This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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From the course by University of Houston System

Preparing for the AP Physics 1 Exam

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This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

From the lesson

Circular Motion and Gravitation

Topics include kinematics and dynamics of circular motion, Newton’s law of universal law of gravitation, and applications of topics. You will watch 2 videos, complete 2 sets of practice problems and take 2 quizzes. Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

The force of gravitation is one of our four fundamental forces.

This is the force responsible for weight, objects with mass attract one another.

These attractive forces will be equal in magnitude but opposite in direction.

So just we saw in Newton's third law.

For instance, the Earth is pulled towards the Sun,

while the Sun experiences an equal force towards the Earth.

We have never observed a scenario in nature where gravity causes two objects,

with mass, to repel.

Compared to other forces,

like electromagnetic interactions, gravity is the weakest force.

We can describe the gravitational force between two objects

with the equation shown here.

The variables M1 an M2 in this equation represent the mass

of the two objects that are interacting through gravitational forces.

The variable R in this equation represents the distance from the center of

one object to the center of the other object.

Notice that this distance is squared in the denominator of our equation.

This means that gravitational force is inversely related to the square

of the distance, what physicists call an inverse-square law.

This is a property of three-dimensional space.

And we will see this behavior again in later modules.

The constant G has a value of 6.67 x10^-11 Newton

meters squared divided by kilograms squared.

It is very small.

This is why we don't observe measurable forces between ordinary objects like

baseballs in our daily lives.

Now we will look at Newton's Law of Universal Gravitation in action.

Calculate the force of gravity on a baseball, 0.14kg at the Earth's surface

if the radiance of the Earth is 6.38 times 10 to the six meters.

And the mass of the earth is 5.98 times ten to the 24th kilograms.

Next, calculate the acceleration of the ball if the only force exerted on it

is this gravitational attraction.

Notice how in the last example our new equation results in the same

calculated force as M multiplied by G.

Remember that MG as gravitational force only works near the Earth's surface

where the gravitational field is uniform.

Whie Newton's Law of Universal Gravitation applies at any distance,

one area in which this relationship is particularly helpful is orbital motion.

In order to solve for gravitational force, I will use my equation F of g equals my

gravitational constant multiplied by M1 M2

divided by R squared.

So let's go ahead and look at what variables we have.

Well, our M1, or at least one of our masses in this case,

is the mass of the baseball, 0.14 kilograms.

The other mass that it's interacting with is the Earth.

So I have the mass of the earth, me or M2,

which is 5.98 times 10 to the 24th power.

I also need the radius.

So in this case, when I'm looking at the radius, I have the center of

the earth that's acting with this baseball that's on the earth's surface.

So my R is going to be the radius of the earth.

The distance between the two.

Which is .38 times ten to the sixth.

You don't have to worry about memorizing these values,

it just the mass of the earth or the radius of the earth,

you will be provided with these constants on the AP exam.

Plugging all of this into my equation.

[SOUND] You have to be really careful with,

gravitational force problems because you have

several numbers that involve exponents.

So you want to be careful with your parentheses.

Also, often times, because the radius involves exponents as well,

students forget to square the radius.

So, notice, I've closed those parenthesis, and I square this.

When I plug this into my calculator, I get my answer as1.37,

or 1.4 newtons for the gravitational force.

Now if I want to solve for acceleration of the ball.

Knowing that the only force exerted is the force of gravity.

I'll go ahead, label this as part b.

F net equals ma.

F of g is the only force that's being exerted.

This will equal ma.

I've solved for F of g.

It's 1.37 or 1.4 newtons.

The mass is 0.14 kg.

From here, when I solve for A, I end up with 9.8 meters per second squared.

There is significance in that answer 9.8 meters per second squared.

That's the value of gravitational acceleration.

>> This problem discusses the interaction between the earth and the sun.

Of course, this is not going to be drawn to scale but here's my sun,

and here is my earth.

And I know several things about this interaction.

The force of gravity is the only force between them,

and it pulls with equal strength on both.

One towards the other.

Gravity only attracts, it never repels.

Even though one object is larger than the other, the force is the same.

The acceleration that those objects may experience will be different.

So this force of gravity here, will equal this force of gravity here.

This, being the sun.

And this, being the earth.

Well, if that's the case, now like we said,

the path that the earth takes around the sun is very nearly circular.

Yes it is an elliptical, just a little, but

we can approximate it very accurately as a circle.

So that'll allow us to use several of our circular motion equations, and

you'll see me use that in a moment.

I want to start analyzing this problem,

since I want the mass of the sun with forces, because we know a lot about

gravitational force with my Newton's law of universal gravitation.

So sum of forces equal ma, we're doing this expression for the earth.

So on the left hand side I need to add all the forces on the Earth.

In this case it's big G as our constant.

That's our gravitational constant.

The mass of the Earth, the mass of the Sun, and

the average distance between their centers.

Which it tells us here.

That's going to be squared.

And on the other side since i'm adding all the forces of the Earth.

This is the mass of the Earth multiplied by the acceleration of the Earth.

Before I do anything else, I notice that the mass of the Earth shows up on both

sides of the equation, so when I divide, they cancel out.

I also know, like we said, that this is perfect circular motion, essentially.

And so, anything in nice uniform circular motion has acceleration to the center.

Our radial acceleration will be squared over R.

So I'm going to sub that into my expression here.

Well, there's also something else I need to recognize.

I know the radius.

That's the distance between their centers.

But I do not know the velocity,

the transitional velocity of the Earth around the Sun.

Well, I'm going to have to use geometry for that.

If this object here's my sun, and this is the earth going around it.

I know that the velocity of the earth around it will

be the distance divided by time.

So that will be the circumference V, would equal two pi R, max of comforts,

divided by the period of the motion.

That is, how long does it take for

the earth to go around the sun, that's one year.

But I'll need to make sure to put that in seconds when we actually sub that in.

So looking at these steps, it's kind of a multi-step substitution process here.

So let me show that here.

Big G, the mass of the sun, which is what we're solving for, that's great,

r squared, my acceleration, v squared over R.

And notice that I have an r on this side, and two r's on that side.

And so when I multiply both sides by r, one of the r's on the left will disappear.

Then I am going to sub in for v.

The potential velocity is the velocity of the Earth around the Sun.

So big G, mass of the Sun over one radius now.

And the velocity, we said,

was the circumference of the circle, divided by the period.

And all of that is squared.

We're solving here for the mass of the Sun.

So let me give myself a little bit more room.

I can see here that if I'm solving the mass of the sun, I'm just reduce this and

rearrange this a little bit just to help me solve,

the mass of the sun would equal four.

Cause that's two squared, pi squared.

I'm going to multiply by the R, that's in the denominator, so

actually it's R cubed in the numerator,

over the period squared, and divide by big G.

That's my expression for the mass of this sun.

I know four pi, those are just constants.

The radius they told us.

T is the period,

which in this case is how many seconds it takes the Earth to go around the Sun.

And G is our gravitational constant.

So, let's sub in our values.

4 pi squared, the radius they gave us,

1.5 times 10 to the 11th cubed.

And then you divide all this by the period, the time it takes for

the Earth to go around the Sun.

We know it takes 365 days.

There are 24 hours in a day.

And there are 60 minutes and 60 seconds.

So 3600 seconds in an hour.

Don't forget that that whole term needs to be squared.

And in the last portion here of gravitational constants,

also an easy number to forget, 6.67 times ten to the negative eleventh.

All of that in the denominator.

So there's a lot there, going on,

very easy to accidentally plug in the wrong numbers into a calculator.

Remember you get a calculator the whole time on the free response and

the multiple choice on the, the exam, so keep that in mind.

You don't have to necessarily do all of this in your head.

In the end, I get the mass of the sun to be about 2

times 10 to the 30th kilograms.

[SOUND] In this problem, we get a chance to apply

Kepler's third law that describes orbital motion.

Rather than showing the full derivation for Kepler's Third Law.

Which we've already done, actually,

in the previous problem where we calculated the mass of the Sun.

I'm just going to skip to that portion of the answer,

which is what we gave in our video just a moment ago.

Which is, that, you'll recall, you can always go back and

check, that when we're calculating the mass of the Sun, we got,

at some point in the equation, an equation expression that looked like this.

[SOUND] Where m is the mass of the Sun,

big G is the constant.

4 and Pi are both constants.

R is the distance from the object to the orbital center,

in this case it was from the Earth to the Sun.

And T is the period, how long it takes for one complete orbit to occur.

We'll notice something very peculiar.

The left hand side of this equation is always constant

as long as you're talking about any object orbiting the same body.

So every planet in our solar system,

the left hand side of this equation would be the same.

They all orbit the sun.

Well this is what Kepler's law tells us,

that as long as you're relating objects that orbit the same center,

I know that they should always equal the same constant.

And so, in a problem like this, it's talking about Saturn, and

it's talking about the Earth.

They both orbit the sun.

I know that this ratio will always equal one another.

And so solving this problem, why don't I set the left hand side to

talk about Saturn and the right hand side to talk about the Earth.

With this relationship we can solve information about one or the other,

knowing a little bit about both.

So I'm going to scroll down for a moment and sub in our numbers.

Calculations like this actually turn out to be quite easy with this relationship.

That means we're looking for the distance Saturn is from the sun.

Which we don't know.

But they do tell us a little bit about its period.

I want do our units in years,

but you could definitely convert these into seconds.

The conversions would just simply end up cancelling out because it's a ratio.

They told me that a Saturn year is 29.7 Earth years.

The distance the Earth is from the Sun,

that's the radius here on the right hand side.

We've seen that already, 1.5 times 10 to the 11th meters.

That's cubed.

And then the period of the Earth, well, that's just 1 Earth year, and

so I'm just going to put in 1.

I'm just making sure the units in the denominators match.

Earth years and Earth years.

Solving this, then, for the radius,

I'm going to move this term

to the right hand side, the 29.7 squared, and

I would make sure to take the cubed root of that number on the right hand side.

When I do that, I get the distance of 1.44

times ten, to the 12th meters.

That's the distance,

on average, that Saturn is from the sun from center to center.

Of course, again, they are mildly elliptical so that does vary some.

But it is very close to being circular and so that's a good estimate for us.

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