Reduction to One Body Problem

Let \(m_1\) be the mass of the satellite, \(m_2\) be the mass of Jupiter, \(r\) be the vector from the satellite to the planet, and \(\theta\) be the angle between the satellite and Jupiter where \(\theta_0=\pi/2.\) We will proceed to use polar coordinates in our calculations. The kinetic energy and potential energy of the system are \(T=\frac{1}{2}m_1(\dot{r}^2+r^2\dot{\theta}^2)\) and \(V=-\frac{Gm_1m_2}{r}\) respectively. Therefore, the Lagrangian is \(\mathcal{L}=\frac{1}{2}m_1(\dot{r}^2+r^2\dot{\theta}^2)+\frac{Gm_1m_2}{r}\). Applying the Euler-Lagrange equation to the \(r\) coordinate, we have

\[\frac{d}{dt}\left (\frac{\partial \mathcal{L}}{\partial \dot{r}} \right )=m_1\ddot{r}= \frac{\partial \mathcal{L}}{\partial r}=m_1r\dot{\theta}^2-\frac{Gm_1m_2}{r^2} \quad \quad (1).\]

Applying the Euler-Lagrange equation on the \(\theta\) coordinate, we have

\[\frac{d}{dt}\left (\frac{\partial \mathcal{L}}{\partial \dot{\theta}} \right )=m_1r^2\ddot{\theta}=\frac{\partial \mathcal{L}}{\partial \theta}=0 \quad \quad (2).\]

From equation \(1\), we have that \(m_1\ddot{r}-m_1r\dot{\theta}^2+\frac{Gm_1m_2}{r^2}=0\). Note that since angular momentum is conserved in this system, \(m_1 r^2\dot{\theta}= l\). Substituting this into equation \(1\), we get \(m_1\ddot{r}-\frac{l^2}{m_1r^3}+\frac{Gm_1m_2}{r^2}=0\). Integrating with respect to time, we have

\[\frac{1}{2}m_1\dot{r}^2+\frac{1}{2}\frac{l^2}{m_1r^2}-\frac{Gm_1m_2}{r}=E,\]

which is a statement of conservation of energy. Solving in terms of \(\dot{r}=\frac{dr}{dt}\), we have

\[\dot{r}= \sqrt{\frac{2}{m_1}\left ( E+\frac{Gm_1m_2}{r}-\frac{1}{2}\frac{l^2}{m_1 r^2}\right )} \quad \quad (3).\]

Note that this is equivalent to

\[dt= \frac{dr}{\sqrt{\frac{2}{m_1}\left ( E+\frac{Gm_1m_2}{r}-\frac{1}{2}\frac{l^2}{m_1 r^2}\right )}} \quad \quad (4).\]

Rewriting the conservation of angular momentum, we have \(dt=\frac{m_1r^2 d\theta}{l}.\) Substituting this into equation \(4\), we have

\[d\theta = \frac{ldr}{m_1r^2\sqrt{\frac{2}{m_1}\left ( E+\frac{Gm_1m_2}{r}-\frac{1}{2}\frac{l^2}{m_1 r^2}\right )}} \quad \quad (5).\]

Integrating both sides, we have

\[\theta = \int \frac{ldr}{m_1r^2\sqrt{\frac{2}{m_1}\left ( E+\frac{Gm_1m_2}{r}-\frac{1}{2}\frac{l^2}{m_1 r^2}\right )}}+\theta_0 \quad \quad (6),\]

where \(\theta_0\) is the initial angle. Evaluating this integral, we have

\[\theta = \theta_0-\arccos{\frac{\frac{l^2}{rm_1Gm_1m_2}-1}{\sqrt{1+\frac{2El^2}{m_1(Gm_1m_2)^2}}}}.\]

We can rewrite this as

\[\frac{1}{r}= \frac{Gm_1^2m_2}{l^2}\left ( 1+\sqrt{1+\frac{2El^2}{G^2m_1^3m_2^2}} \cos(\theta-\theta_0)\right ) \quad \quad (7).\]

Recognizing this as the general equation of a conic, we know that the eccentricity of the orbit is determined by

\[\sqrt{1+\frac{2El^2}{G^2m_1^3m_2^2}} \quad \quad (8).\]

Solution of 2001 AP ® PHYSICS C: MECHANICS FREE-RESPONSE QUESTION Mech 2. Part C

(i) If the speed of the satellite is slightly faster, note that the eccentricity will be \(<1\), implying an elliptical orbit. Additionally, note that \(E<0\). If the satellite is slightly faster, \(E\) will increase and thus \(r\) will increase except for when \(\theta=\theta_0+2\pi n\), where \(n\in \mathbb{Z}\), that is, when the satellite passes the injection point.

(ii) If the speed of the satellite is slightly slower, note that the eccentricity will be \(<1\), implying an elliptical orbit. Additionally, note that \(E<0\). If the satellite is slightly slower, \(E\) will decrease and thus \(r\) will decrease except for when \(\theta=\theta_0+2\pi n\), where \(n\in \mathbb{Z}\), that is, when the satellite passes the injection point.