In **The Feynman Lectures**, 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.

First of all we know that : $W=\Delta T=0$. where $T$ is the kinetic energy of a particle.

The potential energy is given by: $-\frac{GMm}{r}$.

Assume, we have some closed curve, and we observe a particle on it at point $P_1$. It's kinetic energy is $T_{1_i}$. We let it go around in the curve and return to $P_1$ with its final kinetic energy being $T_{1_f}$ . Since $\Delta T=0$ therefore:

$T_{1_i}$= $T_{1_f}=T_{1}$.

Now If we add the potential energy of the particle at $P_1$ which is $-\frac{GMm}{r_{1}}$ to $T_1$, we get:

$T_{1}-\frac{GMm}{r_{1}}=C_1$. Where $C_1$ is a constant.

If we choose another point $P_2$ and apply the same reasoning we get:

$T_{2}-\frac{GMm}{r_{2}}=C_2$. Where $C_2$ is a constant.

My question is:Feynman says in the paragraph that $W=0$ implies $C_1=C_2$, is this implication true? In other words, How does The work done in going around any path in a gravitational field is zero implies conservation of energy($\Delta T=-\Delta U$)?

## Best Answer

The reason for this is that, in your notation, $W=C_1-C_2$. This means that the work performed in moving about a circuit from point $A$ to point $B$ along curve $\mathcal C_1$ and then back along $\mathcal C_2$ is exactly the difference between the work performed in moving the particle along $\mathcal C_1$ versus moving it along $\mathcal C_2$.

This is simply a mathematical fact regarding how we calculate the work done in moving a particle along a curve $\mathcal C$ in a force field $\mathbf F$, which is defined as $$ W=\int_{\mathcal C}\mathbf F\cdot \mathrm d\mathbf r. $$ If $\mathcal C$ is the union of $\mathcal C_1$ and $\mathcal C_2'$, then this is linear, $$ W = \int_{\mathcal C_1\oplus \mathcal C_2'}\mathbf F\cdot \mathrm d\mathbf r = \int_{\mathcal C_1}\mathbf F\cdot \mathrm d\mathbf r + \int_{\mathcal C_2'}\mathbf F\cdot \mathrm d\mathbf r, $$ and if $\mathcal C_2'$ is $\mathcal C_2$ traversed in a reverse fashion then the sign of $\mathrm d\mathbf r$ is reversed, so $$ W = \int_{\mathcal C_1\ominus \mathcal C_2}\mathbf F\cdot \mathrm d\mathbf r = \int_{\mathcal C_1}\mathbf F\cdot \mathrm d\mathbf r - \int_{\mathcal C_2}\mathbf F\cdot \mathrm d\mathbf r .$$

This then means that the statements

and

are equivalent.

To get to Feynman's precise statement, you need two more key facts.

The first is the work-energy theorem, which integrates Newton's second law of motion to tell you that the work performed along a curve $\mathcal C$ is equal to the change in kinetic energy along it.

The second is an explicit integral of $\int_{\mathcal C}\mathbf F\cdot \mathrm d\mathbf r$ for the gravitational inverse-square force.

Both of these are more trivial calculations for which you can probably fill in the gaps if needed. (But tell me if you need more detail on either.)