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Lesson Video: Newton’s Law of Universal Gravitation Mathematics

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

21:44

Video Transcript

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

Two masses, π‘š one and π‘š two, for example, the Earth and the Moon, separated by a distance π‘Ÿ, 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‘Ÿ. This force is a clear example of Newton’s 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.

Newton’s 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 𝐅, equal in magnitude on both bodies, is equal to 𝐺 times π‘š one times π‘š two over π‘Ÿ squared, where π‘š one is the mass of one body, π‘š two is the mass of the other body, π‘Ÿ is the distance between the bodies’ centers of mass, and 𝐺 is the universal gravitational constant. The universal gravitational constant, sometimes referred to as β€œBig 𝐺,” 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.

Let’s 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 Newton’s law of universal gravitation states that two bodies of masses π‘š one and π‘š two will attract each other with a force of magnitude 𝐺 times π‘š one times π‘š two over π‘Ÿ squared, where π‘Ÿ is the distance between their centers of mass and 𝐺 is the universal gravitational constant.

In this case, therefore, we have Big 𝐺, 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 𝐺 and the masses are in SI base units of newtons, meters, and kilograms, we divide by the distance π‘Ÿ 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.

Let’s 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 𝐺 equals 6.67 times 10 to the negative 11 newton-meters squared per kilogram squared.

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

Now inserting the values for all of the variables and constants, for Big 𝐺, we have 6.67 times 10 to the power of negative 11. The masses π‘š one and π‘š two are given in the question as 4.6 and 2.9 in SI units of kilograms. And the force 𝐅 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.

Let’s 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 Earth’s 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 Newton’s law of universal gravitation states that two bodies of masses π‘š one and π‘š two will attract each other with a force of magnitude 𝐺 times π‘š one times π‘š two over π‘Ÿ squared, where π‘Ÿ is the distance between their centers of mass and 𝐺 is the universal gravitational constant. By Newton’s 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 𝐺 is the universal gravitational constant, 6.67 times 10 to the power of negative 11. We can let π‘š one be the mass of the Earth, six times 10 to the power of 24, and π‘š two be the mass of the satellite, 2415. π‘Ÿ squared will require a little more thought. We are given the distance from the satellite to the Earth’s surface and the Earth’s radius. We are not given π‘Ÿ, 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‘Ÿ E, and the distance between the satellite and the surface of the Earth, 𝑑.

It’s 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 𝐺 times π‘š one times π‘š two over π‘Ÿ E plus 𝑑 all squared. So, we have Big 𝐺, 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‘Ÿ E, given as 6360 kilometers, so in SI unit of meters this is 6360 times 10 to the third power, plus the distance 𝑑 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 Newton’s 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: 𝐅 equals π‘š one times π‘Ž, where π‘š 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 Newton’s universal law of gravitation: 𝐅 equals 𝐺 times π‘š one times π‘š two over π‘Ÿ 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‘š one times π‘Ž equals 𝐺 times π‘š one times π‘š two over π‘Ÿ squared. Since a mass will always be nonzero, we can divide through by the common term of π‘š one on both sides. This gives us the acceleration due to gravity of the body: π‘Ž equals 𝐺 times π‘š two over π‘Ÿ 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‘š on a body whose center of mass is a distance π‘Ÿ from it. And this is given by little 𝑔 equals Big 𝐺 times π‘š over π‘Ÿ squared.

Notice that this is in fact the same expression for the acceleration due to gravity mentioned previously. This is because they’re in fact the same quantity. It’s just the other side of the equation of Newton’s 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 𝑔 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 𝑔 equals 9.81 meters per second squared to two decimal places, the standard acceleration due to gravity on the surface of the Earth.

Let’s 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 planet’s 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‘Ž equals 𝐺 times π‘š over π‘Ÿ squared, where π‘š is the mass of the sphere, π‘Ÿ is the radius of the sphere, and 𝐺 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‘Ž E as 𝐺 times the mass of the Earth π‘š E divided by the radius of the Earth π‘Ÿ E squared.

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

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

Let’s 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 𝑔 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‘Ž equals Big 𝐺 times π‘š over π‘Ÿ squared, where Big 𝐺 is the universal gravitational constant, π‘š is the mass of the sphere, and π‘Ÿ is the radius of the sphere. An important thing to note is that the question does not give us the value of Big 𝐺 to use. We could use our knowledge of the constant’s value to solve the problem, but we don’t actually need to. We can rearrange this equation for π‘Ÿ, the quantity we want to find, by multiplying through by π‘Ÿ squared, dividing through by π‘Ž, and taking the square root.

Again, we are not given the value of Big 𝐺, 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€” let’s call this π‘Ÿ P β€” and divide it by the expression for the radius of the Earth, which we’ll call π‘Ÿ E, we obtain π‘Ÿ P over π‘Ÿ E equals the square root of Big 𝐺 times π‘š P over π‘Ž P all divided by the square root of Big 𝐺 times π‘š E over π‘Ž E.

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

Finally, we can multiply through by π‘Ÿ E. And we now have the radius of the planet π‘Ÿ 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‘Ž 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 𝑠, initial velocity 𝑒, final velocity 𝑣, constant acceleration π‘Ž, and time 𝑑. 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 𝑠 equals 𝑒𝑑 plus one-half π‘Žπ‘‘ squared. Since the object starts from rest, 𝑒 equals zero, so 𝑒𝑑 is also equal to zero. We can then rearrange for π‘Ž by multiplying through by two and dividing by 𝑑 squared, giving π‘Ž equals two 𝑠 over 𝑑 squared. We can then replace π‘Ž P in our equation for π‘Ÿ P with this expression. And we can then simplify by multiplying the numerator and denominator by 𝑑 squared, giving π‘Ÿ P equals π‘Ÿ E times the square root of π‘š P times π‘Ž E times 𝑑 squared over two times 𝑠 times π‘š E. Now, π‘Ÿ 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 planet’s radius is exactly four times that of the Earth’s, equal to 2.536 times 10 to the power of seven meters.

Let’s 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 𝐅 equals Big 𝐺 times π‘š one times π‘š two over π‘Ÿ squared, where 𝐅 is the force measured in newtons. π‘š one and π‘š two are the masses of the two bodies measured in kilograms. π‘Ÿ is the distance between the centers of mass of the two bodies measured in meters. And Big 𝐺 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‘š one towards another body of mass π‘š two is given by big 𝐺 times π‘š two over π‘Ÿ squared and is independent of the mass of the first body, π‘š one. For two bodies of masses π‘š one and π‘š two and radii π‘Ÿ one and π‘Ÿ two, the ratio of their accelerations due to gravity is given by π‘Ž one over π‘Ž two equals π‘š one over π‘š two times π‘Ÿ two squared over π‘Ÿ one squared.

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