# Time-Averaging the Potential Energy

## Homework Equations

Gravitational potential energy

## The Attempt at a Solution

Isn't this the solution?

So Is <U> = U(r) where a = r? How do I incorporate the ##\frac {1}{\tau}## and the given "useful definite integral" from zero to 2π? Thanks

I tried taking the integral but I got undefined anwer
Gravitational potential energy = ##U = \frac {-GMm}{r^2}##
##-GMm \int_{0}^{a} \frac{1}{r^2}dr##
##-GMm (\frac {-1}{a} + \frac{-1}{0})##

I tried taking the integral but I got undefined anwer
Gravitational potential energy = ##U = \frac {-GMm}{r^2}##
##-GMm \int_{0}^{a} \frac{1}{r^2}dr##
##-GMm (\frac {-1}{a} + \frac{-1}{0})##
I thought the limits are final on top and initial on the bottom. We're going from the surface (0) to a distance a away from the planet. Why should the bottom limit be infinity?

I thought the limits are final on top and initial on the bottom. We're going from the surface (0) to a distance a away from the planet. Why should the bottom limit be infinity?
I think I understand now, in the second picture of the first post, they're bringing an object from infinitely far away to a radius of r. To clarify, I think ##F = \frac {GMm}{r^2}## is negative because it is an attractive force. Can this reasoning be justified? So for post #2, I'll just replace 0 with a , and the top with infinity. ##\int_{a}^{\infty}##. That would give me a positive value because I must put in work to move the object from a distance of a to infinity.

I think I understand now, in the second picture of the first post, they're bringing an object from infinitely far away to a radius of r. To clarify, I think ##F = \frac {GMm}{r^2}## is negative because it is an attractive force. Can this reasoning be justified? So for post #2, I'll just replace 0 with a , and the top with infinity. ##\int_{a}^{\infty}##. That would give me a positive value because I must put in work to move the object from a distance of a to infinity.
As a side note this seems to indicate that the formula for potential energy breaks down if I cross the Earth's core?

More þan one way to do it?

Now I have the solutions guide but it confuses me even more. Where do the marked definitions come from?

This is an online solutions guide, so I don't know what the teacher's lecture is.

Nathanael
Homework Helper
I think I understand now, in the second picture of the first post, they're bringing an object from infinitely far away to a radius of r.
Yes, this is because that is how gravitational potential energy is (typically) defined.
To clarify, I think ##F = \frac {GMm}{r^2}## is negative because it is an attractive force. Can this reasoning be justified?
Yes that is right. It's truly a vector equation. The force (from A on B) points along the radius vector (from A to B), but in the opposite direction (because it's attractive), and so there is the negative sign.
'll just replace 0 with a , and the top with infinity. ##\int_{a}^{\infty}##. That would give me a positive value because I must put in work to move the object from a distance of a to infinity.
You could do that. Now your distance of zero potential energy is "a." The reason it is typically defined the other way around (from infinity to a) is so that the distance of zero potential energy is infinity.
As a side note this seems to indicate that the formula for potential energy breaks down if I cross the Earth's core?
It breaks down before then, namely once you go below the surface of Earth. The reason is that the force equation assumes the Earth to be a point-object. Once you go below the surface, some of the mass will be above you, and so the force equation must be modified.

Isn't this the solution?
No it's not. That is the way to get the potential energy function from the law of gravity, but that is not what the problem asked for. Re-read the definition of the time average <f> of a function f. You want to apply this operation <U> to the potential energy U.

Calpalned
Here is the second page to the solution

It's frustrating when I don't understand the solutions guide

Nathanael
Homework Helper
I missed your post #7 at first.

The first equation you marked can be understood algebraically. ##\frac{1}{\frac{d\theta}{dt}}=\frac{dt}{d\theta}##.
Then they just used the dot notation (it's just shorthand convention for writing time derivatives) ##\dot \theta \equiv \frac{d\theta}{dt}##

The second equation you marked can be understood geometrically. h is (twice) the speed at which area is swept out by the position vector (sometimes called "areal velocity").
The understanding is geometrical, but I will try to give you the picture with words. In a short amount of time dt, the position vector sweeps out a thin triangle. The length of the long side will be R and the length of the short side will be Rdθ. The formula for the area of a triangle is half the product of these sides. So then the area per time dt is 0.5R2dθ/dt=0.5h
The fact that h is constant is one of Kepler's laws. It can also be understood in terms of angular momentum. (That law of Kepler's is equivalent to saying angular momentum is conserved.)
I hope you have been taught one of these two things (Kepler's law or that angular momentum is conserved in orbit). That is where that equation comes from.

It's frustrating when I don't understand the solutions guide
It's more fun to not check the solution and be confused than it is to check the solution and be frustrated

Calpalned
Just to clarify, it seems that time average <f(x)> is very similar to the average value of an integral

The second equation you marked can be understood geometrically. h
I can't find anything about h in my textbook

I hope you have been taught one of these two things (Kepler's law or that angular momentum is conserved in orbit). That is where that equation comes from.
Yes, I have been taught both, but I can't apply it to this context.
I know that the second law says "...a line between the sun and the planet sweeps equal areas in equal times."
Angular momentum is conserved if there is no net torque and it is ##L = I \times \omega ##

So then the area per time dt is 0.5R2dθ/dt=0.5h
So it looks like h is defined as #h = R^2 d\theta ## But if we let more time pass, then the angle theta will be greater, so shouldn't h be greater to? Thus h changes and isn't constant.

angular momentum is conserved
Kepler's second law is just another way to express angular momentum?

Nathanael
Homework Helper
Just to clarify, it seems that time average <f(x)> is very similar to the average value of an integral
Yes, they are exactly the same in logic. The only difference of course, is that the time average uses time as the independent variable.
I can't find anything about h in my textbook
So it looks like h is defined as ##h = R^2 d\theta ## But if we let more time pass, then the angle theta will be greater, so shouldn't h be greater to? Thus h changes and isn't constant.
You forgot the dt in there, ##h=R^2 \frac{d\theta}{dt}##
Kepler's second law is just another way to express angular momentum?
Yes, Kepler's law of areas implies conservation of angular momentum in orbit. The speed at which area is swept out is ##\frac{1}{2}R^2\frac{d\theta}{dt}=\text{constant by Kepler's law}## (I explain this formula in post #11). The equation for the angular momentum is ##mR^2\frac{d\theta}{dt}## which you can now see is 2m times the speed at which areas are swept out. Since this areal-speed is constant, and m is constant, that means that the angular momentum is constant.

Kepler's law is from observations, though. The theoretical reason angular momentum is conserved is because the force of gravity always acts along the radius vector, and so ##\text{torque}=\vec R \times \vec F= 0##

distance of zero potential energy is infinity
I thought the further away the object is from Earth, the more potential energy it has. For example, a rock held 10 feet above the ground has more potential energy that one held only one foot.

Is the solution from your textbook?
No it is not

Nathanael
Homework Helper
I thought the further away the object is from Earth, the more potential energy it has. For example, a rock held 10 feet above the ground has more potential energy that one held only one foot.
How much potential energy something has is really arbitrary and meaningless. How much potential energy something has is just an artifact of where you choose the zero to be. If you're dealing with earthly problems, you might choose the surface to be the location of zero potential energy. Typically with astronomical problems (like this one) you will choose the zero to be at infinity.

If it doesn't matter the magnitude of the potential energy, then what does matter? The important detail about potential energy functions is how they change. Changes in potential energy are independent of where you choose the zero to be.
Choosing a different zero just shifts the potential energy function, say from U to U+C where C is constant, but you see ##(U_f+C)-(U_i+C)=U_f-U_i##