### Video Transcript

In this video, we will learn how to apply Newtonβs law of universal gravitation to
find the gravitational force between two masses.

Two masses, π one and π two, for example, the Earth and the Moon, separated by a
distance π, will exert a force of attraction on each other called a gravitational
force. This force follows an inverse square law, meaning it decreases in strength with the
square of the distance π. This force is a clear example of Newtonβs third law of motion, which states that when
a body exerts force on a second body, the second body simultaneously exerts a force
equal in magnitude and opposite in direction on the first body.

Newtonβs law of universal gravitation specifically states that two bodies exert
gravitational forces on each other, where the direction of the force on either body
is toward the center of mass of the other body. This force π
, equal in magnitude on both bodies, is equal to πΊ times π one times
π two over π squared, where π one is the mass of one body, π two is the mass of
the other body, π is the distance between the bodiesβ centers of mass, and πΊ is
the universal gravitational constant. The universal gravitational constant, sometimes referred to as βBig πΊ,β is a
fundamental constant of the universe, approximately equal to 6.67 times 10 to the
power of negative 11 newton-meters squared per kilogram squared.

Letβs look at an example of how the gravitational force between two bodies is
determined.

Determine the gravitational force between two identical balls, each of mass 3.01
kilograms, given that the distance between their centers is 15.05 centimeters and
the universal gravitational constant is 6.67 times 10 to the power of negative 11
newton-meters squared per kilogram squared.

Recall that Newtonβs law of universal gravitation states that two bodies of masses π
one and π two will attract each other with a force of magnitude πΊ times π one
times π two over π squared, where π is the distance between their centers of mass
and πΊ is the universal gravitational constant.

In this case, therefore, we have Big πΊ, 6.67 times 10 to the power of negative 11,
multiplied by the mass of the first ball, 3.01, multiplied by the mass of the second
ball, also 3.01. And since Big πΊ and the masses are in SI base units of newtons, meters, and
kilograms, we divide by the distance π squared measured in meters, which is 15.05
times 10 to the power of negative two all squared. Performing this calculation, we get the gravitational force between the balls: 2.668
times 10 to the power of negative eight, and the unit is newtons.

Letβs now look at an example in which the gravitational force between two bodies is
used to determine the distance between their centers of mass.

Given that the gravitational force between two bodies of mass 4.6 kilograms and 2.9
kilograms was 3.2 times 10 to the power of negative 10 newtons, find the distance
between their centers. Take the universal gravitational constant πΊ equals 6.67 times 10 to the negative 11
newton-meters squared per kilogram squared.

Recall that Newtonβs law of universal gravitation states that two bodies of masses π
one and π two will attract each other with a force of magnitude πΊ times π one
times π two over π squared, where π is the distance between their centers of mass
and πΊ is the universal gravitational constant. The question asks us to find the distance between the bodiesβ centers. So, we need to rearrange this equation for this distance π. We can do this by multiplying through by π squared, dividing through by π
, and
taking the square root.

Now inserting the values for all of the variables and constants, for Big πΊ, we have
6.67 times 10 to the power of negative 11. The masses π one and π two are given in the question as 4.6 and 2.9 in SI units of
kilograms. And the force π
is also given in the question as 3.2 times 10 to the power of
negative 10, also in SI units of newtons. Performing this calculation gives us the distance between the centers of the two
bodies, 1.6675 exactly, and the unit is the SI unit meters. In centimeters, this comes to 166.75 centimeters.

Letβs now look at an example of determining the gravitational force between two
bodies when one of the bodies is the Earth.

A satellite of mass 2415 kilograms is orbiting Earth 540 kilometers above its
surface. Given that the universal gravitational constant is 6.67 times 10 to the power of
negative 11 newton-meters squared per kilogram squared and Earthβs mass and radius
are six times 10 to the power of 24 kilograms and 6360 kilometers, determine the
gravitational force exerted by Earth on the satellite.

Recall that Newtonβs law of universal gravitation states that two bodies of masses π
one and π two will attract each other with a force of magnitude πΊ times π one
times π two over π squared, where π is the distance between their centers of mass
and πΊ is the universal gravitational constant. By Newtonβs third law, the force exerted on the satellite by the Earth is equal and
opposite to the force exerted on the Earth by the satellite. And both are given by this equation.

Big πΊ is the universal gravitational constant, 6.67 times 10 to the power of
negative 11. We can let π one be the mass of the Earth, six times 10 to the power of 24, and π
two be the mass of the satellite, 2415. π squared will require a little more thought. We are given the distance from the satellite to the Earthβs surface and the Earthβs
radius. We are not given π, the distance between the centers of mass of the two bodies. The distance between the center of mass of the Earth and the satellite is equal to
the radius of the Earth, π E, and the distance between the satellite and the
surface of the Earth, π.

Itβs worth noting that we are modeling the satellite as a point mass. We are ignoring the size of the satellite and assuming that the distance between its
center of mass and the surface of the Earth is 540 kilometers.

So, we have Big πΊ times π one times π two over π E plus π all squared. So, we have Big πΊ, which is given as 6.67 times 10 to the power of negative 11,
multiplied by the mass of the Earth, given as six times 10 to the power of 24,
multiplied by the mass of the satellite, given as 2415, all divided by the radius of
the Earth π E, given as 6360 kilometers, so in SI unit of meters this is 6360 times
10 to the third power, plus the distance π between the surface of the Earth and the
satellite, given as 540 kilometers, which in meters is 540 times 10 to the third all
squared. Performing this calculation, we get 20300, and the unit is newtons.

The force on a body induces an acceleration via Newtonβs second law, which states
that the acceleration of an object is directly proportional to the force applied to
it, where the mass of the object is the constant of proportionality. This can be expressed with the equation force equals mass times acceleration: π
equals π one times π, where π one is the mass of the body. The acceleration induced by the gravitational force is called the acceleration due to
gravity. The force in this case is given by Newtonβs universal law of gravitation: π
equals
πΊ times π one times π two over π squared.

To find the acceleration of the first body due to gravity, we equate these two
expressions for the force on the body. We therefore have π one times π equals πΊ times π one times π two over π
squared. Since a mass will always be nonzero, we can divide through by the common term of π
one on both sides. This gives us the acceleration due to gravity of the body: π equals πΊ times π two
over π squared.

Notice that this acceleration is independent of the mass of the body and only depends
on the mass of the second body. This is why two objects, in the absence of drag, will fall at the exact same
rate. This leads into one final concept, the gravitational field strength of a point
mass. This is the gravitational force per unit mass exerted by a mass π on a body whose
center of mass is a distance π from it. And this is given by little π equals Big πΊ times π over π squared.

Notice that this is in fact the same expression for the acceleration due to gravity
mentioned previously. This is because theyβre in fact the same quantity. Itβs just the other side of the equation of Newtonβs second law after a little
rearranging.

On the left-hand side, we have the force divided by the mass, in other words, the
force per unit mass. And on the right-hand side, we have the acceleration. So, gravitational field strength and acceleration due to gravity are in fact the same
quantity, just expressed differently. And indeed, we often use little π to denote the acceleration due to gravity as well
as the gravitational field strength. For the Earth, the mass is 5.97 times 10 to the power of 24.

Provided the body is on or above the surface of the Earth, we can treat the Earth as
a point mass. And the distance from the surface to the center of mass is the radius of the Earth,
6.37 times 10 to the power of six. Performing this calculation gives us little π equals 9.81 meters per second squared
to two decimal places, the standard acceleration due to gravity on the surface of
the Earth.

Letβs now look at an example in which the ratio of the acceleration due to gravity on
Earth to another planet is determined.

Given that a planetβs mass and diameter are three and six times those of Earth,
respectively, calculate the ratio between the acceleration due to gravity on that
planet and that on Earth.

Recall that the acceleration due to gravity at the surface of a uniform sphere is
given by π equals πΊ times π over π squared, where π is the mass of the sphere,
π is the radius of the sphere, and πΊ is the universal gravitational constant. We can model both the Earth and the planet as uniform spheres. We can express the acceleration due to gravity on the surface of the Earth π E as πΊ
times the mass of the Earth π E divided by the radius of the Earth π E
squared.

Likewise, we can express the acceleration due to gravity on the surface of the planet
π P as πΊ times the mass of the planet π P divided by the radius of the planet π
P squared. The question tells us that the mass of the planet π P is three times the mass of the
Earth π E. So, π P equals three π E. The question also tells us the diameter of the planet, and therefore also the radius
of the planet π P, is six times the radius of the Earth π E. So, π P equals six π E.

We can therefore reexpress the acceleration due to gravity on the planet π P as πΊ
times three π E over six π E all squared. Expanding the parentheses on the denominator gives us πΊ times three π E over 36π E
squared. And simplifying by dividing the numerator and denominator by three gives πΊ times π
E over 12π E squared. This is the same expression as the expression for the acceleration due to gravity on
the surface of the Earth π E, but with an extra factor of one twelfth. Therefore, π P is equal to one twelfth π E. Therefore, the ratio between the acceleration due to gravity on the surface of the
planet and that on Earth, π P to π E, is one to 12.

Letβs now look at an example in which the radius of a planet is determined from the
acceleration due to gravity.

An astronaut dropped an object from a height of 2352 centimeters above the surface of
a planet, and it reached the surface after eight seconds. The mass of the planet is 7.164 times 10 to the power of 24 kilograms, while that of
Earth is 5.97 times 10 to the power of 24 kilograms, and the radius of Earth is 6.34
times 10 to the six meters. Given that the gravitational acceleration of Earth is π equals 9.8 meters per second
squared, find the radius of the other planet.

Recall that the acceleration due to gravity on the surface of a uniform sphere is
given by π equals Big πΊ times π over π squared, where Big πΊ is the universal
gravitational constant, π is the mass of the sphere, and π is the radius of the
sphere. An important thing to note is that the question does not give us the value of Big πΊ
to use. We could use our knowledge of the constantβs value to solve the problem, but we donβt
actually need to. We can rearrange this equation for π, the quantity we want to find, by multiplying
through by π squared, dividing through by π, and taking the square root.

Again, we are not given the value of Big πΊ, but we are given the mass, radius, and
acceleration due to gravity of the Earth. This is a hint that we can use these quantities to find the radius of the other
planet. We can do this by using a ratio. If we take this expression for the radius of the planet β letβs call this π P β and
divide it by the expression for the radius of the Earth, which weβll call π E, we
obtain π P over π E equals the square root of Big πΊ times π P over π P all
divided by the square root of Big πΊ times π E over π E.

We can use a surd rule to simplify this. The square root of π₯ over the square root of π¦ is equal to the square root of π₯
over π¦. We can therefore reexpress the ratio π P over π E as the square root of πΊ times π
P over π P over πΊ times π E over π E. We have a common factor of Big πΊ in the numerator and the denominator under the
square root, so these will cancel. And we have eliminated Big πΊ from the equation. We can also multiply the numerator and denominator by both π P and π E. This gives π P over π E equals the square root of π P times π E over π E times
π P.

Finally, we can multiply through by π E. And we now have the radius of the planet π P expressed entirely in quantities that
are given in the question, except one. We do not yet know the acceleration due to gravity on the other planet, π P. But we can find another expression for this from other quantities given in the
question using the kinematic or SUVAT equations.

Recall that these are a set of equations that link the quantities of displacement π ,
initial velocity π’, final velocity π£, constant acceleration π, and time π‘. Since the object is falling only a very short distance in comparison to the size of
the planet, we can assume that the acceleration due to gravity is constant for the
duration of its fall. The question gives us the displacement, 2352 centimeters; the initial velocity, zero,
because we can assume the object was at rest when the astronaut dropped it; and the
time taken, eight seconds. We do not have the final velocity, and we wish to find the acceleration.

The kinematic equation we need therefore is π equals π’π‘ plus one-half ππ‘
squared. Since the object starts from rest, π’ equals zero, so π’π‘ is also equal to zero. We can then rearrange for π by multiplying through by two and dividing by π‘
squared, giving π equals two π over π‘ squared. We can then replace π P in our equation for π P with this expression. And we can then simplify by multiplying the numerator and denominator by π‘ squared,
giving π P equals π E times the square root of π P times π E times π‘ squared
over two times π times π E. Now, π P is expressed entirely in quantities given by the question, and we can plug
these numbers in.

Substituting these values into the equation, we get 6.34 times 10 to the power of six
times the square root of 7.164 times 10 to the power of 24 times 9.8 times eight
squared divided by two times 23.52 times 5.97 times 10 to the power of 24. The 23.52 comes from expressing the distance for them, 2354 centimeters, in SI units
of meters. Calculating the square root, this comes to exactly four. Therefore, the planetβs radius is exactly four times that of the Earthβs, equal to
2.536 times 10 to the power of seven meters.

Letβs finish this video by recapping some key points. Two bodies exert gravitational forces on each other in the direction towards the
center of mass of the other body. The magnitudes of the forces are given by π
equals Big πΊ times π one times π two
over π squared, where π
is the force measured in newtons. π one and π two are the masses of the two bodies measured in kilograms. π is the distance between the centers of mass of the two bodies measured in
meters. And Big πΊ is the universal gravitational constant, approximately equal to 6.67 times
10 to the power of negative 11 newton-meters squared per kilogram squared.

The acceleration due to gravity of one body of mass π one towards another body of
mass π two is given by big πΊ times π two over π squared and is independent of
the mass of the first body, π one. For two bodies of masses π one and π two and radii π one and π two, the ratio of
their accelerations due to gravity is given by π one over π two equals π one over
π two times π two squared over π one squared.