A confusion regarding an example in The Feynman Lectures 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?
 A: 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.
A: 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 $(0, 0)$.

Discussion
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".
A: SECTION A : The example in Feynman's Lectures

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

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.  


A: 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.
