0
$\begingroup$

In chapter 13 of volume 1 of The Feynman Lectures, Feynman is discussing how the work done in going around any path in a gravitational field is zero. Here's the link: http://www.feynmanlectures.caltech.edu/I_13.html

The troubling text is as follows:

'Thus we see that the work done in going along the sides of a small triangle is the same as that done going on a slant, because scosθ is equal to x. We have proved previously that the answer is zero for any path composed of a series of notches like those of Fig. 13–3, and also that we do the same work if we cut across the corners instead of going along the notches (so long as the notches are fine enough, and we can always make them very fine); therefore, the work done in going around any path in a gravitational field is zero.

This is a very remarkable result. It tells us something we did not previously know about planetary motion. It tells us that when a planet moves around the sun (without any other objects around, no other forces) it moves in such a manner that the square of the speed at any point minus some constants divided by the radius at that point is always the same at every point on the orbit. For example, the closer the planet is to the sun, the faster it is going, but by how much? By the following amount: if instead of letting the planet go around the sun, we were to change the direction (but not the magnitude) of its velocity and make it move radially, and then we let it fall from some special radius to the radius of interest, the new speed would be the same as the speed it had in the actual orbit, because this is just another example of a complicated path. So long as we come back to the same distance, the kinetic energy will be the same. So, whether the motion is the real, undisturbed one, or is changed in direction by channels, by frictionless constraints, the kinetic energy with which the planet arrives at a point will be the same.'

I don't get what quantity is the speed squared minus some constants divided by r. Is it energy? If so, how? I also don't get what corners or notches he is talking about in the first paragraph. Then, he talks about letting the planet fall from a special radius to the radius of interest. What is that supposed to mean? What does he mean by special and interest radius? How would the speed be same if the planet is moving closer to the Sun? Shouldn't it be faster?

I seem to be asking a lot of questions related to the Feynman Lectures. Thank you !

$\endgroup$
  • $\begingroup$ velocity squared is proportional to energy per unit mass $\endgroup$ – JEB Aug 19 '18 at 16:41
0
$\begingroup$

I don't get what quantity is the speed squared minus some constants divided by r. Is it energy? If so, how?

The Author is talking about the total Energy of the planet moving under gravitational force field of the Sun. Its the ( Kinetic energy + Potential Energy) and as potential energy is negative and equal to a constant divided by r the radial distance from the Sun , its described in the above manner.

Moreover, He wishes to emphasize that the total energy will not change even if the body is made to take a loop in the conservative field (an alternative closed path) as no work will be done in traversing the said path.

I also don't get what corners or notches he is talking about in the first paragraph. Then, he talks about letting the planet fall from a special radius to the radius of interest. What is that supposed to mean? What does he mean by special and interest radius? How would the speed be the same if the planet is moving closer to the Sun? Shouldn't it be faster?

In the above, he takes a triangle as an example- but remembers that the triangle contains discontinuity at the corners so the work done cannot be integrated so he makes the path continuous by rounding off the corners to prove his point.

In another example, he asks to pick up the body and bring it to another radial position and complete the loop back to its position where he picked it up and says that the energy will be conserved and the body will have the same speed.

If however one makes another complicated loop/closed path and arrives back at the special radial position the speed will be same- meaning thereby that the speed is being guided by the 'r' value as the factor governing it is

-constant/r

The famous physicist used to ask a variety of questions and always tried to convince himself as well as his readers by doing thought experiments and at least he has succeeded in drawing you to ask questions and get satisfied.

$\endgroup$
1
$\begingroup$

Gravitational fields are conservative. This means that the work required to bring an object from one point to another does not depend on the path taken through the field. Thus, the net change in energy by going in any closed loop is zero. If not, energy would not be conserved as you could go to another point and come back to your starting point with extra kinetic energy.

The quantity that you are mentioning is probably the total energy, which is equal to the kinetic energy plus the potential energy: $$E = \frac{1}{2} mv^2 - \frac{GMm}{r}$$ By the arguments above, this quantity is conserved, so it remains constant at all points along the orbit.

$\endgroup$
  • $\begingroup$ Thought so. But, if r were to decrease, then v should increase, right? So why does Feynman say that the speed remains the same? $\endgroup$ – Esrom Diaso Aug 19 '18 at 9:54
  • $\begingroup$ @Esrom Diaso actually you are right. r and v are constant only if the orbit is exact circle. Overwise, if r is changing than v will be changing too, but the total energy will be constant. $\endgroup$ – Heopps Aug 19 '18 at 10:44
  • $\begingroup$ @Esrom Diaso: If a planet starts falling radially towards the sun with a certain velocity, its velocity will still increase as it falls closer and closer. This should be obvious due to the gravitational force. $\endgroup$ – user7777777 Aug 19 '18 at 12:04
1
$\begingroup$

In his characteristic way Feynman refers to kinetic plus potential energy, hence the binding energy if the system. The potential energy is negative hence the minus. The binding energy is what you have overcome to take the system apart.

The binding energy is the same in every point of the orbit. Many orbits exist with the same binding energy but with different angular momentum. When you change the direction but not the magnitude of the velocity you change the angular momentum but not the binding energy. From the binding energy you can find the velocity in any point of the orbital.

$\endgroup$
  • $\begingroup$ But Feynman says that if we change the direction, and let the planet fall to some other radius, the speed would remain the same. How? Shouldn't it be going faster? $\endgroup$ – Esrom Diaso Aug 19 '18 at 10:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.