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.
