Example

Question:

Let the speed of the planet at the perihelion \(P\) in Fig. 7.1(a) be \(v_P\) and the Sun-planet distance \(SP\) be \(r_P\). Relate \(\{r_P, v_P\}\) to the corresponding quantities at the aphelion \(\{r_A, v_A\}\). Will the planet take equal times to traverse \(BAC\) and \(CPB\)?

Solution:

The magnitude of the angular momentum at \(P\) is \(L_P = m_P r_P v_P\), since inspection tells us that \(r_P\) and \(v_P\) are mutually perpendicular. Similarly, \(L_A = m_P r_A v_A\). From angular momentum conservation:

\[ m_P r_P v_P = m_P r_A v_A \] \[ \frac{v_P}{v_A} = \frac{r_A}{r_P} \]
Since \(r_A > r_P\), \(v_P > v_A\).
The area \(SBAC\) bounded by the ellipse and the radius vectors \(SB\) and \(SC\) is larger than \(SBPC\) in Fig. 7.1. From Kepler’s second law, equal areas are swept in equal times. Hence, the planet will take a longer time to traverse \(BAC\) than \(CPB\).