, it's an extended object. When dealing with the mass of any extended object in classical mechanics—whether it be a rod, disk, ball, or any other geometrical shape—we can think of the entire shape of that object as being built up by an infinite number of point-masses of mass \(dm\). Given how I defined Equations (2) and (3) as being in terms of only particles, we can use Equation (3) to compute the gravitational force exerted by one of the particles of mass \(dm\) exerted on the particle of mass \(m\) as

$$F_g=Gm\frac{1}{r^2}dm,\tag{4}$$

where \(dm\) and \(m\) are the mass of each particle, and \(r\) is their separation distance. As you can see from Figure 1, if \(x\) represents the position of the mass \(dm\) on the \(x\)-axis, then the separation distance between \(dm\) and \(m\) must be \((L+d)-x\). Thus, we can represent Equation (4) as

$$F_g=Gm\frac{1}{((L+d)-x)^2}dm.\tag{5}$$

Equation (5) represents the gravitational force exerted by any particle in the rod on the particle as horizontal distance \(d\) away from the rod. To find the total gravitational force exerted by the rod, we must "add up" (indeed, "add up" an infinite number of time) the gravitational forces, \(Gmdm/x^2\), exerted by every particle \(dm\) on the mass \(m\):

$$F_{rod,m}=Gm\int{\frac{1}{((L+d)-x)^2}dm}.\tag{6}$$

Equation (6) does indeed give the magnitude of the gravitational force exerted by \(M_{rod}\) on \(m\)—but the only problem is that we cannot calculate the value of this force since we cannot evaluate the integral in Equation (6). To calculate the integral in Equation (60, the integrand and limits of integration must be represented in terms of the same variable. If we assume that the mass density \(λ\) of the rod is constant, then

$$λ=\frac{dm}{dx}$$

and

$$dm=λdx.\tag{7}$$

Substituting Equation (7) into (6), we have

$$F_{rod,m}=Gmλ\int_{-L/2}^{L/2}\frac{1}{((L+d)-x)^2}dx.\tag{8}$$

As you can see, after representing everything in the integral in terms of \(x\), we have ended up with an integral that is fairly straightforward to calculate. If we let \(u=L+d-x\), then

$$\frac{du}{dx}=-1$$

and

$$dx=-du\tag{9}$$

Substituting \(u=L+d-x\) and Equation (9) into (8), we have

$$F_{rod,m}=-Gmλ\int_{?_1}^{?_2}\frac{1}{u^2}du.\tag{10}$$

When \(x=-L/2\), \(u=\frac{3L}{2}+d\) and when \(x=L/2\), \(u=\frac{L}{2}+d\). Substituting these limits of integration into Equation (10), we have


$$F_{rod,m}=Gmλ\int_{\frac{L}{2}+d}^{\frac{3L}{2}+d}\frac{1}{u^2}du.\tag{11}$$

Solving the integral in Equation (11), we have

$$Gmλ\int_{\frac{L}{2}+d}^{\frac{3L}{2}+d}\frac{1}{u^2}du=Gmλ\biggl[\frac{-1}{u}\biggr]_{\frac{L}{2}+d}^{\frac{3L}{2}+d}=Gmλ(\frac{1}{d}-\frac{1}{L+d}).$$

Thus,

$$F_{rod,m}=Gmλ(\frac{1}{d}-\frac{1}{L+d}).\tag{12}$$

Equation (12) allows us to calculate the magnitude of the gravitational force exerted by a rod on a particle. Since each mass \(dm\) in the rod is pulling on the mas \(m\) in the \(-x\) direction, the entire rod pulls on \(m\) in the \(-x\) direction. If we multiply the magnitude of the gravitational force, \(F_{rod,m}\), by \(-\hat{i}\), this will give us a gravitational force with a magnitude of \(F_{rod,m}\) and a direction of \(-\hat{i}\) in the negative \(x\) direction. Thus, the gravitational force exerted by the rod on the mass \(m\) is given by

$$\vec{F}_{rod,m}=Gmλ(\frac{1}{d}-\frac{1}{L+d})(-\hat{i}).\tag{13}$$

The entire problem we just solved would also apply to a problem from electrostatics which deals with finding the electric force exerted by a charged rod on a charged particle. This is because the law specifying the electric force (namely, Column's law) is completely analogous to Newton's law of gravity. What I mean by this is that both laws have analogous mathematical expressions and both involve a universal constant, two parameters describing two particles, and both laws are inverse-square laws.


This article is licensed under a CC BY-NC-SA 4.0 license.