Newton's Law of Gravity: Study with Video Lessons, Practice Problems & Examples

The universal law of gravitation, formulated by Newton, states that all objects with mass attract each other through gravitational forces. The gravitational force (F_G) can be calculated using the equation FG = G(m₁⁢m₂) / r2 , where G is the universal gravitational constant (6.67 × 10⁻¹¹ N m² / kg² ) and r is the distance between the centers of mass. Understanding this principle is crucial for calculating gravitational forces in various contexts, including satellites orbiting planets.

Universal Law of Gravitation

Hey, guys. So in this video, we're going to talk about the universal law of gravitation. So the story is that Newton was sitting under a tree one day, an apple fell and hit his head, and he realized that the force that causes the apple to fall to the earth is the same force that keeps the moon in orbit around the earth. So basically, what Newton figured out was that all objects in the universe attract each other. They all exert gravitational forces on each other. So for instance, in this diagram, I've got these 2 spheres, these 2 blue spheres here. If they both are massive, so they have mass 1 and mass 2, then the gravitational force between them is known as Newton's gravitational law. And that equation is G ( m 1 * m 2 ) r 2 where G, that capital G is just a number. It's known as the universal gravitational constant. It's 6.67 ∙ 10 -11 . Here are the units for that number. You may or may not need to know those. Just check with your professor if you're unsure. And that little r that's in the denominator there is the distance between the objects' centers of mass. So if you have these 2 massive objects here, and they are separated by some distance r, Newton's law of gravitation says that mass 2 will pull on mass 1 in this direction, but because of action-reaction and because everything attracts each other, mass 1 also pulls on mass 2 in that direction. The magnitude of those forces is given by this equation. And as for the direction, those forces are directed along a line that connects the two objects. So these forces always act on a line that connects the two objects' centers. And that's basically it. That's the whole equation. The last thing I want to mention is that this capital G here is known as a universal constant, which means anywhere you go in the universe that number is always going to be the same. Don't confuse that with little g like we've seen in forces and kinematics. That 9.8 meters per second squared . That's what's known as a local constant. A local constant means that anywhere on earth that you go, that number is going to be the same. But as soon as you go off of the earth or you go to a different planet, that number is going to change. So just make sure that you know the difference between those 2 and you don't confuse them. Alright, guys. That's basically it. Let's go ahead and check out this example I've got here. So I've got 2 30-kilogram spheres, and they're separated by 5 meters. So let me go ahead and draw that out. So I've got 2 spheres and there have a mass of 30 kilograms. So I've got mass 1 equals 30, and I've got mass 2 equals 30, and they're separated by a distance of 5 meters. So then I know that that is equal to 5, and I need to find out what the gravitational force is between them. So in other words, I need to find out what FG is equal to, right? So that's not a 6, that's supposed to be a G. Great. So we know the formula for Newton's law of gravitation, that's FG equals capital G, constant, that's just a constant, that's just a number, and I've got this mass here that's 30 kilograms. The other mass, so I've got m 1 and m 2, and I also have the distance between them, that's r equals 5 meters. So I have all of those variables. I can go ahead and plug this stuff into my calculator. So just setting everything up, I've got 6.67 ∙ 10 -11 , then I've got the 2 masses, 30 and 30, and I've got to divide them by the distance between them squared, that 5. And so to do this, just make sure I plugged in everything correctly, I actually have a little calculator that's going to go and help us. So I'm going to plug all of this stuff in. So I've got 6.67 ∙ 10 -11 , and then I'm going to multiply it by the 2 masses, 30 and 30. And now we've just got to divide it by, in parentheses, the distance squared. Make sure the denominator is in parentheses because you don't want to mess up the order of operations. So if you go ahead and plug that in, we get the correct answer, which is that the force of gravity equals 2.4 ∙ 10 -9 , and the unit for that is in newtons. So that's basically the gravitational force between these two spheres. Let me know if you guys have any questions and if not, we're just going to keep moving.

 Two spheres are separated by 10m. If the lighter 40kg sphere feels a gravitational force of 1.6 × 10^-9 N, what is the mass of the heavier sphere?

• A. 6 kg
• B. 9.6 kg
• C. 60 kg
• D. 10 kg

Where is Net Force on middle object zero?

Alright, guys. Let's check out this problem. It is a classic one in gravitation, so we're going to work it out together. So we've got 2 spheres, and we're going to position them along a line like this. I've got 2 spheres like this, I know what the masses are and I know the distance between them. So, I'm going to call the left mass the 10-kilogram mass, mass a, and the 25-kilogram one, mass b. And I know what the distance between them is. I'm told that the distance between them is 5 meters, that's little r. So the whole point of this problem is that I'm going to position a mass somewhere along the line between them, and I know what the I'm just going to call that mass, mass c. And I don't actually know what that mass c is, but I want to figure out where I have to place it so that the gravitational forces, the net gravitational force, is equal to 0. So it means I need to find out what this distance is, and I'm going to call that r_a. So that's my target variable, really. That's what I'm trying to solve for. So I've got r_a as my target variable and I want the net gravitational force to equal 0. Well, we know that Newton's law or Newton's gravitational law says that there are forces between any two objects. So there is a force from this guy, which I'm going to call F_AC, and there's a force from that guy over there, and that's F_BC. And I know how to express that as the equation. And so we want these things to basically cancel out, which means that the magnitude of F_AC needs to be the magnitude of F_BC, but they have to be equal and opposite in direction. So I'm actually going to write out the gravitational forces for those. We're going to start off with Newton's law of gravity. So I've got gm_am_c/r_a2. And then over here, you've got gm_bm_c/r_b2. Where that r_b is just the distance right here between m_a or m_c and m_b, so that's r_b. Alright? Cool. So we can take a look at this equation. We can actually cancel some terms out. I noticed that g pops up on both sides and then also the m_c also pops up on both sides. So it's actually good. We didn't need to know the mass of that center thing in there. And then the r_a is my target variable. Now I know what mass a and mass b both are, so all I need to do is just figure out what this r_b is. That's my only unknown variable. I'm not told what that distance is. So I actually have to go here and figure it out. So what is the distance between this thing and m_b? Well, we don't know what the individual distances are between r_a and r_b. We don't know what those things individually are, but we do know that the whole entire distance between the two spheres is 5. So we can come up with another equation for this. We know that r_a + r_b is equal to 5. So now what happens is, in this equation, where we've come up with 2 unknowns, we've got this other equation that we're going to use to help solve it. So if I can figure out what r_b is, then I can plug it back into that equation. r_b is just, if I move this guy over, just going to be 5 - r_a. So now I'm just going to substitute this equation back into that guy right there, and now we're only going to have one variable. So now we've got mass a divided by r_a² equals mass b divided by (5 - r_a)². Notice how we've gone from 2 unknown variables to now only 1, and that's the one that I need to find out. So because I need to figure out what that variable is, I want to start getting everything over to one side. So what I'm going to do is I'm going to take this expression right here and move it up to the other side on the left and then the m_b is going to come down and trade places with it. So when I do that, I get (5 - r_a)² / r_a² equals m_b / m_a. Now I actually know what these two masses are, right? This mass b was equal to 25, so 25. And I've got this mass a was equal to 10. So how do we get rid of this whole, like, squared thing? Right? We've got this, like, expression here that's squared. You might be tempted to, like, foil it out and start like basically multiplying everything out, but that's actually going to be way too complicated. Notice how both of these things on the top and bottom are both squared. So we can actually take the square root of both sides. So don't go don't don't go ahead and foil it because it's actually going to be, like, way more complicated. So instead, what we can do is basically square roots_of_i sides of this thing. And when we do that, we get that the squares cancel, 5 - r_a / r_a. And then this is just a number. Right?

Two spheres of mass 300 kg and 500 kg are placed in a line 20 cm apart. If another sphere of mass 200 kg is placed between them, 8 cm from the 300 kg sphere, what is the net gravitational force on the 200 kg sphere?

• A. -7.32×10^-4 N (toward 300kg sphere)
• B. -1.62×10^-4 N (toward 300kg sphere)
• C. 7.32×10^-4 N (toward 500kg sphere)
• D. 1.62×10^-4 N (toward 500kg sphere)

Center of Mass Distance

Hey, guys. So we saw from the universal law of gravitation that we could calculate the force between 2 objects. Now if these objects are relatively small, like we have here in this diagram, we call those objects point masses because we can treat them as points. So in this case, if these point masses have mass 1 and mass 2 and they are separated by some distance r, the universal law of gravitation says that we could calculate the gravitational force between these two things. And that's given by this equation up here, F=Gm1m2r2. But let's say we have a different situation in which we have something huge like a planet, and then we're calculating the force on a person that's orbiting some distance above that. So the universal law of gravitation still says that we need G times the product of both of the masses. But in this situation, we're working with large objects. Instead of using m ones and m twos, the big object, it's a capital M, and then you'll see the small object has a lowercase m. So it's still the product of the 2 masses, big M and little m over r squared. This is really the same equation. It's the same thing. I'm not telling you anything new. It's just for a separate, situation.

Okay, so what is this little r distance between the centers of mass? For point masses, it was just this distance between the two objects. But in this situation, in this diagram here, it's a little bit more complicated. Because now if we want to get from the center of mass of the earth to the center of mass of the astronaut, because this little r is the center of mass distance between the two objects. First, we have to go from the core or the center of the Earth out to the surface and then we have to go some extra distance here above that surface. So really, the whole center of mass length is this little r here and that's made up of 2 lengths. The first length from the core out to the surface is called the radius. If you think about the earth just a giant sphere, then the core to the surface is capital R, which is the radius of the earth. And this extra distance here above the surface is called the height. So that little h is called the height. And we can see from this diagram that this little r is really just the sum of both of the radius and the height. So for large objects, the little r is gonna be capital R+H.

Now, I want to point out that I have 2 different kinds of variables here or 2 different kinds of letters. I have an uppercase letter or a capital letter and a lowercase letter. So in physics, a capital letter is always used to represent constants. So for instance, the radius of the earth, a constant, never changes. Whereas this lowercase letter h here represents a variable because you can go any distance away from the surface of the earth, but the radius always stays the same. So that's basically it, guys. Let's go ahead and start working out this example. So in this example, we have the height above the earth. If we're asked at what height above the earth is the gravitational force on a satellite equal to 1000 newtons, let's go ahead and draw a little diagram here and figure out what's going on. So we have the earth represented as that sphere, and I'm just gonna go ahead and represent the satellite as a dot. Don't wanna show you my bad drawing skills.

Okay. So we have, let's see. We've got the force of gravity, so we got the force, the gravitational force is equal to 1000 newtons, and we have the mass of the satellite is equal to 1000 kilograms. And because we're working with the gravitational force here, I wanna go ahead and write out that equation. So I've got Fg=GMmr2. Remember, because we're working with a planet and a small mass, we're going to use that. And really, what am I looking for? Well, I'm actually looking for the height above the surface, so I'm looking for H. But I'm gonna use this gravitational force to solve for that.

Okay, so what you might be tempted to do is replace this formula with r +h, capital R +H squared, and then use that to solve for little h. But I want to warn you against doing that because you're actually going to make the math a little bit more complicated than it needs to be. So instead of doing this here, I've got a pro tip for you guys. If you ever have a problem that asks you to solve for capital R or H, first go ahead and solve for little r first using Newton's law of gravity, and then using this equation, you can solve for whatever you want. So don't do this. Instead, what you're gonna do is gonna solve for Fg=GmM1r2. Go ahead and solve for little r, and then we can use this equation, r equals big R+h, in order to figure out the variable that we're looking for. In this problem, we're looking for this H variable here. So let's go ahead and solve for that gravitational force and see if we can find the little r distance.

Okay. So let me go ahead and write out all of my knowns here. Right. So we've got the Well, in this diagram, we've got the center of the earth. And then if I wanted to find the little r distance, that's going to be 2 things. I've got the radius of the earth and then I've got a height above the center. So that height is really what I'm looking for. And I've got this little r here, that is that distance. We're going to be solving for that first. Okay. So I've got that the mass of the satellite is equal to 1000. What about capital M? Because we need to know what capital M is. Well, I've got I've got G, which is the gravitational constant. I've got capital M, which is actually given over here, the mass of the earth. Remember, that's a capital letter, so it's a constant. I have the mass of the satellite, and I also have the gravitational force between them. So I can go ahead and use and solve for little r. Let me go ahead and write all that stuff out. So I've got that capital M is equal to 5.97 times 10 to the 24th. And, I know what G is, and then Yeah. That's basically it. Awesome. So if I go ahead and rearrange this equation right here, so this, this gravitational equation, I can come up with this expression. R squared, if I move that to the other side and then move the Fg down, is equal to G times capital M, lowercase m over the force of gravity. So because this is a square, I can take the square root of both sides, and I'm gonna get that r equals the square root of Gmlittlem over the force of gravity. I'm gonna go ahead and start moving this over here. So I can actually just go ahead and start plugging in values for this. So I've got 6.67 times 10 to the minus 11. Then I've got the mass of the earth, 5.97 times 10 to the 24th. And then I've got the mass of the satellite, which is 1000. And then I've got the force of gravity, which

is also 1000. If you go ahead and plug all this stuff into your calculator, you should notice that, well, you should get 2 times 10 to the 7th meters. So we're done, right? Well, no, because remember this number here only represents the full center of mass distance, not the H, which is what we're really looking for. So our last step is we're basically just gonna have to solve, using the r equals R +H equation. So if I wanted to figure out what H is, I can go ahead and use this equation and figure out that H is equal to little r minus big R, which is the radius of the earth. But what is that value? What is that capital R? We haven't been given a value for that yet. Well, if I look here at my gravitational constants, that capital R is just represents the radius of the earth, which is given right here as this number. So I've got that for my final answer, I've got H is equal to little r, which is 2 times 10 to the 7th, minus big R, which is 6.37 times 10 to the 6th, and that's gonna be in meters. So if you go ahead and work this out for the final answer, you get 1.36 times 10 to the 7th, and that's gonna be in meters. So that's about 13,600 kilometers above the surface. So that is the answer for this. Let me know if you guys have any questions with this.

Find force between spheres with radii

Okay, guys. Let's check this one out. So we have 2 identical spheres. Let me go ahead and draw out these spheres right here. And we're told the masses of those 2 spheres, and we're told that their surfaces are just barely in contact with each other. So we go ahead and write this stuff out. So I've got \( m \) equals 10 kilograms for \( m_1 \), and \( m_2 \) also equals 10 kilograms. Right? I'm also told that these spheres have a physical size which the diameter is equal to 60 centimeters, which is 0.6 meters. Right? Just be careful that's in centimeters there.

Okay. So in order to figure out what the gravitational attraction is, let's start off with Newton's law of gravitation. We have \( F_g = G \times (\frac{{\text{the mass of both objects}}}{\text{the center of mass distance between them}})\). But if these two things have a physical size, what's that center of mass distance? Well, remember if you have 2 objects here, that the gravitational force always acts between the centers of mass and it's, proportional or it's it's divided by the center of mass distance between them. So the center of mass between these two objects, little \( r \), is actually just 2 times the radius of each individual sphere, since they are identical. Right? So that's just 2 times the 2 times the radius, which is the diameter of the sphere, which is just 0.6 meters. So now we have big \( G \), we have both masses and we have little \( r \). Let's go ahead and just plug everything in. Right?

So we've got \( F_g=6.67 \times 10^{-11}, \text{we've got both spheres' mass } 10 \times 10, \text{and then we've got the center of mass distance } 0.6 \text{, and that's gonna be squared}\). Go ahead and plug that in and you should get \( 1.85 \times 10^{-8} \) Newto_ns.

Alright, guys. Let me know if you have any questions about this.

A 2,000-kg spacecraft is blasting away from the surface of an unknown planet the same size as the Earth. At 1500km above the surface, an instrument onboard reads the gravitational force to be 18000 N. What is the planet's mass?

• A. 8.36×10²⁴kg
• B. 2.89×10¹²kg
• C. 1.67×10²⁸kg
• D. 5.78×10¹⁵kg

Here’s what students ask on this topic:

What is Newton's Law of Universal Gravitation?

Newton's Law of Universal Gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula is given by:

$F_G=G\frac{m_1{m}_2}{r^2}$

where $G$ is the universal gravitational constant (6.67 × 10⁻¹¹ N m² / kg²), $m_1$ and $m_2$ are the masses of the objects, and $r$ is the distance between their centers of mass.

How do you calculate the gravitational force between two objects?

To calculate the gravitational force between two objects, use Newton's Law of Universal Gravitation:

$F_G=G\frac{m_1{m}_2}{r^2}$

Here, $G$ is the universal gravitational constant (6.67 × 10⁻¹¹ N m² / kg²), $m_1$ and $m_2$ are the masses of the objects, and $r$ is the distance between their centers of mass. Plug in the values for the masses and the distance to find the gravitational force.

What is the difference between the universal gravitational constant (G) and the local gravitational constant (g)?

The universal gravitational constant, $G$, is a constant value of 6.67 × 10⁻¹¹ N m² / kg² that applies everywhere in the universe. It is used in Newton's Law of Universal Gravitation. The local gravitational constant, $g$, is approximately 9.8 m/s² and represents the acceleration due to gravity at the Earth's surface. While $G$ is universal, $g$ can vary depending on the location, such as on different planets or altitudes.

How does the distance between two objects affect the gravitational force between them?

The gravitational force between two objects is inversely proportional to the square of the distance between their centers of mass. This relationship is described by Newton's Law of Universal Gravitation:

$F_G=G\frac{m_1{m}_2}{r^2}$

As the distance $r$ increases, the gravitational force $F_G$ decreases rapidly. For example, if the distance between two objects is doubled, the gravitational force between them becomes one-fourth as strong.

What is the significance of the universal gravitational constant (G)?

The universal gravitational constant, $G$, is a fundamental constant in physics that quantifies the strength of the gravitational force between two masses. Its value is 6.67 × 10⁻¹¹ N m² / kg². This constant is crucial for calculating gravitational forces in various contexts, from the attraction between small objects to the gravitational pull between celestial bodies. It ensures that the gravitational force can be consistently calculated regardless of the masses involved or their separation distance.