How does The work done in going around any path in a gravitational field is zero implies conservation of energy?

Question

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($\Delta T=-\Delta U$)?

Answer 1

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

The work performed around a closed loop is zero

and

The work performed between point $A$ and point $B$ is independent of the path

are equivalent.

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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.)

Answer 2

Starting from the work around a close curve $$W = \oint \vec F(r)d\vec s = 0$$

now be stokes throem:

$$ \oint \vec F(r)d\vec s = \iint_A \vec \nabla \times\vec F(r)d\vec A = 0$$

where $A$ is the area surrounded by the closed curve.

Now if you do the math you will see that $ \vec \nabla \times \vec \nabla U( r) = 0 $ for any field $U( r)$ thus we can write $\vec F( r)=-\vec \nabla U( r)$ for some $U(r)$ which we call the potetial. Because writing $\vec F(r)$ like this directly implies the first equation.

We don't need to know $U(r)$ explicitly to do the next argument:

$$T(P_2) - T(P_1 )= W = \int_{P_1}^{P_2} \vec F(r) d\vec s = -\int_{P_1}^{P_2} \vec \nabla U( r) d\vec s = U(P_1) - U(P_2)$$

It follows that $T(P_1) + U(P_1)= T(P_2) +U(P_2)$ thus the quantity $E = T+U$ is conserved for every point in a curve. We call $E$ the Energy. Similarly if we go around a a closed path: $U(P_1) - U(P_2) = -\Delta U = T(P_2) - T(P_1 ) = \Delta T = 0 $

Answer 3

If a mass M is at a point travelling at a given velocity (= speed and direction vector) then it has a given amount of energy (potential and kinetic energy summed) relative to being stationary* at the same point with respect to a given frame of reference. ie solely by knowing position and velocity vector the PE & KE are defined. (* or at some zero reference velocity).

This applies whether the mass is in a gravitational field or in a (local) gravity free field. (The energies will be different in the two cases but in each case velocity and position define the potential and kinetic energy for that case).

If the mass now leaves that point and "wanders about" in some manner and then returns to the same point with the same velocity vector (and with nothing else changed) then the energy will be identical to what it was before. ie just as PE & KE were determined solely by position and velocity vector previously they will again be now. If during the process of "wandering around" no energy was taken from or added to the mass by known and/or measurable mechanisms then, as the energy is unchanged, energy must have been conserved.

In a gravitational field or even in "free space" with no gravity, most things that we do to cause "wandering about" to happen, will require energy input or output from the mass. Usually, if we wish to "turn right"or slow or accelerate or "do a U turn" etc then we need to apply a force and to do so work will be done.

What Feynman is pointing out is that in this instance, although the velocity vector rotates through 360 degrees in direction (in the "orbital plane") and even though velocity increases to some maximum (aka perigee) and decreases to some minimum (aka apogee) and even though the mass travels in an elliptical path (which to our minds might suggest it is being acted on by forces which vary constantly in magnitude and direction), we find that when it passes through the original reference point it has identical velocity as originally, and it can do this time after time after time after ... forever (in the absence of the universe growing old etc) with exactly the same velocity at the same point every time. There is no "very small amount of energy lost or altered" - we know that in the absence of other influences a basic "2 body system" works just the same after 1 or 100 billion "orbits".

ie the mass starts with a given energy, undergoes changes in acceleration, velocity and position and returns to an exactly identical state as before with the same defined energy. So, energy is conserved in the process of completing an orbit.

The strangest thing about completing an "orbit" is, as Feynman notes, that we do not realise that it is strange.

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As a bonus - it happens that for this to work a result drops out of the basic math which shows that attractive forces must be proportional to 1/(distance^N)
where N = 2.000000.... exactly.
ie the inverse square law must be EXACT.
Not 2.0000000000 ...1... or any other value.

This is another way of putting what Feynman put as

  " 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"