- Good morning.
Today we are going to review universal gravitation as a part of the AP Physics C, mechanics curriculum.
Flipping Physics
Bobby, what is the equation for the Force of Gravity or the weight acting on an object?
- Force of gravity equals mass times acceleration due to gravity.
- Yes. However, I do want to point out there is an assumed subscript in this equation.
I have added an “o” to the mass to identify this is the mass of the object on which the force of gravity is acting
And remember this is the force of gravitational attraction between the object and a planet.
For us the planet is usually the Earth.
And it is important we remember, we actually have another equation for the force of gravity.
This is Newton’s Universal Law of Gravitation, or what I like to call The Big G Equation.
[Class]: Big G.
- And it is the force of gravitational attraction between any two masses.
And big G is the Universal Gravitational Constant.
And it is 6.67 times ten to the negative eleven newton meters squared divided by kilograms squared.
And remember, this force of gravitational attraction is between any two masses.
So, we need the mass of object 1 and the mass of object 2.
And then there is this r squared on the bottom.
Class r is not....
[Class]: The radius.
- r is the distance between....
[Class]: The centers of mass of the two objects.
- This can be especially confusing because sometimes r is....
[Class]: The radius.
By definition r is the distance between the centers of mass of the two objects
and yes sometimes r is the radius.
And remember,
this equation gives you the magnitude of the force of gravitational attraction between the two objects.
The direction is always toward the other object.
These two equations are both for the force of gravity, so we can set them equal to one another.
And Let’s assume this force of gravity is for an object which is resting on the surface of the planet Earth.
That means it will be the mass of the object times the acceleration due to gravity
is equal to big G times the mass of the object times the mass of the earth all divided by r-squared.
[Billy and Bobby]: Everybody brought the mass to the party!
Bo: Everybody brought the mass of the object
You gotta be care of that.
- And we are left with the acceleration due to gravity here on planet earth
equals big G times the mass of the earth all divided by r squared.
And again r is the distance between the centers of mass of the two objects,
so it is equal to the radius of the
Earth plus the altitude of the location we are talking about that quantity squared.
And note that this means that...
As long as the radius of the Earth is much much greater than all the altitudes we are talking about,
the acceleration due to gravity can be considered to be constant.
In other words, at a local level, we live in a constant gravitational field
with a value of 9.81 meters per second squared.
And the gravitational potential energy associated with a constant gravitational field
is something we have already talked about.
It is equal to the mass of the object times the acceleration due to gravity, which is constant,
times the vertical height above the horizontal zero line.
When viewed from a frame of reference which is not on the surface of the planet.
In other words, the altitude is not much smaller than the radius of the earth,
the acceleration due to gravity is not constant, in other words,
it is not a constant gravitational field and we need a different equation for
gravitational potential energy.
Billy what is that equation?
- Universal gravitational potential energy equals the negative of the universal gravitational constant
times mass 1 times mass 2 all divided by r squared.
- I thought it was just divided by r.
Ya, it is just divided by r not r squared. I always do that.
- Don't worry Billy, it's a common mistake.
- Let's see.
Unlike gravitational potential energy in a constant gravitational field where we get to pick the horizontal zero line,
the zero reference line is already in the universal gravitational potential energy equation,
it is infinitely far away, or where the r value is infinitely large.
Oh, I should mention r is the same as in the universal gravitational force equation.
Uh,.....
because the zero reference line is at r equals infinity,
which means the two objects would be infinitely far away from one another,
Universal Gravitational Potential Energy can never be positive.
Also, it is important to remember this equation requires two objects. In other words,
a single object by itself cannot have
Universal Gravitational Potential Energy.
Remember, the Universal Gravitational Potential Energy which exists between an object and the earth
as a function of distance r from the center of the earth looks like this...
Notice that when the object is at the surface of the planet, the planet-object system has its lowest
gravitational potential energy of
The negative of Big G times the mass of the object times the mass of the Earth divided by the radius of the Earth.
Now if the object moves away from the surface of the planet,
the universal gravitation potential energy of the planet object system
decreases as 1 over r
and as r approaches infinity that gravitation potential energy approaches zero.
If instead, we move the opposite direction from the surface of the planet to the middle of the planet, and yes,
I know we would have to drill a tunnel all the way through the planet in order for this to work
When we do that though, the gravitational potential energy of the planet object-system
decreases linearly until it reaches zero when the object reaches the center of the planet.
This does assume a constant density for the earth, which is of course not correct,
but it's a good thought exercise.
Okay, so, unfortunately what I just said about the gravitational potential energy
when r is less than the radius of the planet was wrong.
That's right, I have been teaching that incorrectly for roughly 15 years.
Sorry everybody. So, here is an easy way to understand why
the gravitational potential energy graph I have drawn is not a line with a constant negative slope starting
from zero at the center of the planet.
Notice the change in gravitational potential energy from the center of the planet
to infinitely far away would then be zero.
That would mean it would take zero energy to move an object from the center of the planet
to infinitely far away from the planet.
This cannot be true.
You have to add energy to an object to completely remove one object from another object.
This is called binding energy, and we are about to derive it in this video.
In other words, the binding energy of an object cannot be zero.
So the change in gravitational potential energy from r equals zero to r equals infinity cannot be zero.
Again, I am sorry about that error and Mr. Becke, thank you for letting me know about that error.
Also, I have made an entire video to derive the gravitational potential energy inside the planet,
feel free to enjoy that video.
The link is below in the description.
And now, back to your regularly scheduled program.
Let's go through and do some standard universal gravitation example problems.
Bo, please read and do the first one.
- Okay,
the question is...
What is the minimum amount of work necessary to completely remove an object from a planet
if the object is resting on the surface of the planet?
assume no friction
Yeah, okay. Yeah, this is called the binding energy.
We have to assume the object is moved infinitely far away and has zero velocity when it gets there
Let's start with the most basic energy equation; the change in energy of the system
equals the sum of the energy transferred into or out of the system
the change in energy of the system
is the sum of a change in mechanical energy of the system plus the change in internal energy
And the energy transferred to the system is the work done by the force applied
That means the work done by the force applied equals the mechanical energy final
minus the mechanical energy initial
Plus....
Well the minimum amount of work means that the change in internal energy of the system will be zero
And the final mechanical energy will also be zero
because the object will have stopped it will be infinitely far away from the planet.
The only energy the planet object system started with was the universal gravitational potential energy
So, we now have zero minus negative big G mass 1 times mass 2 divided by r
Therefore, the minimum work done by the force applied to remove an object from a planet entirely
or what is called the binding energy
equals big G times the mass of the object times the mass of the planet divided by the radius of the planet
[Mr. P]: Bo, why is r the radius of the planet?
- r is the radius of the planet in this case because the object starts out on the surface of the planet
and the distance between the centers of mass of the two objects would then be the radius of the planet.
- Yes, this is called the binding energy.
If anything less than this amount of energy is given to the object, the object will return back
to the surface of the planet.
Now, let's leave the planet.
Bobby. Please read and do the next problem.
- Sure.
What is the minimum velocity to launch an object off the earth and have it never return?
This is called Escape Velocity.
Assume no atmosphere and no Earth rotation.
Well, ....
we have already launched the object and there is no atmosphere,
so there is no work done by a force applied on the object
and no work done by a force of friction on the object. Therefore, mechanical energy is conserved.
Initially the object is on the surface of the planet and moving, so
Initially, there is universal gravitational potential energy and kinetic energy.
Finally the object is infinitely far away and not moving, so there is no final mechanical energy.
Add universal gravitational potential energy to both sides and....
Everybody brought the mass of the object to the party!!
- Bobby, please continue.
- Multiply the equation by two and take the square root as well and the escape velocity
equals the square root of
two times the universal gravitational constant times the mass of the Earth divided by the radius of the Earth.
- Yes, that is the escape velocity of an object from the surface of the Earth.
In other words, the minimum velocity to give an object
such that it will never return back to the surface of the planet.
Let's do another one.
Billy, please determine the total mechanical energy of an object in circular orbit.
- What is the total mechanical energy of an object in circular orbit?
[Mr. P]: Yes.
- Okay, well the object will have universal gravitation potential energy and kinetic energy.
That's it, right?
- Yeah, but let's actually work out an answer which does not have the velocity of the object in it, please.
- I have to substitute out the velocity of the object?
[Mr. P] Yes, please.
- Okay.
Let's try to solve for the velocity of the object.
Uh, the only force acting on the object is the force of gravity and ...
that is ...
It's acting in the in-direction because the object is moving in a circle. So we can sum the forces in the
in-direction. So the net force in the in-direction equals the force of gravity and the net force
also equals the mass of the object times the acceleration in the in-direction
which is centripetal acceleration
Substituting in equations gives us Big G times the mass of the object times the mass of the planet divided by r
squared which equals the mass of the object times....
oh instead of centripetal acceleration we have the tangential velocity of the object squared
divided by r and everybody brought mass to the party!
[Mr. P]: Actually, let’s not solve it that way.
Instead let's multiply the whole equation by one half
And, actually everybody brought one over r to the party
And you can see that the kinetic energy of the object in orbit equals one-half
the magnitude of the universal gravitational potential energy of the planet object system
- Yeah.
- Billy, please continue.
- Okay.
We can substitute what you just found back into the equation for total mechanical energy and …
factor out every letter in the equation which leaves us with negative one plus one-half in the parentheses
which is negative 1/2. So the total mechanical energy associated with an object in circular orbit is
negative universal gravitational constant times the mass of the object
times the mass of the planet all divided by 2r.
- Wouldn’t it be two times the radius of the planet?
- No, r would be the radius of the planet plus the altitude of the orbital object.
- Yeah. Right.
- Okay.
Now let’s talk about Kepler’s Three Laws.
As far as the AP Physics C Mechanics exam is concerned,
in my experience, having a basic understanding of Kepler's first two laws is adequate,
however, you do need to be able to derive Kepler’s 3rd law.
Okay, Kepler's first law is often called the law of orbits.
Kepler's first law states that all orbits are elliptical and defined
by the figure which we are currently working on drawing.
There are two foci for
this ellipse; one located at F
one and one located at F two, and both of those foci are located a distance
C from the center of the ellipse.
And, the object which is being orbited, for example the Sun which is currently being orbited by the earth, is
located at one of the two foci. So the Sun is located at one of those two foci.
- What is a foci?
- Foci is plural for focus. So when he says foci, he is referring to the two foci.
- The semimajor axis has a length a, which means the major axis has a length 2a.
The semiminor axis has a length b, which means the minor axis has a length 2b.
The sum of the distances from the foci to the orbiting object,
r one and r two, always add up to the magnitude of the major axis, 2a.
The equation for the elliptical path and orbital objects follows is..
the semi-major axis squared is equal to the semi minor axis squared
plus the distance each foci is from the center of the ellipse
squared. That is the basic idea of Kepler's first law.
You should also know about eccentricity, e, which is equal to the distance each foci is from the center
of the ellipse divided by the semi major axis.
Notice that a circle has a C
value of 0 the distance each foci is from the center of the ellipse is actually zero for a circle.
Therefore the eccentricity of a circle is zero.
And the eccentricity of the earth is 0.017 which means the orbital path of the earth is almost a circle.
Kepler's second law is often called the law of areas.
Kepler's second law states that a line between the sun and an orbiting planet
sweeps out an equal area in an equal time interval.
Which means these two areas will have the same value, which means that,
the closer and orbital object is to what it is orbiting, the faster the orbital object's tangential speed will be.
Okay, Kepler's third law is often called the law of periods.
This is where we determine the relationship between the period of the orbital object and its radius.
It begins by recognizing that the only force acting on the orbital object is the force of gravity
and that force of gravity acts
inward toward the center of the and again, this is a circular orbit.
Therefore, we can sum the forces in the in-direction on the object which equals the force of gravity
which equals the mass of the object
times the object's centripetal acceleration
The force of gravity equals big G times the mass of the object times the mass of the planet divided by r squared.
And that equals the mass of the object times the object's centripetal acceleration
which is equal to the radius times the angular
velocity of the object squared.
[Billy and Bobby]: Everybody brought mass to the party!!
Bo: of the object.
And because the distance between the centers of mass and the two objects in this particular case is the
radius we can combine the r values to get on the left hand side r cubed.
The equation for angular velocity is the change in angular position over change in time and for one full
revolution the angular displacement is 2 pi radians, and the change in time is the period, capital T.
Squaring two PI over the period and solving for period squared
Gives us that the square of the period of an object in circular orbit is equal to 4pi squared
divided by the universal gravitational constant times the mass of the planet
that quantity multiplied by the radius of the circular orbit cubed.
Notice everything in the parentheses is a constant. Therefore,
the square of the orbital period is proportional to the cube of the orbital radius.
That completes my review of universal gravitation.
Next feel free to enjoy my review of simple harmonic motion or visit my AP Physics C review webpage.
Thank you very much for learning with me today.
I enjoyed learning with you.