Stochastic Gauss equations

Abstract

We derived the equations of Celestial Mechanics governing the variation of the orbital elements under a stochastic perturbation, thereby generalizing the classical Gauss equations. Explicit formulas are given for the semimajor axis, the eccentricity, the inclination, the longitude of the ascending node, the pericenter angle, and the mean anomaly, which are expressed in term of the angular momentum vector $\mathbf{H }$ per unit of mass and the energy E per unit of mass. Together, these formulas are called the stochastic Gauss equations, and they are illustrated numerically on an example from satellite dynamics.

Appendix: Proof of the stochastic Gauss equations

In what follow, we always simplify the computations in terms of orbital elements using the formulas from (4) to (8). Moreover, we denote by $\tilde{\mathbf{E}}=v\tilde{\mathbf{R}}+rw\tilde{\mathbf{T}}$ and $\tilde{\mathbf{H}}=r\tilde{\mathbf{T}}$ the stochastic part of the variation of the energy and the angular momentum [see Eq. (33) and (34)]. In the same way, we define for all the orbital elements and the angular momentum vector components, the quantities $\tilde{\mathbf{a}},\tilde{\mathbf{e}},\tilde{\mathbf{i}},\tilde{{\varvec{\Omega }}},\tilde{{\varvec{\omega }}},\tilde{\mathbf{H}}_x,\tilde{\mathbf{H}}_y$ and $\tilde{\mathbf{H}}_z$ to be the stochastic part in their variation.

1.1 Semimajor axis a

We use the relation (13) linking the energy E and the semimajor axis a in order to have

$$\begin{aligned} a=-\frac{\mu }{2E}. \end{aligned}$$

(53)

Using Itô’s formula on the previous equation gives

$$\begin{aligned} \mathrm{d}a = \frac{\mu }{2 E^2}\mathrm{d}E - \mu \frac{\tilde{\mathbf{E}}\cdot \tilde{\mathbf{E}}}{2 E^3}\mathrm{d}t. \end{aligned}$$

Using the expression of the variation of the energy E we obtain the result for $\mathrm{d}a$.

1.2 Eccentricity e

Using Itô’s formula on Eq. (14), we obtain

$$\begin{aligned} \mathrm{d}e=\frac{2HE}{e\mu ^2}\mathrm{d}H+\frac{H^2}{e\mu ^2}\mathrm{d}E + \left( \frac{E}{e^3\mu ^2}\tilde{\mathbf{H}}\cdot \tilde{\mathbf{H}}- \frac{H^4}{2e^3\mu ^4}\tilde{\mathbf{E}}\cdot \tilde{\mathbf{E}}+ \frac{2H(H^2E+\mu ^2)}{e^3\mu ^4}\tilde{\mathbf{H}}\cdot \tilde{\mathbf{E}}\right) \mathrm{d}t. \end{aligned}$$

First, notice that

$$\begin{aligned} 1+e\cos f-\frac{(1-e^2)}{1+e\cos f}=e\left( \cos f + \frac{e+\cos f}{1+e\cos f}\right) \end{aligned}$$

then

$$\begin{aligned} \frac{2HE}{e\mu ^2}\mathrm{d}H+\frac{H^2}{e\mu ^2}\mathrm{d}E =&\left[ \sqrt{\frac{a(1-e^2)}{\mu }}\left( \sin f \bar{R} + \left( \cos f + \frac{e+\cos f}{1+e\cos f}\right) \bar{T} \right) \right. \\&\left. +\frac{a(1-e^2)}{2e\mu } \left( \tilde{\mathbf{R}}^2+\tilde{\mathbf{T}}^2\right) \right] \mathrm{d}t \\&+\sqrt{\frac{a(1-e^2)}{\mu }}\left( \sin f \tilde{\mathbf{R}}+ \left( \cos f + \frac{e+\cos f}{1+e\cos f}\right) \tilde{\mathbf{T}}\right) \cdot \mathrm{d}\mathbf{B} . \end{aligned}$$

Second, using the expression of dH and dE we obtain

$$\begin{aligned} \tilde{\mathbf{H}}\cdot \tilde{\mathbf{H}}&= \frac{a^2(1-e^2)^2}{(1+e\cos f)^2}\tilde{\mathbf{T}}^2, \\ \tilde{\mathbf{E}}\cdot \tilde{\mathbf{E}}&= \frac{\mu e^2\sin ^2 f}{a(1-e^2)}\tilde{\mathbf{R}}^2 + \frac{\mu (1+e\cos f)^2}{a(1-e^2)}\tilde{\mathbf{T}}^2+\frac{2e\mu \sin f (1+e\cos f)}{a(1-e^2)}\tilde{\mathbf{R}}\cdot \tilde{\mathbf{T}},\\ \tilde{\mathbf{H}}\cdot \tilde{\mathbf{E}}&= \sqrt{\mu a(1-e^2)}\tilde{\mathbf{T}}^2+\frac{e\sin f \sqrt{\mu a(1-e^2)}}{1+e\cos f}\tilde{\mathbf{R}}\cdot \tilde{\mathbf{T}}. \end{aligned}$$

Finally, after simplifications we obtain the result for de.

1.3 Inclination i and ascending node $\varOmega$

In what follows, we assume that i is not equal to zero. The variation of the inclination and the ascending node are related to the variation of the angular momentum vector $\mathbf{H }$. We compute firstly the variation of the vector $\mathbf{H }$. Using Itô’s formula, we obtain

$$\begin{aligned} \mathrm{d}\mathbf{H } = \mathrm{d}\mathbf{r } \times \mathbf{v } + \mathbf{r } \times \mathrm{d}\mathbf{v } + \mathrm{d}\mathbf{r } \times \mathrm{d}\mathbf{v }. \end{aligned}$$

Then, using the perturbed equations of motions (31)–(32) we obtain

$$\begin{aligned} \mathrm{d}\mathbf{H } = \mathbf{r } \times \mathrm{d}\mathbf{v }_P . \end{aligned}$$

(54)

Finally,

$$\begin{aligned} \mathrm{d}\mathbf{H } = -r(\bar{N}dt+\tilde{\mathbf{N}}\cdot d\mathbf{B}) \mathbf{e }_T + r(\bar{T}dt +\tilde{\mathbf{T}}\cdot d\mathbf{B} ) \mathbf{e }_N. \end{aligned}$$

(55)

The expression of $d \mathbf{H }$ in the inertial frame is obtained as using three rotations (see Fig. 1)

$$\begin{aligned} \mathrm{d} \mathbf{H } =&\,r \bigg [\bigg ( \sin i \sin \varOmega (\bar{T}\mathrm{d}t + \tilde{\mathbf{T}}\cdot \mathrm{d}\mathbf{B} ) \nonumber \\&+ (\bar{N}\mathrm{d}t + \tilde{\mathbf{N}}\cdot \mathrm{d}\mathbf{B} ) (\cos i \sin \varOmega \cos (f+\omega )+\cos \varOmega \sin (f+\omega ))\bigg )\mathbf{e }_x \nonumber \\&-\bigg ( \sin i \cos \varOmega (\bar{T}\mathrm{d}t + \tilde{\mathbf{T}}\cdot \mathrm{d}\mathbf{B} ) \nonumber \\&+ (\bar{N}\mathrm{d}t + \tilde{\mathbf{N}}\cdot \mathrm{d}\mathbf{B} ) (\cos i \cos \varOmega \cos (f+\omega )-\sin \varOmega \sin (f+\omega ))\bigg )\mathbf{e }_y \nonumber \\&+\bigg ( \cos i (\bar{T}\mathrm{d}t + \tilde{\mathbf{T}}\cdot \mathrm{d}\mathbf{B} )- \sin i \cos (f+\omega ) (\bar{N}\mathrm{d}t + \tilde{\mathbf{N}}\cdot \mathrm{d}\mathbf{B} )\bigg )\mathbf{e }_z \bigg ] \nonumber \end{aligned}$$

with

$$\begin{aligned} \tilde{\mathbf{H}}_x&=r \left( \sin i \sin \varOmega \tilde{\mathbf{T}}+ \tilde{\mathbf{N}}\left( \cos i \sin \varOmega \cos (f+\omega )+\cos \varOmega \sin (f+\omega )\right) \right) , \\ \tilde{\mathbf{H}}_y&=r\left( -\sin i \cos \varOmega \tilde{\mathbf{T}}- \tilde{\mathbf{N}}\left( \cos i \cos \varOmega \cos (f+\omega )-\sin \varOmega \sin (f+\omega )\right) \right) , \\ \tilde{\mathbf{H}}_z&=r\left( \cos i\tilde{\mathbf{T}}- \sin i \cos (f+\omega )\tilde{\mathbf{N}}\right) . \end{aligned}$$

Now we can compute the variation of the inclination i. Using Itô’s formula on Eq. (15), we obtain

$$\begin{aligned} -\sin i\mathrm{d}i-\frac{1}{2} \cos i (\tilde{\mathbf{i}}\cdot \tilde{\mathbf{i}}) \mathrm{d}t= \frac{H_z}{H^2}\mathrm{d}H-\frac{\mathrm{d}H_z}{H}+ \left( \frac{\tilde{\mathbf{H}}_z \cdot \tilde{\mathbf{H}}}{H^2}-\frac{H_z(\tilde{\mathbf{H}}\cdot \tilde{\mathbf{H}})}{H^3}\right) \mathrm{d}t \end{aligned}$$

and so

$$\begin{aligned} \mathrm{d}i = -\frac{H^3 \cos i (\tilde{\mathbf{i}}\cdot \tilde{\mathbf{i}})-2 H (\tilde{\mathbf{H}}\cdot \tilde{\mathbf{H}}_z )+2 H_z (\tilde{\mathbf{H}}\cdot \tilde{\mathbf{H}})}{2 H^3 \sin {i}} \mathrm{d}t + \frac{H_z}{H^2 \sin {i}}\mathrm{d}H -\frac{1}{H \sin {i}} \mathrm{d}H_z. \end{aligned}$$

Using the expression of $\mathrm{d}H_z$ and $\mathrm{d}H$, we finally obtain the result for $\mathrm{d}i$. Next, we compute the variation of the ascending node $\varOmega$. Using Itô’s formula on Eq. (16), we obtain

$$\begin{aligned} \frac{\mathrm{d}\varOmega }{\cos ^2 \varOmega } + \frac{ (\tilde{{\varvec{\Omega }}}\cdot \tilde{{\varvec{\Omega }}}) \tan \varOmega }{\cos ^2 \varOmega }\mathrm{d}t = -\frac{\mathrm{d}H_x}{H_y}+\frac{H_x}{H_y^2}\mathrm{d}H_y + \left( \frac{\tilde{\mathbf{H}}_x\cdot \tilde{\mathbf{H}}_y}{H_y^2} -\frac{ H_x (\tilde{\mathbf{H}}_y\cdot \tilde{\mathbf{H}}_y)}{H_y^3}\right) \mathrm{d}t \end{aligned}$$

and so

$$\begin{aligned} \mathrm{d}\varOmega =&\left( \frac{-H_x \cos ^2\varOmega (\tilde{\mathbf{H}}_y \cdot \tilde{\mathbf{H}}_y)}{H_y^3}+ \frac{ \cos ^2\varOmega (\tilde{\mathbf{H}}_x \cdot \tilde{\mathbf{H}}_y)}{H_y^2} -\tan \varOmega (\tilde{{\varvec{\Omega }}}\cdot \tilde{{\varvec{\Omega }}}) \right) \mathrm{d}t \\&-\frac{\cos ^2\varOmega }{H_y}\mathrm{d}H_x + \frac{H_x \cos ^2\varOmega }{H_y^2} \mathrm{d}H_y . \end{aligned}$$

Using the expression of $H_x$,$H_y$ and $\mathrm{d}H_x$,$\mathrm{d}H_y$, we can simplify the expression as

$$\begin{aligned} \frac{\cos ^2\varOmega }{H_y^2}(H_x \mathrm{d}H_y - H_y \mathrm{d}H_x) = \frac{r \sin (f+\omega )}{H\sin i}\bar{N}\mathrm{d}t + \frac{r \sin (f+\omega )}{H\sin i}\tilde{\mathbf{N}}\cdot \mathrm{d}\mathbf{B}. \end{aligned}$$

After simplifications, we obtain the result for $\mathrm{d}\varOmega$.

1.4 Pericenter $\omega$

In order to derive the variation of the pericenter location, we compute firstly the variation of the true anomaly f and secondly the variation of the position angle $\theta$. Using Itô’s formula on Eq. (4), we obtain

$$\begin{aligned}&\frac{2 H}{\mu r}\mathrm{d}H +\left( \frac{\tilde{\mathbf{H}}\cdot \tilde{\mathbf{H}}}{\mu r}-\frac{H^2 v}{\mu r^2}\right) \mathrm{d}t\\&\quad = \cos fde -e \sin f\mathrm{d}f + \left( -\frac{1}{2} e \cos f (\tilde{\mathbf{f}}\cdot \tilde{\mathbf{f}}) -\sin f (\tilde{\mathbf{E}}\cdot \tilde{\mathbf{f}})\right) \mathrm{d}t \end{aligned}$$

and so

$$\begin{aligned} \mathrm{d}f= \left( \frac{H^2 v }{e \mu r^2 \sin {f}}-\frac{\tilde{\mathbf{H}}^2 }{e \mu r\sin {f}}-\frac{\tilde{\mathbf{E}}\cdot \tilde{\mathbf{f}}}{e}-\frac{1}{2}\cot f (\tilde{\mathbf{f}}\cdot \tilde{\mathbf{f}})\right) \mathrm{d}t + \frac{\cot f}{e} \mathrm{d}e -\frac{2 H }{e \mu r \sin {f}} \mathrm{d}H. \end{aligned}$$

Using the expression of the variation of the angular momentum $\mathrm{d}H$, we obtain

$$\begin{aligned} \mathrm{d}f =&\left( -\frac{2 H \bar{T} }{e \mu \sin {f}} -\frac{\tilde{\mathbf{E}}\cdot \tilde{\mathbf{f}}}{e}+\frac{H v w }{e \mu \sin {f}}-\frac{r \tilde{\mathbf{T}}^2 }{e \mu \sin {f}} -\frac{1}{2} \cot f (\tilde{\mathbf{f}}\cdot \tilde{\mathbf{f}})\right) \mathrm{d}t \\&-\frac{2 H}{e \mu \sin {f}}\tilde{\mathbf{T}}\cdot d\mathbf{B} + \frac{\cot f}{e} \mathrm{d}e. \end{aligned}$$

Finally, using the expression of de and after simplifications, we obtain

$$\begin{aligned} \mathrm{d}f =&\bigg [ \sqrt{\frac{a(1-e^2)}{\mu }}\frac{1}{e}\left( \cos f \bar{R}- \sin f \left( \frac{2+e\cos f}{1+e\cos f} \right) \bar{T}\right) + \frac{\sqrt{\mu }}{(a(1-e^2))^{3/2}}(1+e\cos f)^2 \nonumber \\&+ \frac{a(1-e^2)}{\mu e^2}\bigg ( -\frac{\sin 2f}{2} \tilde{\mathbf{R}}^2 + \left( e+\cos f(2+e\cos f)^2 \right) \frac{\sin f}{(1+e\cos f)^2} \tilde{\mathbf{T}}^2 \nonumber \\&- \left( \frac{2+e\cos f}{1+e\cos f}\right) \cos 2f \tilde{\mathbf{R}}\cdot \tilde{\mathbf{T}}\bigg )\bigg ]\mathrm{d}t \nonumber \\&+\sqrt{\frac{a(1-e^2)}{\mu }}\frac{1}{e}\left( \cos f \tilde{\mathbf{R}}- \sin f \left( \frac{2+e\cos f}{1+e\cos f} \right) \tilde{\mathbf{T}}\right) \cdot \mathrm{d}\mathbf{B} . \end{aligned}$$

(56)

In order to compute the variation of the position angle, we use the z-component of the vector $d\mathbf{r }$ and we use the Itô’s formula on the z-component of $d\mathbf{r }$. We have

$$\begin{aligned} \mathrm{d}(r \sin i \sin \theta )=(r w \sin i \cos \theta +v \sin i \sin \theta )\mathrm{d}t \end{aligned}$$

which leads to

$$\begin{aligned}&r \cos i \sin \theta \mathrm{d}i +r \sin i \cos \theta \mathrm{d}\theta \\&\qquad +\left( r \cos i \cos \theta (\tilde{\mathbf{i}}\cdot \tilde{{\varvec{\theta }}})-\frac{1}{2} r \sin i \sin \theta (\tilde{\mathbf{i}}\cdot \tilde{\mathbf{i}})-\frac{1}{2} r \sin i \sin \theta (\tilde{{\varvec{\theta }}}\cdot \tilde{{\varvec{\theta }}})+v \sin i \sin \theta \right) \mathrm{d}t \\&\quad = (r w \sin i \cos \theta +v \sin i \sin \theta )\mathrm{d}t. \end{aligned}$$

So we obtain

$$\begin{aligned} \mathrm{d}\theta = \left( w-\cot i (\tilde{\mathbf{i}}\cdot \tilde{{\varvec{\theta }}})+\frac{1}{2}\tan \theta (\tilde{\mathbf{i}}\cdot \tilde{\mathbf{i}}+\tilde{{\varvec{\theta }}}\cdot \tilde{{\varvec{\theta }}}) \right) \mathrm{d}t -\cot i \tan \theta \mathrm{d}i. \end{aligned}$$

Using the expression of di and after simplifications, we obtain

$$\begin{aligned} \mathrm{d}\theta =&\bigg [ -\sqrt{\frac{a(1-e^2)}{\mu }}\frac{\sin (f+\omega )\cot i}{1+e\cos f} \bar{N}+ \frac{\sqrt{\mu }}{(a(1-e^2))^{3/2}}(1+e\cos f)^2 \nonumber \\&+\frac{a(1-e^2)}{2\mu (1+e\cos f)^2}\frac{\tan (f+\omega )}{\sin ^2 i} \left( \cos ^2 (f+\omega ) \left( 1 + \cos ^2 i\right) + \cos ^2 i \right) \tilde{\mathbf{N}}^2 \nonumber \\&+\frac{a(1-e^2)}{\mu (1+e\cos f)^2} \cot i \sin (f+\omega ) \tilde{\mathbf{T}}\cdot \tilde{\mathbf{N}}\bigg ]\mathrm{d}t \nonumber \\&-\sqrt{\frac{a(1-e^2)}{\mu }}\frac{\sin (f+\omega )\cot i}{1+e\cos f} \tilde{\mathbf{N}}\cdot \mathrm{d}\mathbf{B} . \end{aligned}$$

(57)

Remarking that

$$\begin{aligned} \mathrm{d}\theta =\frac{\sqrt{\mu }(1+e\cos f)^2}{(a(1-e^2))^{3/2}}\mathrm{d}t +\frac{a(1-e^2)\sin (2(f+\omega ))}{4\mu (1+e\cos f)^2} \tilde{\mathbf{N}}^2 -\cos i \mathrm{d}\varOmega , \end{aligned}$$

(58)

we can deduce the variation of the pericenter location from the equation (6).

1.5 Mean anomaly M

Using Itô’s formula on Eq. (11), we obtain

$$\begin{aligned} \mathrm{d}M =&\left[ \frac{(6 \cos f+e (5+\cos (2 f))) \sin f}{4 \sqrt{1-e^2} (1+e \cos f)^3} \left( \tilde{\mathbf{e}}\cdot \tilde{\mathbf{e}}\right) + \frac{e \left( 1-e^2\right) ^{3/2} \sin f}{(1+e \cos f)^3} \left( \tilde{\mathbf{f}}\cdot \tilde{\mathbf{f}}\right) \right. \nonumber \\&\left. -\frac{\sqrt{1-e^2} \left( 3 e+\left( 2+e^2\right) \cos f\right) }{(1+e \cos f)^3} \left( \tilde{\mathbf{e}}\cdot \tilde{\mathbf{f}}\right) \right] \mathrm{d}t \nonumber \\&-\frac{\sqrt{1-e^2}\sin f (2+e \cos f)}{(1+e \cos f)^2}\mathrm{d}e-\frac{\left( 1-e^2\right) ^{3/2}}{(1+e \cos f)^2} \mathrm{d}f \end{aligned}$$

(59)

with

$$\begin{aligned} \tilde{\mathbf{e}}\cdot \tilde{\mathbf{e}}=&\frac{\left( a \left( 1-e^2\right) \sin ^2f\right) }{\mu }\tilde{\mathbf{R}}^2+\frac{\left( a \left( 1-e^2\right) (e+\cos f)^2\right) }{\mu (1+e \cos f)^2}\tilde{\mathbf{T}}^2 \\&+\frac{\left( 2 a \left( 1-e^2\right) (e+\cos f) \sin f\right) }{\mu (1+e \cos f)}\tilde{\mathbf{R}}\cdot \tilde{\mathbf{T}}\\ \tilde{\mathbf{f}}\cdot \tilde{\mathbf{f}}=&\frac{a \left( 1-e^2\right) \cos ^2f}{\mu e^2}\tilde{\mathbf{R}}^2+\frac{a \left( 1-e^2\right) (2+e \cos f)^2 \sin ^2f}{\mu e^2 (1+e \cos f)^2}\tilde{\mathbf{T}}^2\\&-\frac{a \left( 1-e^2\right) (2+e \cos f) \sin (2 f)}{\mu e^2 (1+e \cos f)}\tilde{\mathbf{R}}\cdot \tilde{\mathbf{T}}\\ \tilde{\mathbf{e}}\cdot \tilde{\mathbf{f}}=&\frac{a \left( 1-e^2\right) \cos f \sin f}{\mu e}\tilde{\mathbf{R}}^2-\frac{a \left( 1-e^2\right) (e+\cos f) (2+e \cos f) \sin f}{2 \mu e (1+e \cos f)^2}\tilde{\mathbf{T}}^2 \\&+\frac{a \left( 1-e^2\right) \left( 3 \cos (2 f)+2 e \cos ^3f-1\right) }{2 \mu e (1+e \cos f)}\tilde{\mathbf{R}}\cdot \tilde{\mathbf{T}}. \end{aligned}$$

Remarking that

$$\begin{aligned} \mathrm{d}f=\frac{\sqrt{\mu }(1+e\cos f)^2}{(a(1-e^2))^{3/2}}\mathrm{d}t +\frac{a(1-e^2)\sin (2(f+\omega ))}{4\mu (1+e\cos f)^2} \tilde{\mathbf{N}}^2 -(\mathrm{d}\omega +\cos i \mathrm{d}\varOmega ), \end{aligned}$$

(60)

we obtain after simplifications the result for dM.