# Time-Averaging the Potential Energy

1. Oct 16, 2015

### Calpalned

1. The problem statement, all variables and given/known data

2. Relevant equations
Gravitational potential energy
3. 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

2. Oct 16, 2015

### Calpalned

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

3. Oct 16, 2015

### Calpalned

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?

4. Oct 16, 2015

### Calpalned

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.

5. Oct 16, 2015

### Calpalned

As a side note this seems to indicate that the formula for potential energy breaks down if I cross the Earth's core?

6. Oct 16, 2015

### Calpalned

More þan one way to do it?

7. Oct 16, 2015

### Calpalned

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.

8. Oct 16, 2015

### Nathanael

Yes, this is because that is how gravitational potential energy is (typically) defined.
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.
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.
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.

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.

9. Oct 16, 2015

### Calpalned

Here is the second page to the solution

10. Oct 16, 2015

### Calpalned

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

11. Oct 16, 2015

### Nathanael

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 more fun to not check the solution and be confused than it is to check the solution and be frustrated

12. Oct 16, 2015

### Calpalned

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

13. Oct 16, 2015

### Calpalned

I can't find anything about h in my textbook

14. Oct 16, 2015

### Calpalned

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 $L = I \times \omega$

15. Oct 16, 2015