-1
$\begingroup$

This is a portion from Feynman Lectures:

...............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.

What is it that always stays the same?
Can anyone please tell me what is that constant(the line in bold)?

And what does the last half of the paragraph(the highlighted part) mean?

$\endgroup$
0

1 Answer 1

5
$\begingroup$

“The square of the velocity minus some constants divided by radius”.

Let’s write down the law of conservation of energy: kinetic plus potential energy is constant.

$$K.E. = \frac12 m v^2$$

$$P.E. = -\frac{G m M}{r}$$

Where $m$ is the mass of the planet, $M$ the mass of the sun it is orbiting, $G$ the gravitational constant, and $r$ the distance.

Add them, together and divide by $\frac{m}{2}$:

$$v^2-\frac{2GM}{r} = \rm{const}$$

And there it is - the thing Feynman described.

$\endgroup$
14
  • $\begingroup$ I'm having some difficulty understanding the process in the last half of the paragraph that Feynman is doing for finding out how fast a planet is moving when it's closer to the sun. Can you explain this too please? $\endgroup$ Commented Jan 12, 2021 at 12:07
  • $\begingroup$ But, @Floris, didn't Feynman tell to change the planet from Sun's orbit to some other? Then how can you put Some mass over there? Just a doubt. You know I'm not so big a science geek. $\endgroup$
    – lee
    Commented Jan 12, 2021 at 12:11
  • 1
    $\begingroup$ The last part talks about changing the orbit in a “frictionless manner”... change the direction but not the velocity (as though the planet was briefly “in a channel”, on a rail car with curved rails, if you like). Since no energy was added or taken away, the conservation equation continues to hold, and now we can know the velocity at any radius. $\endgroup$
    – Floris
    Commented Jan 12, 2021 at 12:37
  • $\begingroup$ @Floris if it falls from one radius to another, how can the new speed be the same as the old speed of the actual orbit? $\endgroup$ Commented Jan 12, 2021 at 14:55
  • 1
    $\begingroup$ @ShadmanSakib Feynman explains the gravity is a conservative force field. The energy you have after moving from point A to point B in such a field depends ONLY on the positions A and B (in this case it depends on the radial distance to the mass) and not the PATH used to get there. He first shows this for a path that is either always perpendicular to the force (no work done) or parallel to the force (easy math), then shows that any path can be cut into segments that are either one or the other. If you do a complete loop, you have moved the same distance AWAY and TOWARDS the sun - net work zero. $\endgroup$
    – Floris
    Commented Jan 13, 2021 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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