Comet at perihelion and aphelion

Problem

A comet is at perihelion at distance $d_p = 3\cdot 10^{10}\ \text{m}$ from the Sun with velocity $v_p = 5\cdot 10^4\ \text{m/s}$. At aphelion it is at $d_a = 6\cdot 10^{12}\ \text{m}$. Find $v_a$ using conservation of angular momentum.

At perihelion and aphelion the velocity is perpendicular to the line joining the comet and the Sun.

Solution

At the extreme points of the orbit (perihelion and aphelion) the velocity is perpendicular to the radius vector, so the lever arm of the momentum is the distance itself and the angular momentum is $L = m\,v\,d$.

Angular momentum about the Sun is conserved (the force is always directed along the line joining the two bodies, so its torque is zero): $m\,v_p\,d_p = m\,v_a\,d_a \;\Rightarrow\; v_p\,d_p = v_a\,d_a$

Solving for $v_a$: $v_a = v_p\,\frac{d_p}{d_a} = 5\cdot 10^4\cdot\frac{3\cdot 10^{10}}{6\cdot 10^{12}} = 5\cdot 10^4\cdot 5\cdot 10^{-3}$

$\ev{v_a = 250\ \text{m/s}}$

At aphelion, where it is furthest away, the comet is much slower: closer to the Sun $\Rightarrow$ faster.

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Topics: Gravitazione Concepts: Momento angolare · Conservazione del momento angolare · Orbite Skills: Conservazione del momento angolare