- Good morning.
We have already used Newton's universal law of gravitation
to determine the equation for universal gravitational potential energy.
Flipping Physics
We did that by using the equation that relates a conservative force
to the potential energy associated with that conservative force.
The force equals the negative of the derivative with respect to position
of the potential energy associated with that force.
Please remember that this equation is NOT on the AP Physics C equation sheet.
We are not going to walk through that derivation again right now.
Our goal is to graph the force of gravity and universal gravitation potential energy
which exists between an object and a planet all the way from where the object is in the center of the planet
to where the object is infinitely far away from the planet. So, mass 1 is the object and mass 2 is the planet.
Let's start with the functions where the object is on the surface of the planet and beyond.
In other words, where r is greater than or equal to the radius of the planet.
On the y-axes, we have force of gravity and universal gravitational potential energy.
On the x-axes, we have r; the distance between the centers of mass of the two objects.
Bo, what do you notice about the difference between the shapes of these two graphs?
- Well they are both concave down graphs.
The force of gravity graph is proportional to the inverse of r squared,
but the gravitational potential energy graph is proportional to the inverse of just r.
Therefore, the force of gravity decreases more rapidly than gravitational potential energy as r increases.
- Correct. Now, we need to move the object to inside the planet where r is less than the radius of the planet.
And here is where it starts to get a bit tricky, and we need to assume a few things.
We need to assume the planet has a constant density.
- Constant density?! - That makes no sense!
- Yeah, so I know. No planet has a constant density; however,
it is a good thought experiment and helpful for learning.
We need to determine the force of gravity acting on an object that can move without friction
through a tunnel we have drilled all the way through the center of the planet.
In order to do this, we also need to assume that the planet is not rotating.
- Uh!
-Still makes no sense! -What?!
- So, we are simplifying the problem to make it easier to work with and it is still a useful tool for our learning.
We need to assume that the only mass of the planet that exerts a net force of gravity on our object when r is less
than the radius of the planet
is the mass of the planet that is inside a hypothetical sphere created by our variable r;
the radius of the current location of the object.
In other words, while all the mass outside of r still causes a force of gravity on the object,
due to symmetry and the fact that the force of gravity is proportional to the inverse square of the distance,
all of the forces of gravitational attraction caused by the mass of the planet
with r greater than or equal to the radius of the location of the object inside the planet.
All of those forces of gravity caused by that mass cancel out.
This is called Newton’s Shell Theorem, in case you were curious,
and it is actually a proven theorem, however, we are not going to prove it today.
So, really, I get no guff for that assumption?
- No. - No.
- Okay, Bobby, starting with the equation for density, please determine the mass which is inside the -
hypothetical sphere created by the variable r in terms of the density of the planet.
- Well, okay. So, um....
Density equals mass divided by volume.
So, so that could be the total mass of the planet divided by the total volume of the planet.
Oh, but the density of the planet is constant, so it also equals the mass inside r divided by the volume inside r
and we solve for the mass inside r.
- And the volume of a sphere is four thirds pi r cubed.
- And we get the mass inside r equals four-thirds pi times the density of the planet times r cubed.
And Billy, please use that
to determine the magnitude of the force of gravity acting on the object while it is inside the planet.
- The magnitude of the force of gravity inside the planet equals the universal gravitational constant
times the mass of the object
times the mass inside the radius of the variable r all divided by r squared.
r in that equation is the distance between the center of mass of the planet and the center of mass of the object,
which is the same as our variable r.
We can substitute in what Bobby just worked out for the density of the planet.
r squared cancels out and we are left with r.
Therefore, the magnitude of the force of gravity acting on an object inside a planet equals
four thirds pi times the universal gravitational constant times the mass of the object
times the density of the planets times the variable r.
- This is the magnitude of the force.
We know the force of gravity acting on the object is directed toward the center of the planet,
so the vector form of this equation is.....
Notice everything inside the parentheses is constant.
Which means the function for a force of gravity acting on an object inside the planet is linear,
has a negative slope, and a y-intercept of 0.
It is basically y equals mx plus b,
where the slope is the negative of everything inside the parentheses, and b, the y-intercept is 0.
That completes the force of gravity graph.
Let's now work on the universal gravitational potential energy
inside the planet where r is less than or equal to the radius of the planet.
Bo, please start with the equation which is not on the AP Physics C equation sheet, which relates a
conservative force to the potential energy associated with that conservative force.
- Sure, that equation is the force in the x-direction equals the negative of the derivative with respect to position
of the potential energy associated with that force.
In this case, that will be the force of gravity acting on an object
inside the planet was the negative of the derivative with respect to position.
But, the position is the radial position and for radial position, we use the letter r not x.
Oh and that is the derivative of the universal gravitation potential energy inside the planet.
Multiply both sides by dr and then take the integral of both sides.
That means gravitation potential energy equals the negative...
Well, we can substitute in the force of gravity equation we just solved for.
Negative times a negative is a positive, and we can bring all the constant values outside of the integral.
The integral of r with respect to r is r squared divided by 2.
Therefore, the gravitational potential energy function for moving an object from the center of the planet
to the surface of the planet equals
two thirds pi times the universal gravitational constant times the mass of the object
times the density of the planet times the square of the variable r,
which is the distance from the center of mass of the planet to where the object is located.
All that plus the constant C, because this is an indefinite integral.
- Again, notice how everything inside the parentheses is a constant value.
That means the gravitational potential energy function inside the planet is a square function,
which is shifted on the access via the constant C.
But, we can actually figure out how it is shifted
because we know the values of the two gravitational potential energy functions
have to be the same at one particular location.
Does anybody know where that is?
- Oh, at the surface of the planet, where r equals the radius of the earth.
- Right. The two gravitational potential energy
equations have to have the same value when the object is on the surface of the planet.
- Therefore at r equals the radius of the planet, we can set the two gravitational potential energy equations
equal to one another and solve for the constant C.
Again, notice the constant C equals a bunch of letters;
however, every single one of those letters represents a constant value
which they should because C is a constant.
And then, we can plug all those letters back into the equation
for the universal gravitational potential energy for an object inside the planet.
And I want to reiterate that this equation is just a parabola, a square function shifted down on the y-axis
by the value of the constant C.
Everything other than the variable r in this equation is a constant number.
So, there we have our answers.
We have equations and graphs for the force of gravitational attraction between an object and a planet
both when the object is outside the planet and when the object is inside the planet
and we have equations and graphs for
the universal gravitational potential energy between an object and a planet
both when the object is outside the planet and when the object is inside the planet.
Now, I know we've already spent quite a while on this; however,
I think it's worthwhile to take a few moments to look at the graphs specifically.
This is a graph of the force of gravitational attraction which exists between a 1 kilogram object and the earth.
In green is the force of gravity when the 1 kilogram object is outside the earth, and yes,
I plugged in all the real values for the mass and radius of the earth for this graph.
In black is the force of gravity acting on the 1 kilogram object when it is
inside the tunnel we drilled which goes all the way through the planet.
Let's start by pointing out that the value of the force of gravity acting on the 1 kilogram object
at the surface of the earth has a magnitude of
9.81 Newtons as it should be because 1 kilogram times 9.81 meters per second squared equals 9.81 Newtons.
Let's also now add universal gravitational potential energy to the graph.
Notice, we have to divide the universal gravitation potential energy between the 1 kilogram object
and the earth by 10 million in order to make it fit on this graph.
When the object is inside the earth, the graph of the gravitational potential energy is in red.
When the object is outside the earth, the graph is in blue.
Now, remember because the force of gravity is a conservative force,
the force of gravity equals the negative of the derivative, with respect to position,
of the universal of gravitational potential energy.
That means, the derivative of the universal gravitational potential energy on this graph, or the slope of the line
should be the same as the negative of the force of gravity.
Billy, at r equals zero,
what is the initial slope of the universal gravitational potential energy graph?
- Our magic tangent line finder is horizontal at r equals zero.
So, the initial slope of the gravitational potential energy graph is zero,
which matches the initial value of the force of gravity. The force of gravity at r equals 0 is 0.
- And Billy, what happens to the slope of the potential energy graph
as r increases from zero to the radius of the earth.
- The slope of the potential energy curve increases in value as r increases,
but the value of the force of gravity decreases because it equals the negative of the slope of the curve.
Oh, and the slope of the tangent line at r equals the radius of the earth must equal
9.81 newtons because it equals the negative of the value of the force of gravity there. That's cool!
- Bo, what happens as r increases from the radius of the earth to an infinite value?
- Well, as r increases from the radius of the earth, the slope of the potential energy curve decreases.
Which is why the magnitude of the force of gravity decreases.
However, the magnitude of the force of gravity decreases
faster than the magnitude of the potential energy
because the force of gravity is proportional to the inverse of r squared,
but the potential energy is proportional to the inverse of r.
- I think that is enough and we can stop there.
Thank you very much for learning with me today.
I enjoyed learning with you.