Lesson Video: Gravitational Forces near Planetary Bodies

In this video we see how to calculate the acceleration due to a planet’s gravity that acts on masses near the planet, making an approximation to a special case of Newton’s Law of Universal Gravitation.


Video Transcript

In this video, we’re going to learn about gravitational forces near planetary bodies. We’ll see what these forces are, how to calculate them, and how they relate to Newton’s law of universal gravitation.

To start out, imagine it’s far into the future at a time when space travel has become normal. Planet hopping has become the thing to do. And in preparation for an upcoming trip, you’re getting your luggage ready. Since rocket fuel is expensive though, there are weight limits on how much a person’s luggage can weigh. And interestingly, that weight limit is the same regardless of which planet a person is visiting. You won’t have any trouble figuring out how much your luggage weighs on your planet of departure. But what about on all the planets you’ll visit on your trip? To figure this out, it will be helpful to know about gravitational forces near planetary bodies.

Recalling Newton’s law of universal gravitation, we know it says that the gravitational force between two masses—we can call them capital and lowercase 𝑚—is equal to the product of their masses divided by the square of the distance between their centres of mass—we’ve called it 𝑟—all multiplied by the universal gravitational constant big 𝐺. But what if we had a case of a very, very, very large mass with a relatively small mass close to that large mass, just like we are on the surface of the Earth for example? Just like the law of universal gravitation says, we would take these two masses, multiply them, then divide them by a distance between the centre of mass of the Earth and our centre of mass and multiply all that by capital 𝐺.

Looking at this diagram though, something may stand out about this distance between our two masses. If we were to draw in the radius of this planet—and for now we can assume it’s Earth—and call that Earth radius 𝑟 sub 𝐸 compared to our distance between our two masses which we’ll call 𝑟, then we can see by inspection that 𝑟 sub 𝐸 over 𝑟 is very nearly equal to one. And in fact, this will be true anytime we have an object which is near the surface of this much larger planet Earth. We might wonder what does “near” mean exactly? In other words, when can we consider this ratio to effectively be equal to one?

Imagine a layer that goes all the way around the Earth and inside this layer, we can say that the tallest skyscraper on Earth would fit. We can say that within this zone we are effectively near to this planetary body of the Earth. And in this condition, we can say that the radius of the Earth is approximately equal to the distance between the Earth’s centre of mass and the centre of mass of objects in this zone. Here is why all this is important. Let’s look back to the law of universal gravitation.

Looking at this expression, we see that everything in this equation except for the mass small 𝑚 is effectively constant when we’re near the surface of the Earth. The Earth’s mass is a constant and the universal gravitational constant is the same everywhere. So if we can say that the distance between our two objects is simply the radius of the Earth, then this value that we’ve highlighted will stay the same. And in that case, once we know this highlighted value which effectively is a constant, we’ll be able to solve for the gravitational force on an object simply by knowing the mass of that object. We won’t need to know anything else about the second mass—the large planet, the Earth—that it’s near to.

We can write that when we’re near a planet, this value of capital 𝐺 times the large planet’s mass all divided by 𝑟 squared is basically equal to a constant we’ve sometimes called lowercase 𝑔. This value turns out to have units of metres per second squared. And it’s known as the acceleration that’s due to gravity—that is, the force of gravity acting on an object. As we’ve said, once we know the acceleration due to gravity for a certain planet, all we need to know to figure out the force of gravity on a mass on that planet is the value of the mass itself.

Knowing that, what if we calculate the acceleration due to gravity for an object on Earth and near Earth’s surface? That’s equal to the universal gravitational constant times the mass of the Earth divided by the radius of the Earth squared. When we look up those values and plug them in, we find a result of approximately 9.8 metres per second squared. You may have seen that number before in physics calculations or problems. And this is where it comes from: we say that 𝑔 sub 𝐸 is approximately this value because of course, it varies with the distance 𝑟 squared.

Technically, if we have an object which is slightly farther away from the centre of the Earth, the gravitational acceleration on that object will be less. And another interesting factor is that the Earth is actually not a perfect sphere, but rather along the line of the equator, it bulges outward due to its spinning rotation. This means that the radius of the Earth is greater at the equator than at one of the poles. And therefore, the acceleration due to gravity on the Earth is less along the equator than at a polar position. All that said, 9.8 metres per second squared is still a good approximation for the acceleration due to gravity near Earth’s surface—anywhere on that surface.

So far, we’ve been talking about the Earth. But in fact, every planet has its own value for the acceleration due to gravity 𝑔. And often those values are very different from that value on Earth. The acceleration due to gravity for a general planet—we can call this 𝑔 sub 𝑝— is equal to the mass of the planet times the universal gravitational constant divided by the radius of that planet squared. Strictly speaking, when we calculate this value, we’re only calculating the acceleration due to gravity at—exactly at—a planet surface. But under our approximation for objects near the surface, we’re effectively calculating the acceleration due to gravity on anything near that planetary body. Let’s get some practice now with these ideas about the gravitational forces near a planet using an example.

Jupiter has a radius at its equator of 71492 kilometers and the gravitational acceleration at that point is 23.1 meters per second squared. Calculate Jupiter’s mass from its radius and the gravitational acceleration at its equator. What is the ratio of the calculated mass of Jupiter to NASA’s Jupiter fact sheet value of 1898 times 10 to the 24th kilograms?

We can label the mass of the planet Jupiter capital 𝑀 and the ratio of this mass to the fact sheet value mass 𝑀 over 𝑀 sub 𝑐. To solve first for the overall mass of the planet Jupiter based on the information about its radius and its acceleration due to gravity, we can recall that in general the universal gravitational constant times an object’s mass—typically a large object such as a planet—divided by the radius squared of that object is equal to the gravitational acceleration it causes at its surface.

In our example, we’re told the acceleration due to gravity at Jupiter’s equator as well as the radius of the planet at that point. For capital 𝐺, the universal gravitational constant, we’ll use a value of 6.67 times 10 to the negative 11th cubic metres per kilogram second squared. We can now rearrange this expression to solve for capital 𝑀, the mass of Jupiter. We see it’s equal to 𝑟 squared times little 𝑔 over big 𝐺.

And when we plug in for these values, we’re careful to write our radius of the planet in units of metres so that it agrees with the units in the rest of our expression. To three significant figures, the mass of Jupiter is 1.77 times 10 to the 27th kilograms. That’s its mass calculated based on its equatorial radius, the universal gravitational constant, and its acceleration at its radius.

Next, we want to solve for the ratio of this mass we’ve just solved for to the fact sheet value mass of Jupiter which is 1898 times 10 to the 24th kilograms. Entering both these values in in scientific notation, when we calculate this fraction to three significant figures, it’s 0.933. So our calculated value for Jupiter’s mass is within 10 percent of the fact sheet mass value.

Let’s summarize what we’ve learnt so far about gravitational forces near planetary bodies.

We’ve seen that when we’re near a planet surface, 𝐺 times the planet’s mass 𝑀 all over the planet’s radius squared can be treated as a constant value. We often abbreviate capital 𝐺 times 𝑀 over 𝑟 squared with a lowercase 𝑔, known as the acceleration due to gravity for a particular planet. Once we’ve calculated little 𝑔, when we multiply that by the mass value of an object near the surface of that planet, we found the weight of that mass 𝑚 on that planet. And finally, we saw that precisely this acceleration due to gravity in fact changes with an object’s elevation above a planet or it changes with the planet’s radius.

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