Sign In | Subscribe

Enter your Sign on user name and password.

Forgot password?
  • Follow us on:
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of AP Physics 1 & 2
  • Discussion

  • Study Guides

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Lecture Comments (10)

1 answer

Last reply by: Professor Dan Fullerton
Mon May 23, 2016 9:38 AM

Post by El Einstein on May 21, 2016

In example 5, Im slightly confused on the second part of this question "what speed does the shuttle travel to maintain this orbit?". And now my question is, Why did you use the mass of the earth instead of the mass of the shuttle?
On the first part of the question. My logic was, since we are finding the gravitational field strength DUE TO THE EARTH then we must not cancel out the mass of the earth when equating both gravitational forces. So i used this logic with the second part of this question and then got stuck because the mass of the shuttle was not given. So again, Why did you use the mass of the earth instead of the mass of the shuttle? to find the speed of the shuttle. I hope this makes sense.

1 answer

Last reply by: Professor Dan Fullerton
Sat Nov 15, 2014 11:07 PM

Post by Scott Beck on November 15, 2014

Hi on example 10 where the force of gravity was calculated

(6*10^24)(3*10^4)^2/1.5*10^11= if we were to use the laws of significant figures would the answer be 4*10^22 N? How much should we round on the AP exam?

1 answer

Last reply by: Professor Dan Fullerton
Mon Jun 2, 2014 7:30 PM

Post by Thivikka Sachithananthan on June 2, 2014

Hello Prof Fullerton,
Can you help me with this question: When two other solid spheres of radius 3R that are made of this same steel are placed in contact, what is the
magnitude of the gravitational force each exerts on each other?

1 answer

Last reply by: Professor Dan Fullerton
Wed Jan 29, 2014 7:18 AM

Post by Roy Wayne Aipperspach on January 29, 2014

Thank you so much for you lectures Prof. Dan.

It helped me realized how important, though simple, was UNIVERSAL GRAVITATION.

I will finish your course.

From Philippines

1 answer

Last reply by: Professor Dan Fullerton
Wed Jun 5, 2013 6:07 AM

Post by Jay Gill on June 4, 2013

Your lectures are fantastic for the following reasons:

(1) Very concise
(2) Right to the point
(3) enjoy the humor tidbits
(4) You seem passionate about teaching

I am writing the Mcat in two weeks and I am using your lectures for review.

What have your lectures done for me?

After watching your lectures I now don't feel the need to memorize a plethora of equations, I actually have a strong understanding of the concepts and my formula recall is almost perfect. What helped from your lectures the most was how you began with simple equation applications and then provided graphical interpretations.

Thanks Dan....your a physics stud in my eyes haha


  • Newton's Law of Universal Gravitation describes the gravitational force of attraction between any two bodies with mass. Fg=G*m1*m2/(r^2)
  • Gravity can only attract, not repel.
  • Gravitational field strength describes the force of gravity a mass would feel in the vicinity of another mass due to gravitational forces. It also describes the acceleration the object would undergo due to the gravitational force.
  • Gravitational Potential Energy describes the energy an object has due to its position in a gravitational field. Generally, PE=G*m1*m2/r, or, in a constant gravitational field, PE=mgh. The zero point of potential energy is set at an infinite distance from all masses.
  • Objects in orbit remain under the influence of gravity.
  • Objects in orbit are not weightless. Rather, they are in a specific type of free fall such that they are constantly falling.


Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:08
  • Universal Gravitation 0:29
    • The Bigger the Mass the Closer the Attraction
    • Formula for Gravitational Force
  • Calculating g 2:43
    • Mass of Earth
    • Radius of Earth
  • Inverse Square Relationship 4:32
  • Problem Solving Hints 7:21
    • Substitute Values in For Variables at the End of the Problem Only
    • Estimate the Order of Magnitude of the Answer Before Using Your Calculator
    • Make Sure Your Answer Makes Sense
  • Example 1: Asteroids 8:20
  • Example 2: Meteor and the Earth 10:17
  • Example 3: Satellite 13:13
  • Gravitational Fields 13:50
    • Gravity is a Non-Contact Force
    • Closer Objects
    • Denser Force Vectors
  • Gravitational Field Strength 15:09
  • Example 4: Astronaut 16:19
  • Gravitational Potential Energy 18:07
    • Two Masses Separated by Distance Exhibit an Attractive Force
    • Formula for Gravitational Field
  • How Do Orbits Work? 19:36
  • Example5: Gravitational Field Strength for Space Shuttle in Orbit 21:35
  • Example 6: Earth's Orbit 25:13
  • Example 7: Bowling Balls 27:25
  • Example 8: Freely Falling Object 28:07
  • Example 9: Finding g 28:40
  • Example 10: Space Vehicle on Mars 29:10
  • Example 11: Fg vs. Mass Graph 30:24
  • Example 12: Mass on Mars 31:14
  • Example 13: Two Satellites 31:51

Transcription: Gravitation

Hi, folks. I am Dan Fullerton and I would like to welcome you back to

Today's lesson -- gravity and gravitation.0004

Our objectives are going to be to utilize Newton's Law of Universal Gravitation to determine the gravitational force of attraction between two objects.0007

We are going to determine the acceleration due to gravity near the surface of the earth, calculate gravitational field strength and explain apparent weightlessness for objects in orbit.0016

So with that, why not dive right in?0027

Universal gravitation -- All objects that have mass attract each other with a gravitational force.0029

For example, right now you are attracted to me.0036

Yes, I know, that is kind of creepy, but any two objects that have mass, no matter how far apart they are, all have some level of attraction.0039

The bigger the masses, the more the attraction and the closer the masses are to each other, the closer the attraction, which is why we have a very, very, very, very tiny amount of attraction between us at the moment...0047

...or probably a long way away, our masses are relatively small and there is not much gravitational force there.0059

Between you and the earth, for example, the earth has a very big mass and you are relatively close to it, so you have a very measurable gravitational force of attraction there.0065

If we wanted to look at this in terms of our math, the force of gravity is gm1m2/r2 in the direction of our hat and the negative just says that it is an attractive force.0076

If we have one object over here -- let us call this mass 1 -- over here, we have some object, mass 2, and the distance between their centers of mass, we are going to call (r).0090

In this case, (r) is not specifically a radius; it is a distance between the two centers of mass.0094

Then you are going to have a force of Object 2 on Object 1 and you are going to have a force of Object 1 on Object 2 and they will be equal in magnitude and opposite in direction.0109

We know that because of Newton's Third Law.0124

If we wanted to get just the magnitude of the force, which is typically how this relationship is used, we say that the force of gravity is equal to mass 1 × mass 2/r2.0128

If you do that and your masses are in kilograms and your distance is in meters, the units do not work out to anything overly useful, so we put in this fudge factor, this universal gravitational constant (G). 0152 That is equal to 6.67 × 10-11 Nm2/kg2.0139

It is there to make the units work out.0158

So, how do we calculate g, the acceleration due to gravity?0162

Let us see if we cannot use what we know to find out.0168

The mass of the earth is approximately 6 × 1024 kg and its radius is about 6.38 million meters.0171

The force of gravity -- we typically write -- is mg in a constant gravitational field.0179

Universal Gravitational Law says that G times the first mass times the second mass divided by the square of the distance between them.0186

Over here, we are assuming the second mass is already the mass of the earth.0194

Let us rearrange this a little bit.0200

What we can do is realize that we have the mass of the object here.0202

That is a mass of the object, so we are left with the mass of the earth, therefore g equals G, that constant, times the mass of Object 2, the earth, divided by the square of the distance between the objects between their centers of mass.0207

Therefore, g = 6.67 × 10-11Nm2/kg2 × the mass of the earth -- 6 × 1024 kg divided by the square of the distance between them -- 6.38 × 106 m -- roughly the radius of the earth.0223

Do not forget to square that. That is a big mistake that students make.0244

Go through that and you should get an answer right around 9.8 m/s2 or 9.8N/kg; the units are equivalent.0247

Of course, that is what we expect. That is the acceleration due to gravity we have been using here on earth.0259

For the AP test, we typically round that to 10 to make the math a little simpler but you can see that it works out.0264

The force of gravity decreases with the square of the distance between the centers of the masses. 0277 This is called an inverse square law.0273

The force of gravity is gm1m2/r2.0279

We are going to see lots of relationships in physics that have this inverse square relationship based on the distance between them.0281

As the distance gets bigger, the force gets smaller and it gets smaller by the square of that distance between the objects.0289

Graph of force versus distance is distance gets bigger -- the force tails off very, very quickly. 0304 That distance is an important factor because it is squared.0298

So this graph would be proportional to 1/r2.0308

So then the question then, what happens to the force of gravity if you double the distance from the centers of mass?0313

Let us take a look at how we could answer that.0340

If the initial force of gravity Fgiinitial is gm1m2/r2, the final is going to be gm1m2 over...0344

We are going to double that distance, so this becomes 2 times whatever our initial was squared and that becomes gm1m2/initial r2, but the 2 is squared there too over 4.0361

So if you rewrite this a little bit, you could write this as 1/4 × gm1m2/r2, but notice this is the force of gravity initial.0383

So the final gravitational force is 1/4th the initial gravitational force.0402

If you double the distance, you get 1/4th the gravitational force.0412

If you halve the distance, you get 4 times the gravitational force.0417

If you triple the distance between objects -- 1/9th the gravitational force.0420

If you cut the distance between them into 1/3rd -- 9 times the gravitational force.0426

Whatever that factor is that you change the distance by, you square it in order to find out what happens with that new force.0430

Here are some problem-solving hints as we go through a lot of these gravity problems.0442

Try and substitute values in for variables at the end of the problem only.0446

Because you oftentimes have some pretty unwieldy numbers, the longer you can keep the formula in terms of variables, the fewer opportunities there are to make mistakes.0450

Secondly, before using your calculator to find an answer, it is oftentimes valuable to try and estimate the order of magnitude of the answer.0458

We will have to go through and calculate the whole thing but try and get a guess as to roughly where your answer is going to be and that way, if you make a goofy calculator error, it is pretty easy to pick up.0470

Finally, once your calculations are complete, take a second to make sure your answer makes sense by comparing your answer to some sort of known or similar quantity where you can.0475

If your answer does not make sense, stop, take just a second and see if you made a goofy calculator error or math mistake because lots of the problems I see are not with the physics here, it is with making goofy mistakes on calculators and calculations.0483

Example 1 -- What is the gravitational force of attraction between two asteroids in space if each has a mass of 50,000 kg and they are separated by a distance of 3800 m?0500

The force of gravity -- we are going to worry about the magnitude -- is equal to gm1m2/r2, where g is 6.67 × 10-11Nm2/kg2...0513

...that is given to you for the exam -- × the first mass, m1 (50,000 kg) × the second mass, also 50,000 kg divided by the square of the distance between their centers of mass, 3800 m2.0527

When I go through and do this, I get an answer of around 1.15 × 10-8N.0544

Why so small a force? You need a very, very, very big mass in order to have an appreciable gravitational force.0555

If we wanted to take this problem and do a quick order of magnitude estimation -- just to show you how you have done that -- what I do, is I would look at this expression here and try and estimate it quickly.0564

We have 10-11. We have -- that is something times 104, so I would say times 104, that is times 104 divided by...0576

...Well, those are 103, so 1032. Okay, 108. Then, 10-11, 10-3.0586

And you have 106 down here, so I would say you are roughly talking in the order of magnitude of something in the 10-9 and look you are only off by a factor of 10.0597

You are in the ballpark. You probably did not make a really goofy calculator error.0607

So that is how I would do an order of magnitude estimation here.0611

Example 2 -- Meteor and earth -- As a meteor moves from a distance of 16 earth radii to a distance of 2 earth radii from the center of earth, the magnitude of the gravitational force between the meteor and the earth becomes...0618

We have a couple of different solutions to choose from.0630

The biggest problem I see students having with questions like this has to do with reading the question and understanding what it is talking about.0634

Let us draw Earth here.0642

The meteor starts at a distance of 16 earth radii away, so it is going to be way over there.0643

There is its initial position -- 16 r's away.0650

Now if this is one (r) right there, then when it is 2 earth radii away from the center of the earth, there is 1 (r), there is the second (r), so it is moving from 16r to 2r.0656

The distance (r) is going from 16r to 2r -- the distance is 1/8th -- that's great.0668

The first thing I do here, say, is the force going to get bigger or smaller? As it gets closer together, you expect a bigger force. Right away, we can make answer 1 go away.0677

Because we have got that inverse square law with distance, our factor is not going to be 1/8th, it is going to be 1/8th squared, which is 164 and we said this is going to be bigger.0690

Because the distance is in the denominator, it is going to be 64 × that's great.0701

Another way you could do this is you could say the initial gravitational force is gm1m2/16 r2, which will be gm1m2/256r2.0709

What I am going to do is I am just going to take this gm1m2/r2 and I am going to call that x.0728

So my initial force is going to be 1/256x.0734

Now the final gravitational force is gm1m2/2r2 which is gm1m2/4r2.0741

I am going to pull the same trick again and call that x. So that is 1/4th x.0757

If we want to know the ratio then -- what happens -- we will take the final gravitational force over the initial gravitational force, which is 1/4th x/256x or 256/4 which is a factor of 64 times larger.0764

Which diagram best represents the gravitational forces (Fg) between a satellite (s) in the earth?0780

First thing -- gravity only attracts, it never repels.0801

So over here in number 1, the satellite is being attracted, but earth is being repelled.0804

Nope, that does not work.0809

Number 2 -- they are both being repelled.0811

Number 3 -- they are both being attracted -- that is looking promising and they are both being attracted with the same force.0813

Even more promising, Newton's Third Law says that the force on one must be equal in magnitude to the force on the other just opposite in direction, so 3 must be our answer.0816

Let us talk for a minute about gravitational fields.0831

Gravity is what is known as a non-contact or a field force.0834

We cannot see it. We cannot go touch it. We cannot detect it with a special scope.0838

We just know it is there by putting an object there and then seeing what happens to it -- observing the force on some test particle that we would put out in space to see if there is a field there.0844

The closer objects are to large masses, the more gravitational force they experience and the denser the force vectors, as shown here, the force that you would see on a test object, the stronger the gravitational force.0854

So we could say that the gravitational field is weaker the further away you are if the lines are less dense and stronger as you get closer, where the lines are closer together.0867

Now, you can use that the gravitational force or the weight of an object is mg when you are close to earth -- where the change in the radius is negligible or really what we are talking about is a constant gravitational field strength.0879

Universally, this one always works -- gm1m2/r2, which is why it is called Newton's Law of Universal Gravitation.0899

Going a little bit further into this gravitational field strength concept, if the magnitude of the gravitational force is gm1m2/r2 and that is equal to m1g, assuming that we do not have a big change in that distance -- that we are in a constant gravitational field...0909

...then in that instance, we could take a look and say that g therefore must equal gm2/r2 and the units of that are going to be N/kg or m/s2.0927

This is what we call gravitational field strength.0945

Wait -- you might say -- We have been calling g the acceleration due to gravity.0952

Yes, they are the same thing.0957

N/kg, m/s2, gravitational field strength, acceleration due to gravity -- they are the same thing, just different ways, different approaches of looking at the same phenomenon.0959

So those are equivalent -- the acceleration due to gravity and gravitational field strength.0971

Let us take a look at an example.0979

Suppose we have 100 kg astronaut feeling a gravitational force of 700N when placed in the gravitational field of a planet. What is the gravitational field strength at the location of the astronaut?0981

The force of gravity is mg, therefore, we could find gravitational field strength -- the force of gravity divided by the mass or 700N/100 kg, should be 7N/kg or 7 m/s2.0996

What is the mass of the planet if the astronaut is 2 × 106 m from its center?1018

To do that, let us go to the Universal Law of Gravitation -- Fg = gm1m2/r2.1023

If we want the mass of the planet, that is going to be the force of gravity times the square of the distance between their centers of mass divided by G times the mass of our astronaut.1032

Our force is 700N.1048

Our distance is going to be 2 x 106 -- do not forget to square that-- divided by G, 6.67 × 10-11Nm2/kg2 × the mass of our astronaut, 100 kg...1051

...therefore, I come out with a mass of the planet of about 4.2 × 1023 kg.1067

Now, what happens if we talk about gravitational potential energy?1085

Two masses separated by some distance exhibit an attractive force on each other.1091

They want to move closer together because that gives them gravitational potential energy.1095

In a uniform gravitational field, the gravitational potential energy can be found by mg -- the weight of the object times the height, and we will talk about that more when we get to energy and work and a couple of other topics.1101

If the height is varying significantly to where we are not looking at a uniform gravitational field, we need something more general, a Universal Law for Gravitational Potential Energy.1112

That is -gm1m2/r. What does that minus mean?1123

Typically, we assume that potential energy equals 0 when you are infinitely far away from all other objects -- a long, long, ways away, you do not have any other influences.1130

Practically, you cannot get there; theoretically, you can.1139

If you were to take -- and we have a planet here and we have an object infinitely far away and we bring it closer and closer and closer and closer and closer, it wants to get sucked in -- gravity attracts.1143

If it had 0 potential energy way out there -- well, to get it back to the point where it is completely free of this planet's influence, you would have to add energy to free it.1153

It is almost like it is in energy debt before it is free, while it is trapped in the gravitational field here.1164

That is where the negative sign comes from.1168

Let us take a look at how orbits work.1176

This is a very interesting discussion problem because lots of folks have seen videos of astronauts and the space shuttle and they are floating around and the question often comes up, "Why are they floating around? They must be weightless."1180

No, they are not weightless and to understand that, you really have to know how orbits work.1197

We are going to go back to a thought experiment that Isaac Newton proposed many years ago.1201

He said, "Let us imagine that we have this hypothetical mountain, huge mountain, so high that at the very top of it, you are above the atmosphere of the earth."1207

You do not have any friction because there is no air to slow anything down.1216

At the top of this mountain, we are going to place a cannon.1219

I know the cannon is not going to work without an atmosphere, but just hang with me for the purposes of the thought experiment.1222

While we are up there, if we were to shoot a cannon ball, it is going to follow some projectile path down to the earth.1228

But if we shot it a little bit faster, it is going to travel a little bit further as it follows that parabolic trajectory.1237

Give it a little bit more velocity, it is going to travel even further, but eventually you are going to come to a point where you shoot it fast enough that at the rate it is falling, it is also falling around the earth because the earth is a circular path.1247

Yes, it is constantly falling. It is falling all the time, but it is moving so fast horizontally that by the time it falls, the earth has moved underneath it and it stays at the same altitude above the earth.1262

That is what happens in orbit.1275

They are not weightless. They are falling.1278

They are just moving so fast horizontally that by the time they fall and the earth has moved around underneath them and because the earth is a sphere, they maintain the same altitude.1280

Let us take a look and see if we cannot prove that a little bit.1295

If the space shuttle orbits the earth at an altitude of 380 km above the surface of the earth, what is the gravitational field strength due to earth at that altitude?1298

At what speed does the shuttle have to travel to maintain that orbit?1306

Let us start with the gravitational field strength.1311

The force of gravity is mg, which equals gm1m2/r2.1315

Therefore, the gravitational field strength (g) must be g times mass, which is going to be the mass of the earth divided by r2, where g, we know is that constant 6.67 times10-11Nm2/kg2.1323

The mass of the earth is 6 × 1024 kg over the distance between their centers.1339

To find the distance between their centers -- if this is 380 km above the surface of the earth, we also have to account for the radius of the earth.1347

The radius of the earth is 6.37 × 106 m roughly + 380,000 m2 or about 8.78 m/s2 or 8.78N/kg.1356

Compare that to 9.8, what we have here on the surface of the earth.1377

That is not a huge reduction. There is still an awful lot of gravitational field out there where they are orbiting.1381

What speed does the shuttle travel to maintain that orbit?1389

To do that one, let us take a look at the force of gravity, which is gm1m2/r2 = mv2, mv2/r because it is moving in a circular path -- centripetal force.1393

Therefore, the square of our velocity if we rearrange these is going to be -- we have rgm1m2/mr2 and I can do a little bit of simplifying here.1410

We have rn and r2, we have a mass and a mass, so that will leave me with g times the mass of the earth divided by r.1428

If that is v2, then v itself must be g times the mass of the earth over (r) square root.1440

When I substitute in my values, that is 6.67 × 10-11Nm2/kg2, -- mass of the earth is about 6 × 1024 kg and the distance between their centers, 6.37 × 106 radius of the earth + 380,000 m above the surface of the earth.1449

The square root of all that and I come up with a velocity of about 7700 m/s or that is greater than 17,000 miles per hour (mph).1473

To put that in perspective, that is more than 23 times the speed of sound at sea level.1494

That is fast!1505

Let us take a look at another example.1513

Calculate the magnitude of the centripetal force acting on earth as it orbits the sun, assuming a circular orbit of radius 1.5 × 1011 m in an orbital speed of 3 × 104 m/s.1515

Use that to determine the mass of the sun.1528

Let us start out with the magnitude of the centripetal force.1532

Centripetal force is mv2/r or 6 × 1024 × our velocity, 3 × 104)2/1.5 × 1011...1536

... which gives me a value of about 3.6 × 1022N.1552

Let us use that to determine the mass of the sun.1562

If that is the force, we know gravitational force is gm1m2/r2, where one of those is mass of the sun -- one is mass of the earth and that is equal to 3.6 × 1022N.1565

Therefore, we could say the mass of the sun is equal to 3.6 × 1022N × r2/G × the mass of the earth.1580

Or 3.6 × 1022 or 1.5 × 10112/G, 6.67 × 10-11Nm2/kg2 × the mass of the earth, about 6 × 1024 kg.1593

If I plug that all into my calculator very carefully and I find that the mass of the sun is right around 2 × 1030 kg.1616

So you can see we are using the same equations and relationships over and over again.1628

The tricky part is keeping all of your values well taken care of, being careful with the calculator -- very fastidious in your calculations.1632

Example 7 -- The diagram shows two bowling balls, A and B.1645

Each has a mass of 7 kg and they are 2 m apart.1649

Find the magnitude of the gravitational force exerted by ball A on ball B.1653

The gravitational force is gm1m2/r2 where 6.67 × 10-11 × mass 1 (7) mass 2 (7)/the square of the distance between them -- 2 m2) or about 8.2 × 10-10N.1660

Example 8 -- A 2 kg object is falling freely near earth's surface.1680

What is the magnitude of the gravitational force that earth exerts on the object?1693

If it is near earth's surface, we can do this one a simple way.1698

Force of gravity or the object's weight is mg, which is going to be 2 kg; g is 9.8 or let us round that to 10 to make it easy -- about 20 N.1702

Nice, simple, straightforward because it is near the earth's surface.1714

Let us do an example finding g.1719

What is the acceleration due to gravity at a location where a 15 kg mass weighs 45N?1722

Weight, mg = 45N, therefore, g must equal 45N/mass (15 kg) or 3 m/s2.1728

Just some very simple interpretation problems.1744

Let us take a look at a space vehicle on Mars.1749

A 1200 kg space vehicle travels at 4.8 m/s along the level surface of Mars.1753

If the magnitude of the gravitational field strength on the surface of Mars is 3.7 N/kg -- that is g -- find the magnitude of the normal force acting on the vehicle.1759

When I see normal force, right away I start thinking FBD.1771

We have the weight down (mg) -- normal force which we will call Fn -- pointing up -- and they must be balanced -- we call that +y direction.1774

It is not accelerating spontaneously up off the surface of the planet or going down through it. 1790 Therefore, the net force in the y direction must be 0 and the normal force and mg must be matched, therefore net force in the y direction is the normal force minus mg must equal 0.1784

Therefore, the normal force equals the object's weight (mg) or its mass (1200 kg) × g (3.7 N/kg) for a force of around 4,440N.1803

Let us take a look at a graphical analysis problem.1825

This graph represents the relationship between gravitational force and mass for objects near the surface of the earth.1828

What does the slope represent? The slope is rise/run.1834

Rise is going to be change in gravitational force and our run is going to be change in mass.1844

Change in gravitational force, as long as we are near the surface of the earth is δmg/δm and that is just going to give us g.1851

Then the slope is the acceleration due to gravity.1861

All right. Let us go back to Mars. A 2 kg object weighs 19.6N on Earth.1874

If the acceleration due to gravity on Mars is 3.71 m/s2, what is the object's mass on Mars?1878

I love these questions! They are so simple but meant to trick you and it is so easy to fall into the trap.1886

It asks you what is the object's mass on Mars. The mass has not changed.1891

The weight may have changed, but its mass is still 2 kg. Do not get suckered into those tricks!1897

Your find is the same as your given.1905

One more -- Here we have two satellites.1910

The diagram shows the two satellites, both of equal mass, A and B, in circular orbits around a planet here.1913

Compare the magnitude of the gravitational force of attraction between A and the planet.1919

Find the magnitude of the gravitational force of attraction between B and the planet.1925

First thing -- since B is further away, it should be pretty obvious that it is going to have a smaller force. Okay?1929

Right away -- twice as great, four times as great we can eliminate.1938

Because of that Inverse Square Law, we are going from radius (r) to 2r as we are doubling the distance and we must have 1/4th the force.1942

The answer is number 3: Inverse Square Law.1950

There are lots of different ways you can go through and solve that.1953

You can go through and do it analytically or you can make up numbers for them, but the easiest way is if you understand the Inverse Square Law, you can realize right away if the distance doubles, the force becomes 1/4th.1956

Hopefully, that gets you a great start on gravity and Newton's Law of Universal Gravitation.1968

Thank you so much for your time and make it a great day!1973