mr.p: Good morning.
Today are going to review the universal gravitation topics
covered on the AP physics one exam.
Bo: Hey guys.
Billy: Hey Bo.
Bobby: Hi Bo.
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mr.p: Let's start with Newton's Universal law of Gravitation
which I like to call the big G equation.
The force of gravity that exists
between any two objects is equal to big G,
which is the universal gravitational constant,
times the mass of one object times the mass
of the other object divided by r squared.
Where big G is a constant and it is given as...
Big G is 6.67 times ten to the negative 11 newtons,
times meters squared, divided by kilograms squared.
And m is the mass of the two objects in kilograms.
It is r that usually gives students trouble.
So class, r is not?
All: The radius.
mr.p: r is?
All: The distance between
the centers of mass of the two objects.
mr.p: Which can be confusing because sometimes r is?
All: The radius.
mr.p: Again, r is not defined as the radius but rather,
the distance between the centers of mass of the two objects.
And this can be confusing because
sometimes r does work out to be
the radius but not always.
So now that we have two different equations
for the force of gravity,
let's make sure that we understand
the difference between the two.
The force of gravity equals the mass of the object
times the acceleration due to gravity
is a planet specific equation.
We just need to know, or be able to figure out,
the acceleration due to gravity, little g,
on the surface of that planet.
Newton's universal law of gravitation,
unlike the planet specific equation
for the force of gravity, is always applicable.
Notice we can set the magnitudes
of these two equations for the force of gravity
equal to one another to solve for the
acceleration due to gravity on any celestial object.
More specifically, we can figure out
the acceleration due to gravity here on planet earth.
So the force of gravity that exists
between an object and the earth
is equal to the mass of the object times
the acceleration due to gravity here on planet earth,
which is equal to big G times the mass of the object
times the mass of the earth divided by r squared.
Now r, remember is not the radius,
but rather the distance between the center
of the earth and the center of the object,
which is going to be equal to the radius of the earth
plus the altitude of that object
and we can cancel out the mass
of the object from both sides of the equation.
And we have an equation for the acceleration due to gravity
that exists here on planet earth.
On a local level that looks like this.
The gravitational field is represented using a vector field,
which is these arrows, and because our height
is very small compared to the radius of the earth,
the gravitational field can be considered to be constant.
However, when we look at it from a much larger,
global perspective,
the gravitational field is not constant
which is why these gravitational field lines
are not parallel to one another.
The closer those field lines are to one another,
the larger the gravitational field.
Okay, let's take a look at a satellite
in orbit around planet Earth.
How would you solve for the linear velocity
of the satellite? Bo?
Bo: Okay, um, the satellite is moving in a circle,
so we need to sum the forces in the "in" direction.
However, before we can do that
we need to draw a free body diagram
of the forces acting on the satellite.
The only force acting on the satellite
is the force of gravity which is caused by the earth.
Now we know the net force in the in direction acting
on the satellite equals the force of gravity,
which is positive because it's in the in direction
and the net force equals the mass of the satellite times
the acceleration in the in direction,
which is the centripetal acceleration,
we can substitute big G times the mass of the satellite
times the mass of the Earth,
all divided by r squared which then equals
the mass of the satellite times...
Oh, we can substitute tangential velocity squared
divided by r for centripetal acceleration.
The mass of the satellite cancels out,
and the inverse of r cancels out,
and we take the square root of the whole equation to get
tangential velocity equals the square root of big G
times the mass of the Earth, all divided by r.
Oh, and we should substitute in the radius of the Earth
plus the altitude of the satellite in for r
to get our answer for the tangential velocity
of the satellite.
mr.p: Bo, very nice.
Now let's talk about gravitational potential energy.
Our equation for the gravitational potential energy
of an object, which is equal to the mass of the object
times the acceleration due to gravity
times the vertical height above the zero line,
requires a constant gravitational field.
When you have a non-constant gravitational field
we need a different equation.
We need the equation for
universal gravitational potential energy.
Universal gravitational potential energy equals
the negative of big G times the mass of one object,
times the mass of the second object, divided by r.
Now, this looks very similar to
Newton's universal law of gravitation,
and it is that same r value, where r is not the radius,
it is the distance between the centers of mass
of the two objects.
Given the similarities between
the equation for universal gravitational potential energy
and Newton's universal law of gravitation,
it is very common for students
to add a square to this equation.
Please do not, r is not squared
in universal gravitational potential energy.
Okay, let's also talk about this negative.
Universal gravitational potential energy
is always less than or equal to zero.
[mr.p] Yes?
Bobby: But what does it mean that the
gravitational potential energy
is always less than or equal to zero?
mr.p: I know, this is always a source of confusion for students.
So let's go back to the equation
for a constant gravitational field
for gravitational potential energy.
When we use this equation, we always
have to set the horizontal zero line.
When using the equation for
universal gravitational potential energy
in a non-constant gravitational field,
we do not need to set the zero line.
It has already been set for us.
It is set with the two objects infinitely far apart.
Bo, what is anything divided by infinity?
Bo: Anything divided by infinity is zero.
mr.p: Which means the universal gravitational potential energy
between two objects which are infinitely far apart
is going to be equal to zero because
anything divided by infinity equals zero.
That is why universal gravitational potential energy
is always less than or equal to zero.
In all other cases r is going to be
less than infinitely large.
Therefore it's like we are below the zero line
where the universal gravitational potential energy
is negative.
Which brings me to another point.
A single object can not have
universal gravitational potential energy.
Universal gravitational potential energy
is defined as the gravitational potential energy
that exists between two objects, mass one and mass two.
[Billy] Mr. P?
mr.p: Yes Billy?
Billy: What about the regular gravitational potential energy?
Mass times the acceleration due to gravity
times the vertical height above the zero line?
That equation only has one mass in it.
mr.p: Okay, this is the equation
for the gravitational potential energy
that exists between the object and planet Earth.
So there are two objects involved in the equation
for gravitational potential energy
in a constant gravitational field.
We have reached the end of my review
of universal gravitation for AP Physics one.
My next review lesson is about simple harmonic motion.
You are welcome to enjoy that video.
You can also visit my website flippingphysics.com
where you can find all of my AP Physics one
review lessons organized with lecture notes.
Thank you very much for learning with me today.
I enjoyed learning with you.