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 $CPB$?

![Figure 7.1.a

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 $CPB$?

![Figure 7.1.a](/figures/gravitation/ellipse-traced-out-planet-around-sun-closest-fig-7-1-a.svg)

Accepted Answer

> [!TLDR]
> Conservation of angular momentum about the Sun gives $v_P r_P = v_A r_A$, so the planet moves faster at perihelion ($v_P>v_A$ since $r_P<r_A$). Because it moves faster near $P$ and slower near $A$, the planet does NOT take equal times to traverse $BAC$ and $CPB$, even though it sweeps equal areas in equal times.

### Relating perihelion and aphelion
The Sun exerts a central force on the planet — gravity acts along the line joining the Sun and the planet — so the torque about the Sun is always zero, and the planet's angular momentum about the Sun is conserved throughout the orbit.

At perihelion $P$, the planet is at distance $r_P$ and moves with speed $v_P$, which is perpendicular to the radius $SP$. At aphelion $A$, it is at distance $r_A$ with speed $v_A$, again perpendicular to $SA$. Since the velocity is perpendicular to the radius at both extreme points, the angular momentum there is simply $L=mrv$:
$$L_P = mr_Pv_P, \qquad L_A = mr_Av_A$$
Conservation of angular momentum ($L_P=L_A$) gives
$$r_Pv_P = r_Av_A$$
Since $r_A>r_P$ (aphelion is farther from the Sun), this forces $v_P>v_A$ — the planet moves fastest at perihelion and slowest at aphelion.

### Will the times to traverse $BAC$ and $CPB$ be equal?
Suppose the major axis divides the ellipse into two halves: arc $BAC$ (the half containing aphelion $A$) and arc $CPB$ (the half containing perihelion $P$). By Kepler's second law, the radius vector from the Sun sweeps out equal areas in equal times — and since each half of the ellipse (split by the major axis) has exactly half the total area, a naive reading might suggest the two halves take equal time. …

Situation	What happens	Why
Ice skater pulling arms in	Spins faster	$I$ decreases, $\omega$ increases to keep $L$ constant
Diver tucking into a ball	Rotates faster in midair	Same principle — no external torque during flight
Cat falling upside-down	Twists body to land on feet	Changes $I$ of different body parts to rotate without external torque
Planet orbiting the Sun	Speeds up when closer, slows down when farther	Gravitational force is central (torque = 0), so $L$ is constant

Q.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 $CPB$ ?

Concept understanding — Conservation of Angular Momentum

Conservation of Angular Momentum

The Intuition First

The Precise Statement

What Is Angular Momentum?

Why It Works: The Physics

Real-World Examples

Common Mistake to Avoid …

Relating perihelion and aphelion

Will the times to traverse $BAC$ and $CPB$ be equal?

Relating perihelion and aphelion

Will the times to traverse BACBACBAC and CPBCPBCPB be equal?

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Will the times to traverse $BAC$ and $CPB$ be equal?

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