# Considerations on Gravitation

##### 5 minutes • 965 words

Every theory of gravitation has to deal with the problem of the influence, exerted on this force by the motion of the heavenly bodies.

The solution is easily deduced from our equations; it takes the same form as the corresponding solution for the electromagnetic actions between charged particles.[4]

I shall only treat the case of a body A, revolving around a central body M, this latter having a given constant velocity p. Let r be the line MA, taken in the direction from M towards A, x, y, z the relative coordinates of A with respect to M, w the velocity of A’s motion relatively to M, ϑ the angle between w and p, finally p r {\displaystyle p_{r}} {\displaystyle p_{r}} the component of p in the direction of r.

Then, besides the attraction k r 2 (14)

which would exist if the bodies were both at rest, A will be subject to the following actions.

1st. A force k ⋅ p 2 2 V 2 ⋅ 1 r 2 (15)

in the direction of r.

2nd. A force whose components are (16)

A force − k V 2 p ⋅ 1 r 2 d r d t (17)

parallel to the velocity p.

4th. A force k V 2 1 r 2 p w cos ϑ (18)

in the direction of r.

Of these, (15) and (16) depend only on the common velocity p, ( 17) and (18) on the contrary, on p and w conjointly.

It is further to be remarked that the additional forces (15) — (18) are all of the second order with respect to the small quantities

p V

In so far, the law expressed by the above formulae presents a certain analog)- with the laws proposed by Weber, Riemann and Clausius for the electromagnetic actions, and applied by some astronomers to the motions of the planets. Like the formulae of Clausius, our equations contain the absolute velocities, i. e. the velocities, relatively to the aether.

There is no doubt but that, in the present state of science, if we wish to try for gravitation a similar law as for electromagnetic forces, the law contained in (15) — (18) is to be preferred to the three other just mentioned laws.

§ 9.

The forces (15) — (18) will give rise to small inequalities in the elements of a planetary orbit ; in computing these, we have to take for p the velocity of the Sun’s motion through space. I have calculated the secular variations, using the formulae communicated by Tisserand in his Mécanique céleste.

Let:
-`a`

us the mean distance to the sun:

`e`

the eccentricity,`φ`

the inclination to the ecliptic,`θ`

the longitude of the ascending node,`ϖ`

{\displaystyle \varpi } {\displaystyle \varpi } the longitude of perihelion,`ϰ`

{\displaystyle \varkappa ‘} {\displaystyle \varkappa ‘} the mean anomaly at time t=0, in this sense that, if n be the mean motion, as determined by a, the mean anomaly at time t is given by

ϰ ′ + ∫ 0 t n d t {\displaystyle \varkappa ‘+\int \limits _{0}^{t}\ n\ dt} {\displaystyle \varkappa ‘+\int \limits _{0}^{t}\ n\ dt}.

Further, let λ, μ, and ν be the direction-cosines of the velocity p with respect to= 1st. the radius vector of the perihelion, 2nd. a direction which is got by giving to that radius vector a rotation of 90°, in the direction of the planet’s revolution, 3rd. the normal to the plane of the orbit, drawn towards the side whence the planet is seen to revolve in the same direction as the hands of a watch.

Put ω = ϖ − θ {\displaystyle \omega =\varpi -\theta } {\displaystyle \omega =\varpi -\theta }, p V = δ {\displaystyle {\frac {p}{V}}=\delta } {\displaystyle {\frac {p}{V}}=\delta } and n a V = δ ′ {\displaystyle {\frac {na}{V}}=\delta ‘} {\displaystyle {\frac {na}{V}}=\delta ‘} (na is the velocity in a circular orbit of radius a).

Then I find for the variations during one revolution Δ a = 0 Δ e = 2 π 1 − e 2 { λ μ δ 2 ( 2 − e 2 ) − 2 1 − e 2 e 3 − λ δ δ ′ 1 − 1 − e 2 e 2 } Δ φ = 2 π 1 − e 2 ν { [ − λ δ 2 cos ω + δ ( e δ ′ − μ δ ) sin ω ] 1 − 1 − e 2 e 2 + μ δ 2 sin ω } Δ θ = − 2 π 1 − e 2 sin φ ν { [ λ δ 2 sin ω + δ ( e δ ′ − μ δ ) cos ω ] 1 − 1 − e 2 e 2 + μ δ 2 cos ω } Δ ϖ = π ( μ 2 − λ 2 ) δ 2 ( 2 − e 2 − 2 1 − e 2 ) e 4 + 2 π μ δ δ ′ 1 − e 2 − 1 e 3 − − 2 π t g 1 2 φ 1 − e 2 ν { [ λ δ 2 sin ω + δ ( e δ ′ − μ δ ) cos ω ] 1 − 1 − e 2 e 2 + μ δ 2 cos ω } Δ ϰ ′ = π ( λ 2 − μ 2 ) δ 2 ( 2 + e 2 ) 1 − e 2 − 2 e 4 − 2 π δ 2 − 2 π μ 2 δ 2 − 2 π μ δ δ ′ ( 1 − e 2 ) − 1 − e 2 e 3 .