In The Feynman Lectures, In the chapter entitled Work and potential energy, Feynman states:

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.

I understand that $\frac{1}{2}mv^2-\frac{GMm}{r}=\text{constant}$ in a closed path. However, I totally fail to visualize the example of the planet which Feynman is talking about.

So When he says, make the direction of the velocity radial, is it radially inwards or outwards?

And I can't form a mental picture of the shape of the path he's describing and How it's closed?

Overall, I'm confused, can anyone unpack it for me?

  • $\begingroup$ I'm still confused by what he means by "By the following amount:" and then he goes of and illustrates how the velocity only depends on the radius. But does he ever answer his question: $E = T + V = \frac{1}{2}mv^2-\frac{GMm}{r}\Rightarrow v = \sqrt{\frac{2E}{m}+\frac{2GM}{r}}$ $\endgroup$ – john Jul 15 '15 at 23:06

SECTION A : The example in Feynman's Lectures

enter image description here

Let a body P (Planet or Particle or whatever) moving in orbit around a center of attraction called $\:\rm{SUN}$, as in above Figure. Suppose that the attractive force $\:\mathbf{f}\left(r\right)\:$ depends continuously only on the distance $\:r\:$ of the body P from the center $\:\rm{SUN}$. Here it's not necessary this force to obey an inverse square law or to be any special function of $\:r\:$. In other words we would say that $\:\mathbf{f}\left(r\right)\:$ is something like that
\begin{equation} \mathbf{f}\left(r\right)=\;-\;f\left(r\right)\dfrac{\mathbf{r}}{r}=\;-\;f\left(r\right)\mathbf{n}_{r} \tag{01} \end{equation} where $\:f\left(r\right)\left(>0\right)$ the magnitude of $\:\mathbf{f}\left(r\right)\:$, continuous and integrable, and $\:\mathbf{n}_{r}\:$ the unit vector along $\:\mathbf{r}\:$. In the case of Newton's gravitation or Coulomb's electrostatic force

\begin{equation} f\left(r\right)= \dfrac{C}{r^{2}} \;, \quad C=\text{positive real} \tag{02} \end{equation} Now, let the body $\:\rm{P}\:$, moving on its orbit, is located at an instant at point $\:\rm{P}_{1}\:$ with velocity $\:\mathbf{v}_{1}\:$ and at distance $\:r_{1}\:$. Later on the body $\:\rm{P}\:$ is located on its orbit at point $\:\rm{P}_{2}\:$ with velocity $\:\mathbf{v}_{2}\:$ of greater magnitude and at a less distance $\:r_{2}\:$.

A twin body $\:\rm{P}^{\boldsymbol{\prime}}\:$, exact copy of $\:\rm{P}\:$, starts from $\:\rm{P}_{1}\:$ at this same distance $\:r_{1}\:$with velocity $\:\mathbf{w}_{1}\:$, of equal to $\:\mathbf{v}_{1}\:$ magnitude ($\:{w}_{1}=\Vert\mathbf{w}_{1}\Vert=\Vert\mathbf{v}_{1}\Vert={v}_{1}\:$), travelling radially and reaching at point $\:\rm{P}_{2}^{\boldsymbol{\prime}}\:$ with velocity $\:\mathbf{w}_{2}\:$ at this same distance $\:r_{2}\:$. The result to be proved is that the velocity $\:\mathbf{w}_{2}\:$ is of equal magnitude to $\:\mathbf{v}_{2}\:$ : $\:{w}_{2}=\Vert\mathbf{w}_{2}\Vert=\Vert\mathbf{v}_{2}\Vert={v}_{2}\:$.

We'll apply the well-known principle : \begin{equation} \textbf{Change of kinetic energy = Work done by forces} \tag{03} \end{equation}

For the body $\:\rm{P}\:$ on its orbit between points $\:\rm{P}_{1}\:$ and $\:\rm{P}_{2}\:$ above principle yields \begin{equation} \tfrac{1}{2}m\left( v_{2}^{2}- v_{1}^{2}\right)=\int_{\rm{P}_{1}}^{\rm{P}_{2}}\mathbf{f}\left(r\right)\circ d\mathbf{r}=\int_{\rm{P}_{1}}^{\rm{P}_{2}}\left[\;-\;f\left(r\right)\dfrac{\mathbf{r}}{r}\right]\circ d\mathbf{r}= - \int_{r_{1}}^{r_{2}}f\left(r\right)dr \tag{04} \end{equation}

In last to the right equality we use the fact that \begin{equation} \mathbf{r}\circ d\mathbf{r}= \tfrac{1}{2}d\left(\mathbf{r}\circ \mathbf{r} \right)=\tfrac{1}{2}d\left(\Vert\mathbf{r}\Vert^{2}\right)=\tfrac{1}{2}d\left(r^{2}\right)= r dr \tag{05} \end{equation}

So, \begin{equation} \tfrac{1}{2}m\left( v_{2}^{2}- v_{1}^{2}\right)= - \int_{r_{1}}^{r_{2}}f\left(r\right)dr = - \left[\Phi\left(r_{2}\right)-\Phi\left(r_{1}\right)\right] \tag{06} \end{equation} where $\:\Phi\left(r\right)\:$ the following indefinite integral \begin{equation} \Phi\left(r\right)=\int f\left(r\right)dr \tag{07} \end{equation}

For the twin body $\:\rm{P}^{\boldsymbol{\prime}}\:$ travelling radially from point $\:\rm{P}_{1}\:$ to point $\:\rm{P}_{2}^{\boldsymbol{\prime}}\:$ the principle (03) yields of course the same result for the change of kinetic energy \begin{equation} \tfrac{1}{2}m\left( w_{2}^{2}- w_{1}^{2}\right)=\int_{\rm{P}_{1}}^{\rm{P}_{2}^{\boldsymbol{\prime}}}\mathbf{f}\left(r\right)\circ d\mathbf{r}=\int_{\rm{P}_{1}}^{\rm{P}_{2}^{\boldsymbol{\prime}}}\left[\;-\;f\left(r\right)\dfrac{\mathbf{r}}{r}\right]\circ d\mathbf{r}= - \int_{r_{1}}^{r_{2}}f\left(r\right)dr \tag{08} \end{equation} that is \begin{equation} \tfrac{1}{2}m\left( w_{2}^{2}- w_{1}^{2}\right)= - \int_{r_{1}}^{r_{2}}f\left(r\right)dr = - \left[\Phi\left(r_{2}\right)-\Phi\left(r_{1}\right)\right] \tag{09} \end{equation} From (06) and (09) \begin{equation} \tfrac{1}{2}m\left( w_{2}^{2}- w_{1}^{2}\right)= \tfrac{1}{2}m\left( v_{2}^{2}- v_{1}^{2}\right) \tag{10} \end{equation} so if $\: w_{1}= v_{1}\:$ then $\: w_{2}= v_{2}\:$, QED.

But the whole story is not only to prove this but to talk about what is under the table, as Feynman did.

The function $\:\Phi\left(r\right)\:$ is the potential energy and it is a very important tool : think that you have to calculate the work done by a force $\:\mathbf{f}\left(r\right)\:$ like this in equation (01) from point $\:\rm{P}_{1}\:$ to point $\:\rm{P}_{2}\:$ on a curvilinear path of very complicated equation. Instead of being involved in complex and tedious calculations you have immediately the answer using the potential energy :

work done =$\:\Phi\left(r_{1}\right)-\Phi\left(r_{2}\right)\:$.

Equation (06) or (09) may be expressed as

\begin{equation} \tfrac{1}{2}m v_{2}^{2}+\Phi\left(r_{2}\right)=\tfrac{1}{2}m v_{1}^{2}+\Phi\left(r_{1}\right) \tag{11} \end{equation} yielding the energy conservation \begin{equation} \underbrace{\tfrac{1}{2}m v^{2}}_{kinetic\: energy}+\underbrace{\tfrac{}{}\Phi}_{potential\: energy} = \text{ constant} \tag{12} \end{equation} Note that the potential $\:\phi \:$ is the potential energy per unit charge \begin{equation} \phi = \dfrac{\Phi}{\xi} \tag{13} \end{equation} where $\:\xi\:$ is the charge : $\:\xi= m = \text{mass}\:$ in gravitation , $\:\xi= q = \text{electric charge}\:$ in electrostatics.

SECTION B : Conservative Vector Fields

enter image description here

There exists a relation that connects the vector field $\:\mathbf{f}\left(r\right)\:$ of equation (01) and the scalar potential $\:\Phi\left(r\right)\:$ of equation (07). From (07)

\begin{equation} f\left(r\right)=\dfrac{d\Phi}{dr} \tag{14} \end{equation} On the other hand since $\:\mathbf{r}=\left(x,y,z\right)\:$ and $\:r=\Vert\mathbf{r}\Vert=\sqrt{x^{2}+y^{2}+z^{2}}\:$
\begin{equation} \mathbf{n}_{r}=\dfrac{\mathbf{r}}{r}=\left(\dfrac{x}{r},\dfrac{y}{r},\dfrac{z}{r}\right)=\left(\dfrac{\partial r}{\partial x},\dfrac{\partial r}{\partial y},\dfrac{\partial r}{\partial z}\right) \tag{15} \end{equation}

Inserting the expressions (14) and (15) in (01) yields

\begin{equation} \mathbf{f}\left(r\right)=-\dfrac{d\Phi}{dr}\left(\dfrac{\partial r}{\partial x},\dfrac{\partial r}{\partial y},\dfrac{\partial r}{\partial z}\right)=-\left(\dfrac{d\Phi}{dr}\dfrac{\partial r}{\partial x},\dfrac{d\Phi}{dr}\dfrac{\partial r}{\partial y},\dfrac{d\Phi}{dr}\dfrac{\partial r}{\partial z}\right)=-\left(\dfrac{\partial \Phi}{\partial x},\dfrac{\partial \Phi}{\partial y},\dfrac{\partial \Phi}{\partial z}\right) \tag{16} \end{equation} that is \begin{equation} \mathbf{f}\left(r\right)=\;- \;\nabla \Phi \tag{17} \end{equation} where \begin{equation} \nabla = \left(\dfrac{\partial}{\partial x},\dfrac{\partial }{\partial y},\dfrac{\partial }{\partial z}\right) \tag{18} \end{equation} the well-known "gradient", an important differential operator applied to scalar functions of $\:\left(x,y,z\right)\:$.

The gradient $\:\nabla \Phi \:$ is a vector with magnitude equal to the rate of change of $\:\Phi\:$, change per unit length. But it's not only this : its direction is at any point always normal to the surfaces $\:\Phi = \text{constant}\:$, the so-called equipotential surfaces, as shown in above Figure, and is pointing to the direction of the maximum rate of increase per unit length. The field force is pointing in the opposite, to the maximum rate of decrease of the potential (energy).

Note that under the light of the gradient definition, equation (15) reads \begin{equation} \nabla r = \dfrac{\mathbf{r}}{r}= \mathbf{n}_{r} \tag{15'} \end{equation} In this case the equipotential surfaces are surfaces of spheres.

In the Figure below \begin{equation} \int_{\rm{A}}^{\rm{B}}\mathbf{f}\circ d\mathbf{r}=-\int_{\rm{A}}^{\rm{B}}\nabla \Phi \circ d\mathbf{r}=-\int_{\rm{A}}^{\rm{B}}\left(\dfrac{\partial \Phi}{\partial x} dx + \dfrac{\partial \Phi}{\partial y} dy + \dfrac{\partial \Phi}{\partial z} dz \right)=-\int_{\rm{A}}^{\rm{B}}d\Phi \tag{19} \end{equation} so

\begin{equation} \int_{\rm{A}}^{\rm{B}}\mathbf{f}\circ d\mathbf{r}= \Phi_{1}-\Phi_{2} = \text{independent of the path of integration} \tag{20} \end{equation} or \begin{equation} \oint\mathbf{f}\circ d\mathbf{r}= 0 \quad \text{for every closed path of integration} \tag{21} \end{equation}

Note that equations (17), (20), (21) are equivalent : for example, if the curvilinear integral of a vector field is zero on any closed path then it is the gradient of a scalar field and vice versa. These properties characterize what is called conservative vector fields.

enter image description here

  • $\begingroup$ One thing that I'm not seeing in this answer is the actual answer of "what the heck was Feynman talking about?" He was trying to answer "the closer it is, the faster it's going, but by how much?": his answer is, suppose that we use polar coordinates to go from $(r_1,\theta_1)$ to $(r_2,\theta_2)$ , not by the actual path, but instead by first implicitly going to $(r_1,\theta_2)$ by a circular path (not changing $v^2$ as we go), then falling in radially from $(r_2,\theta_2)$ to $(r_1,\theta_2)$ (speeding up $\vec v$), then orienting $\vec v$ in whatever direction it's supposed to be now. $\endgroup$ – CR Drost Jul 15 '15 at 21:50
  • $\begingroup$ @ChrisDrost You meant from $(r_1,\theta_2)$ to $(r_2,\theta_2)$, right? $\endgroup$ – Omar Nagib Jul 15 '15 at 22:08
  • $\begingroup$ Sorry, yes. Those last two points are flipped. I read "make it move radially, and then we let it fall" as saying "we move in a circle of constant radius, and then we vary just the radius." $\endgroup$ – CR Drost Jul 15 '15 at 22:11
  • $\begingroup$ @Chris Drost Thanks for your comment. I believe that the whole story here is not gravitation or planetary motion. It's a Feynman's effort to introduce the student or anybody interested to the concept of conservative fields etc. $\endgroup$ – user82794 Jul 15 '15 at 22:11
  • 1
    $\begingroup$ I agree too; I'm not saying that your answer was wrong, rather that it was good but left out a crucial detail (the literal-answer to the literal question, which is as important as the general-answer informing about the general question) which I opted to add in a comment rather than by proposing an edit. $\endgroup$ – CR Drost Jul 15 '15 at 22:13

The point is that if $\frac{1}{2} mv^2 - GMm/r$ is constant, then $v$ only depends on $r$! This is surprising and very useful; it means that $v$ will be the same no matter what path a planet takes from some $r_1$ to $r_2$.

In this case, the two paths he's using are the planet's usual elliptical orbit, and a path that goes straight toward the sun. You don't have to worry about visualizing exactly what the paths are; the point is that the path doesn't matter at all.

  • $\begingroup$ Yes I agree with you that $v$ depends only on $r$ and I agree that the path doesn't matter. However my question was, what is Feynman doing with this planet? what he's doing when he says, let's change the direction(but not the magnitude) of the velocity and make it radial, and when he says let the planet freely fall into the radius of interest, I don't get what he's doing in this example. $\endgroup$ – Omar Nagib Jul 15 '15 at 9:12
  • 2
    $\begingroup$ Think about an elliptical orbit Omar, see this. At apogee the planet is going very slow. At perigree it's going very fast, and it's effectively fallen from one to the other. It didn't fall vertically towards the Sun, but if it had fallen vertiacally from the apogee elevation down to the perigree elevation, it would exhibit the same speed difference. $\endgroup$ – John Duffield Jul 15 '15 at 12:08
  • $\begingroup$ You have hit the nail on the head, sir. - Ed. (FLP) $\endgroup$ – Michael A. Gottlieb Jul 16 '15 at 10:44

Feynman's trajectory

The trajectory discussed by Feynman is shown below in red for the blue path, which is a hyperbolic deflection of a small particle around a large star centered at $(0, 0)$.

A blue hyperbola arcs from A on the left to B on the right; a constant-radius circle follows its polar angle until B's angle has been achieved, then the path falls radially in towards B.


Feynman's trajectory here trying to answer the question: how much has the speed increased between A and B. He is answering that by saying that there is an alternative, and in many ways simpler trajectory, which answers that question.

This red path starts with an initial deflection: the particle's $\vec v$, which was approximately $[-3 u, u]$ for some $u$, has instead been deflected to approximately $[0, u\sqrt{10}]$. So, $v^2$ has not changed, but the direction of $\vec v$ has, to match a circular orbit.

The red path then contains a circular orbit which is about 240° long. This orbit is not a physical trajectory, in the sense of gravity providing it for you: as discussed, this is a hyperbolic trajectory, which means $v$ is greater than the escape velocity of the star. Accordingly, the particle can only be kept in this orbit by being pulled towards the star, perhaps by a very strong rope or a very massive rollercoaster track, or a carefully tuned rocket engine. Whichever it is, it provides the extra force needed to keep the particle in this orbit, without speeding up or slowing down the particle. This sounds very complicated! But it is "simple" in the sense that the speed $v$ of the particle is held constant, not changing, and only the angle of the particle about the star changes.

After this circular orbit, the point B is directly between the particle and the star. The particle's velocity again changes direction without changing magnitude. It now points directly in towards the star, and the particle falls, with this starting velocity, directly onto point B. This is the only place where the particle speeds up, and it is also a very simple in-falling trajectory. The "meat" of what Feynman is saying is that the speed increase during this abstract in-falling path is exactly the same as the speed increase of the physical path that the particle actually takes.

Finally, the velocity changes direction to something almost parallel to $-\hat y$, to match the direction of the velocity that the particle has at B. The particle can now follow the exact same hyperbolic trajectory away from the star, as it has the same position and velocity as the physical path would have given it.

So Feynman is saying that there is a two-part trajectory which gives the same speed, in which one arc with "some special radius" (which is the distance from the star to A, the "starting radius" in polar coordinates) is used to fall down to "the radius of interest" (the distance from the star to B). The kinetic energy difference between both paths is the same, therefore if the starting speeds on the two different trajectories are the same, the ending speeds must also be the same. This is a general property of any "conservative force law".

  • $\begingroup$ This is the kind of answer I was hoping for. I think you explained Feynman's reasoning in a very nice way, well done! $\endgroup$ – Omar Nagib Jul 17 '15 at 4:40

Here is what I think he means: first we have a planet going around the sun in some orbit, then we change the direction of the velocity to go radially outwards, for example by letting the planet go inside some pipe we put in it's path (notice that a normal planet would never do this, because there are no big pipes in space and also there would be quite a lot of friction). Then we leave the planet alone, and as you can imagine it will decelerate, and reach zero velocity at some special radius. Then as the planet falls back inwards it accelerates and at some point it will cross the original orbit point where the pipe was (we have quickly removed the pipe). Since we are now back at the same position we started at the velocity has to be the same.

  • 1
    $\begingroup$ this is classic feynman :). letting stuff go around in pipes, or having any other unimaginable friction less constraints. $\endgroup$ – john Jul 15 '15 at 22:53
  • $\begingroup$ If the thing with the pipe seem strange, just imagine a tenis ball going around the sun. We put a plane surface in its path, such that the ball elastically deflects and goes radially outwards. $\endgroup$ – john Jul 16 '15 at 14:07

protected by Qmechanic Jul 16 '15 at 6:57

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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