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.
As a side note this seems to indicate that the formula for potential energy breaks down if I cross the Earth's core?
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.
It's more fun to not check the solution and be confused than it is to check the solution and be frustrated
I can't find anything about h in my textbook
Yes, I have been taught both, but I can't apply it to this context.
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.
Kepler's second law is just another way to express angular momentum?
Yes, they are exactly the same in logic. The only difference of course, is that the time average uses time as the independent variable.
Is the problem from your textbook? Is the solution from your textbook? Does your text talk about the polar equation of ellipses?
You forgot the dt in there, ##h=R^2 \frac{d\theta}{dt}##
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.
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.
No it is not
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.