Perturbed planetary orbit

Suppose that the planet is subject to a perturbing force per unit mass, ${\bf F}$. Consequently, the planet's perturbed equation of motion becomes

$$\ddot{{\bf r}} + \mu\,\frac{\bf r}{r^{\,3}} = {\bf F}.$$ (I.14)

We can write

$${\bf F} = F_r\,{\bf e}_r + F_\theta\,{\bf e}_\theta + F_z\,{\bf e}_z.$$ (I.15)

The perturbing force causes the planet's orbital energy per unit mass to evolve in time as

$$\skew{3}\dot{\cal E} = {\bf F}\cdot \dot{\bf r}=F_r\,\skew{5}\dot{r} + r\,\skew{5}\dot{\theta}\,F_\theta,$$ (I.16)

where use has been made of Equations (I.3) and (I.15). Thus, it follows from Equations (I.5), (I.7), (I.9), and (I.10) that

$$\frac{\skew{3}\dot{a}}{a} = \frac{2\,h}{\mu\,(1-e^{\,2})}\left[e\,\sin\theta\,F_r + (1+e\,\cos\theta)\,F_\theta\right].$$ (I.17)

The perturbing force causes the planet's orbital angular momentum per unit mass to evolve in time as

$$\dot{\bf h} = {\bf r}\times \ddot{\bf r} = {\bf r}\times {\bf F} = -r\,F_z\,{\bf e}_\theta + r\,F_\theta\,{\bf e}_z,$$ (I.18)

where Equations (I.2), (I.6), (I.14), and (I.15) have been employed. Consequently,

$$\skew{3}\dot{h}_x = r\,\sin\theta\,F_z,$$ (I.19)

$$\skew{3}\dot{h}_y = -r\,\cos\theta\,F_z,$$ (I.20)

$$\skew{3}\dot{h}_z =r\,F_\theta.$$ (I.21)

Furthermore,

$$\skew{3}\dot{h} = \frac{{\bf h}\cdot\dot{\bf h}}{h} = \skew{3}\dot{h}_z = r\,F_\theta,$$ (I.22)

where use has been made of Equation (I.6).

The perturbing force causes the planet's eccentricity vector to evolve in time as

$$\dot{\bf e} = \frac{\ddot{\bf r}\times {\bf h} +\dot{\bf r}\times \dot{\bf h}}{\mu} + \frac{({\bf r}\cdot\dot{\bf r})\,{\bf r} - r^{\,2}\,\dot{\bf r}}{r^{\,3}}$$

$$= \frac{2\,(\dot{\bf r}\cdot {\bf F}) - ({\bf r}\cdot\dot{\bf r})\,{\bf F} - ({\bf r}\cdot{\bf F})\,\dot{\bf r}}{\mu}$$

$$= \frac{2\,h\,F_\theta\,{\bf e}_r - (h\,F_r + r\,\skew{5}\dot{r}\,F_\theta)\,{\bf e}_\theta - r\,\skew{5}\dot{r}\,F_z\,{\bf e}_z}{\mu},$$ (I.23)

where Equations (I.7), (I.8), (I.14), (I.15), and (I.18) have been employed. It follows that

$$\skew{3}\dot{e}_x = \frac{h}{\mu}\left[\sin\theta\,F_r+(\cos\theta+\cos E)\,F_\theta \right],$$ (I.24)

$$\skew{3}\dot{e}_y = -\frac{h}{\mu}\left[\cos\theta \,F_r-\left(\frac{2+e\,\cos\theta}{1+e\,\cos\theta}\right)\sin\theta\,F_\theta\right],$$ (I.25)

$$\skew{3}\dot{e}_z = -\frac{h}{\mu}\left(\frac{e\,\sin\theta}{1+e\,\cos\theta}\right)F_z,$$ (I.26)

where use has been made of Equation (I.11). Furthermore,

$$\skew{3}\dot{e} = \frac{{\bf e}\cdot\dot{\bf e}}{e} = \skew{3}\do... ..._x = \frac{h}{\mu}\left[ \sin\theta\,F_r+(\cos\theta+\cos E)\,F_\theta \right],$$ (I.27)

where Equation (I.8) has been employed.

Differentiation of Equation (I.9) with respect to time yields

$$\skew{5}\dot{r}=(1-e\,\cos E) \,\skew{3}\dot{a}- a\,\cos E\,\skew{3}\dot{e} +a\,e\,\sin E\,\dot{E}.$$ (I.28)

Making use of Equations (I.4), (I.7), and (I.10), we can rearrange the previous expression to give

$$\dot{E} =\frac{n}{1-e\,\cos\,E} - \frac{(1-e\,\cos E)}{e\,\sin E}\,\frac{\skew{3}\dot{a}}{a} + \frac{\cos E}{\sin E}\,\frac{\skew{3}\dot{e}}{e}.$$ (I.29)

Differentiation of Equation (I.13) with respect to time yields

$$\skew{5}\dot{\cal M} =(1-e\,\cos E)\,\dot{E}-\sin E\,\skew{3}\dot{e}.$$ (I.30)

The previous two equations can be combined to give

$$\skew{5}\dot{\cal M} = n + \frac{1}{e\,\sin E}\left[(\cos E - e)\,\skew{3}\dot{e} - \left(\frac{r}{a}\right)^2\,\frac{\skew{3}\dot{a}}{a}\right]$$

$$= n + \frac{(1-e^{\,2})^{1/2}}{e\,\sin\theta}\left(\cos\theta\,\skew{3}\dot{e} - \frac{r}{a}\,\frac{\skew{3}\dot{a}}{a}\right),$$ (I.31)

where use has been made of Equations (I.9), (I.11), and (I.12). Thus, it follows from Equations (I.11), (I.17), and (I.27) that

$$\skew{5}\dot{\cal M} = n + \frac{h}{\mu}\,\frac{(1-e^{\,2})^{1/2}... ... -\left[1+\frac{1}{(1-e^{\,2})}\,\frac{r}{a}\right]\sin\theta\,F_\theta\right).$$ (I.32)