Derivation of Laplace's Equation for the Speed of Gravity from Newtonian Dynamics

 Derivations of Laplace's formula for the change in period of the moon, that yielded a formula for the speed of gravity, has been derived by a number of authors. The key point is to consider the tangential force acting on the moon due to aberration due to the finite speed at which (supposedly) gravity propagates. To the central force, $\mu/a^2$,  where $\mu$ is the gravitational parameter containing the mass of the earth, $GM$, and $a$ the semi-major axis, the aberration term $v/v_G$ is appended where $v_G$ is the speed at which gravity supposedly propagates. This can be looked upon as a phase which is the angular speed, $\omega$, times the time that the gravitational disturbance travels a distance, $a$, viz. $\theta=\omega\cdot a/v_G$ $v$ is the angular velocity of the moon about the earth, and not, as McDonald has claimed in his note, "Laplace and the Speed of Gravity", is the velocity of the earth,  The velocity of the earth is entirely superfluous to the problem.

Another attempt to derive the Laplace equation,

$$\dot{P}=-6\pi\frac{v}{v_G},$$

 for the rate of change of the period, $P$, where the dot denotes the derivative with respect to time, was carried out by van Flandern who came to the same conclusion as Laplace that $v_G$  must be "at least $100$ million times greater than that of light." He uses the secular variation of the semi-major axis

$$a^2(t)=a_0^2+4\frac{\mu}{v_G}(t-t_0),$$

which he attributes to Danby, Fundamental of Celestial Mechanics, p. 327. Notwithstanding the my inability to find such an equation, that equation is just the same as the rate of change of the angular velocity below under the condition of Kepler's III, $a v^2=\mbox{const}$, or $a^2\omega^3=\mbox{const}$, viz.,

$$\frac{\dot{\omega}}{\omega}=-\frac{\dot{P}}{P}=-\frac{3}{2}\frac{\dot{a}}{a}.$$

Now to Kepler's II, which reduces to $av^2=\mbox{const}$, expressing the conservation of angular momentum, 

$$K=av\sin\alpha,$$

under the condition that $\sin\alpha\sim v$, which can be equivalently expressed by

$$\frac{\dot{a}}{a}=-2\frac{\dot{v}}{v}.$$

For the acceleration, $\dot{v}$, we introduce the tangential force, $\mathcal{T}$,

$$\dot{v}=\mathcal{T}=\frac{\mu}{a^2}\frac{v}{v_{G}},$$

which is the central force field, $\mu/a^2$, and the first-order aberration term $v/v_G$, which takes into account the finite speed at which gravity propagates that is responsible for aberration. 

Laplace's formula follows easily. Eliminating $\dot{a}$ in favor of $\dot{P}$, and using the fact that for uniform acceleration, $v=2\pi a/P$, we easily arrive at Laplace's equation, including the negative sign. This says that drag, which results from aberration, leads to a decrease in the period. 

Laplace was pleased with this since it supposedly supported Halley's claims that the moon would eventually fall to earth. Darwin predicted the contrary: tidal forces would increase the period of the moon and the lengthening of the earth day caused by the increasing distance between the earth and moon. It is surprising that McDonald gets the same numerical estimate by using the angular speed of the earth about the sun which is totally irrelevant. 

Be that as it may, we have just shown that Laplace's formula is a simple consequence of Kepler's II and III laws. These are the perturbation equations of celestial mechanics, so that it is very hard to find fault with Laplace's equation. Plugging in the known values leads immediately to Laplace's conclusion. The speed of light, in comparison, would be far too low to satisfy it.

We can do more. Consider Newton's proof of his inverse square law. The  tangential, or centripetal, force is

$$\mathcal{T}=\mathcal{F}\sin\alpha=\frac{v^2}{\rho},$$

where $\mathcal{F}$ is the central force whose inverse square dependency, Newton sought to determine. $\alpha$ is the complementary angle to $\theta$ which is the angle between the central force and the centripetal force. 

Now rhw radius of curvature $\rho$ is

$$\rho=a\frac{[1+z^2(\alpha)]^{3/2}}{|z^{\prime}(\alpha)|},$$

where the prime means differentiation with respect to $\alpha$, and $z$ is the slope of the curve,

$$z=\frac{1}{r}\frac{dr}{d\theta}=\cot\alpha.$$ The expression in the numerator of the radius of curvature is

$$1+z^2=1/\sin^3\alpha$$

We thus find the curvature given by

$$\rho=a/\sin\alpha.$$

Alternatively, we can find $\rho$ from Newton's equation

$$\rho=\frac{K^2}{\mathcal{F}a^2\sin\alpha^3},$$

where the angular momentum per unit mass, $K$ is given above.  In order that Newton's law of gravitation hold, it is necessary that

$$\rho\sin^3\alpha=a\sin^2\alpha=\frac{av^2}{v_G^2}=\mbox{const}.$$

The latter follows from Kepler's II: $av^2=\mbox{const}$.

Let us consider the last equation more closely. Due to the retardation of  the gravitational force, we compare the position true position of the moon $A_t$, and its linearly extrapolated retarded position $A_c$, as shown in the figure below.

comparison

For small angles $\alpha$, which is indeed small no matter the magnitude of $v_G$, it is the angle measured on earth $B$ through which the moon moves, in gravitational time $a/v_G$, from its past position, $A_r$ to its present position, $A_t$. The arc length is $s=\alpha a=r^2\omega/v_G$, which is the ratio of the specific angular momentum to the speed of gravity, We have constructed the same distance from $A_r$ to $A_c$ if the moon where to fly off its circular path. The angular difference between the two position is $\psi$ which causes the linearly extrapolated position to feel a transverse force that increases continually as the moon makes its way around the earth. 

By simple trigonometry we have

$$\psi=\alpha-\tan^{-1}\alpha\approx\frac{v}{v_G}-\tan^{-1}\frac{v}{v_G},$$

where we used the fact that $\sin\alpha\approx\alpha$. Expanding the arc tangent in powers and retaining only the lowest non-vanishing order give

$$\psi=\frac{1}{3}\left(\frac{v}{v_G}\right)^3.$$

It is this factor that multiplies the radius of curvature in the above formula causing

$$\rho\psi=\mbox{const}.$$

Aberration with therefore cause the curvature $K_G=1/\rho$ to increase due to the effect of the force felt by the finite time of propagation of the gravity.

If $a$ were constant, the radius of curvature $\rho$ would undergo a first-order aberrational effect. But, because of the perturbation that causes $a$ to vary it will undergo a third order aberrational effect, and this ensures Newton's inverse square law. 

To get an idea of the magnitudes we are talking about consider the binary pulsar $PSR 1913+16$ The measured value of $\dot{P}=-2.42\times 10^{-12}$, $a/c=2.342$, and $P=27,907$ both in seconds. This gives a relativity velocity, with respect to light speed $c$ of $v/c=5.26\times 10^{-4}$. Plugging the numbers into Laplace's equation gives a relative speed of propagation of gravity of

$$v_G=4 \times 10^9 c.$$

This is one order of magnitude smaller than that predicted by  van Flandern  ($2\times 10^{10}$) shortly before his death.