ϰ
′
+
∫
0
t
n
d
t
{\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 }
p
V
=
δ
{\displaystyle {\frac {p}{V}}=\delta }
n
a
V
=
δ
′
{\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
.
{\displaystyle {\begin{array}{l}\Delta a=0\\\\\Delta e=2\pi {\sqrt {1-e^{2}}}\left\{\lambda \mu \delta ^{2}{\frac {\left(2-e^{2}\right)-2{\sqrt {1-e^{2}}}}{e^{3}}}-\lambda \delta \delta '{\frac {1-{\sqrt {1-e^{2}}}}{e^{2}}}\right\}\\\\\Delta \varphi ={\frac {2\pi }{\sqrt {1-e^{2}}}}\nu \left\{\left[-\lambda \delta ^{2}\cos \omega +\delta \left(e\delta '-\mu \delta \right)\sin \omega \right]{\frac {1-{\sqrt {1-e^{2}}}}{e^{2}}}+\mu \delta ^{2}\sin \omega \right\}\\\\\Delta \theta =-{\frac {2\pi }{{\sqrt {1-e^{2}}}\sin \varphi }}\nu \left\{\left[\lambda \delta ^{2}\sin \omega +\delta \left(e\delta '-\mu \delta \right)\cos \omega \right]{\frac {1-{\sqrt {1-e^{2}}}}{e^{2}}}+\mu \delta ^{2}\cos \omega \right\}\\\\\Delta \varpi =\pi \left(\mu ^{2}-\lambda ^{2}\right)\delta ^{2}{\frac {\left(2-e^{2}-2{\sqrt {1-e^{2}}}\right)}{e^{4}}}+2\pi \mu \delta \delta '{\frac {{\sqrt {1-e^{2}}}-1}{e^{3}}}-\\\\\qquad -{\frac {2\pi \ tg\ {\frac {1}{2}}\varphi }{\sqrt {1-e^{2}}}}\nu \left\{\left[\lambda \delta ^{2}\sin \omega +\delta \left(e\delta '-\mu \delta \right)\cos \omega \right]{\frac {1-{\sqrt {1-e^{2}}}}{e^{2}}}+\mu \delta ^{2}\cos \omega \right\}\\\\\Delta \varkappa '=\pi \left(\lambda ^{2}-\mu ^{2}\right)\delta ^{2}{\frac {\left(2+e^{2}\right){\sqrt {1-e^{2}}}-2}{e^{4}}}-2\pi \delta ^{2}-2\pi \mu ^{2}\delta ^{2}-2\pi \mu \delta \delta '{\frac {\left(1-e^{2}\right)-{\sqrt {1-e^{2}}}}{e^{3}}}.\end{array}}}
§ 10. I have worked out the case of the planet Mercury, taking 276° and + 34° for the right ascension and declination of the apex of the Sun's motion. I have got the following results:
![{\displaystyle {\begin{array}{l}\Delta a=0\\\\\Delta e=2\pi {\sqrt {1-e^{2}}}\left\{\lambda \mu \delta ^{2}{\frac {\left(2-e^{2}\right)-2{\sqrt {1-e^{2}}}}{e^{3}}}-\lambda \delta \delta '{\frac {1-{\sqrt {1-e^{2}}}}{e^{2}}}\right\}\\\\\Delta \varphi ={\frac {2\pi }{\sqrt {1-e^{2}}}}\nu \left\{\left[-\lambda \delta ^{2}\cos \omega +\delta \left(e\delta '-\mu \delta \right)\sin \omega \right]{\frac {1-{\sqrt {1-e^{2}}}}{e^{2}}}+\mu \delta ^{2}\sin \omega \right\}\\\\\Delta \theta =-{\frac {2\pi }{{\sqrt {1-e^{2}}}\sin \varphi }}\nu \left\{\left[\lambda \delta ^{2}\sin \omega +\delta \left(e\delta '-\mu \delta \right)\cos \omega \right]{\frac {1-{\sqrt {1-e^{2}}}}{e^{2}}}+\mu \delta ^{2}\cos \omega \right\}\\\\\Delta \varpi =\pi \left(\mu ^{2}-\lambda ^{2}\right)\delta ^{2}{\frac {\left(2-e^{2}-2{\sqrt {1-e^{2}}}\right)}{e^{4}}}+2\pi \mu \delta \delta '{\frac {{\sqrt {1-e^{2}}}-1}{e^{3}}}-\\\\\qquad -{\frac {2\pi \ tg\ {\frac {1}{2}}\varphi }{\sqrt {1-e^{2}}}}\nu \left\{\left[\lambda \delta ^{2}\sin \omega +\delta \left(e\delta '-\mu \delta \right)\cos \omega \right]{\frac {1-{\sqrt {1-e^{2}}}}{e^{2}}}+\mu \delta ^{2}\cos \omega \right\}\\\\\Delta \varkappa '=\pi \left(\lambda ^{2}-\mu ^{2}\right)\delta ^{2}{\frac {\left(2+e^{2}\right){\sqrt {1-e^{2}}}-2}{e^{4}}}-2\pi \delta ^{2}-2\pi \mu ^{2}\delta ^{2}-2\pi \mu \delta \delta '{\frac {\left(1-e^{2}\right)-{\sqrt {1-e^{2}}}}{e^{3}}}.\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fbc920870e3136cfc41f4047b32d219c1d95d474)
