# Gravitational Attraction of Extended Bodies

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• 0:01 What Is an Extended Body?
• 1:24 Gravitational…
• 4:57 Example Problem
• 6:25 Lesson Summary

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Lesson Transcript
Instructor: David Wood

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

After watching this lesson, you will be able to explain what an extended body is, understand the derivation for the force of gravity between a point charge and a long, thin bar and solve problems using the result of that derivation.

## What Is an Extended Body?

In another lesson, we introduced Newton's Law of Universal Gravitation. This law says that every body in the universe attracts every other body with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. It is represented by this equation: Fg = GMm / r^2. It means that if you double the mass of an object, you double the gravitational force, and if you double the distance between two objects, you reduce the force to a quarter of what it was.

But this law, like all laws in physics, makes certain assumptions. The biggest of these is that a gravitational body can be treated as a point mass, or in other words, the force of gravity can be said to act at a specific point, usually at the center of the object. It turns out that if you're looking at spherical objects, like planets or stars, this assumption is a great one. It works very well. But not all objects are spherical or even nearly spherical. What happens when the shape is completely random?

Well, in that case, the object is said to be an 'extended object.' Suddenly the law of gravitation doesn't work so well, and things get a whole lot more complicated. In this lesson, we're going to explore one particular example of an extended object and use calculus to figure out an equation that will work.

## Gravitational Attraction Equation

When you're analyzing an extended object, you need to use some calculus. The final gravitation equation you get out of it will vary depending on the shape of the object. The shape we're going to look at today is a uniform cuboid bar - a long, thin, bar-shaped object. And we'll look at the gravitational force between that bar and a regular point mass, M, that happens to be nearby.

We'll put the bar along an axis and say that the bar starts at coordinate (a, 0) and ends at coordinate (L + a, 0), where L is the length of the bar. The way we solve this problem is by looking at the force of gravity created by a tiny element of the bar, which we'll call dm (an infinitesimally small mass). By adding up all the forces from all the tiny mass elements that make up the bar using calculus, we can figure out the total force.

The regular equation for a point mass, Newton's Law of Universal Gravitation, looks like this.

Here, G is the gravitational constant of the universe, which is always 6.67 * 10^-11; F is the force of gravity between the objects, measured in newtons; M is the mass of the point mass on the left of the diagram; and r is the radius, or distance, you are away from the mass, M. This equation contains m, which would normally be a second point mass. But this time we have a bar, so we'll replace that smaller mass, m, with our tiny mass element, dm. This is because each tiny mass element does, after all, act like a point mass.

The dm makes this a differential equation. So, this equation now says that F is equal to the integral of GM times 1 over r^2, dm. But what are the limits of this integral? Well, we're looking at mass elements from one side of the bar to the other: from a radius of a to a radius of L + a - so, those are our limits.

Now we need to solve this integral. The only problem is that we have an equation in terms of the radius r, and we're asked to integrate with respect to m. That isn't going to work. They need to be the same variable. So, the trick we can use here is to define a new term, lambda, which is the mass per unit length, or mass per unit radius, if you'd like. It's mass divided by the radius r, or in calculus terms, it's dm over dr. Rearranging this for dm, we see that dm is equal to lambda dr. We can substitute that into our integral, and we now have an integral that can be solved.

The integral of 1 over r'-squared, with respect to r, comes out as negative 1 over r. Substitute in our limits of L + a and a, and simplify a bit, and we have our equation for the gravitational attraction of the bar, which looks like this:

Finally, to make it a bit simpler, we can remind ourselves what lambda means. Lambda is the mass per unit length. So, lambda times L is the mass per unit length times the length of the bar. So, that's just the mass of the bar, m. So, now by substituting that in, we have our final equation: F = GMm / a(L + a).

The force of gravity between the bar and the point mass M is equal to G times M times m divided by a bracket L plus a. Just as a reminder, M is the mass of the point mass on the left of the diagram, m is the mass of the bar, a is the distance from the point mass to the nearside of the bar and L is the length of the bar.

## Example Problem

Let's go through an example of how to use the equation: a metal bar leftover from an experiment completed in space is floating at a distance from the sun. Assume the sun can be treated as a point mass. If the mass of the metal bar is 100,000 kilograms, the mass of the sun is 2 * 10^30 kilograms, the distance to the nearside of the bar from the sun is 500,000 meters and the length of the bar is 500 meters, what is the force of gravity between the sun and the bar?

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