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I'm trying to follow the formulation of the restricted 3-body problem in Taff's "Celestial Mechanics" book as I'm writing an essay on Lagrange points and am confused by the transformation into the rotating coordinate system. This appears on pages 193-195 but I have paraphrased it as best I can here:

We have masses $M_1$, $M_2$ as the primaries and $m_3$ as the third body. $M_1$ and $M_2$ are in circular orbits about the common center of mass at radii $b$ and $a$ respectively. Let $k^2=G$ and $n$ be the mean motion/angular speed of revolution.

Recreation of the diagram from the book

Let $\mathbf{R}=(X,Y)$ be the location of $m_3$ w.r.t. the inertial reference frame and let $t^*$ denote time. The equations of motion of $m_3$ are \begin{equation} \label{6.16} \mathbf{\ddot{R}}=\nabla{F} \tag{6.16} \end{equation} where $F=\frac{k^2M_1}{R_1}+\frac{k^2M_2}{R_2}$ and $R_i^2=(X-X_i)^2+(Y-Y_i)^2$ with $M_1$ at $(X_1,Y_1)$ and $M_2$ at $(X_2,Y_2)$.

Choose the origin of $t^*$ be such that the primaries are on the x-axis at $t^*=0$. Then, $X_1=bcos(nt^*), Y_1=bsin(nt^*), X_2=-acos(nt^*), Y_2=-asin(nt^*)$.

The explicit form of (6.16) is $$\frac{d^2X}{dt^{*2}}=-k^2 \bigg[M_1 \frac{(X-X_1)}{R_1^3} + M_2 \frac{(X-X_2)}{R_2^3} \bigg]$$

$$\frac{d^2Y}{dt^{*2}}=-k^2 \bigg[M_1 \frac{(Y-Y_1)}{R_1^3} + M_2 \frac{(Y-Y_2)}{R_2^3} \bigg]$$

Now we reach the conversion to a rotating coordinate system, which rotates about the origin with period $P=\frac{2 \pi}{n}$. (This is the part I don't understand), if $$\mathbf{R}=R_3(-nt^*)\mathbf{\bar{r}}$$ then $$R_1=[(\bar{x}-b)^2+\bar{y}^2]^{\frac{1}{2}}, R_2=[(\bar{x}+a)^2+\bar{y}^2]^{\frac{1}{2}}$$

and equation (6.16) becomes, $$\frac{d^2 \bar{x}}{dt^{*2}} - 2n \frac{d \bar{y}}{dt^{*2}} - n^2 \bar{x} = -k^2 \bigg[M_1 \frac{(\bar{x}-b)}{R_1^3} + M_2 \frac{(\bar{x}+a)}{R_2^3} \bigg]$$ $$\frac{d^2 \bar{y}}{dt^{*2}} - 2n \frac{d \bar{x}}{dt^{*2}} - n^2 \bar{y} = -k^2 \bigg[M_1 \frac{(\bar{y})}{R_1^3} + M_2 \frac{(\bar{y})}{R_2^3} \bigg]$$.

In particular, I'm confused by how this transformation yields $\mathbf{R}=R_3(-nt^*)\mathbf{\bar{r}}$ and the left-hand side of (6.16) when transformed into the rotating coordinate system. If someone could give the explicit steps to how the LHS is derived from $\mathbf{\ddot{R}}$ then it'd be much appreciated as I've got nowhere with my attempts

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