Velocity of a comet at perihelion

Problem

A comet has an elliptical orbit around the Sun with perihelion $r_P = 1{,}0\ \text{UA}$ and aphelion $r_A = 5{,}0\ \text{UA}$. Using conservation of energy and of angular momentum, calculate the comet’s velocity at perihelion. Data: $GM_S \approx 1{,}33\cdot 10^{20}\ \text{m}^3/\text{s}^2$, $1\ \text{UA} = 1{,}5\cdot 10^{11}\ \text{m}$.

Solution

The two conservation laws. At perihelion and aphelion the velocity is perpendicular to the radius, so $L = m\,v\,r$. Conservation of angular momentum: $v_P\,r_P = v_A\,r_A \;\Rightarrow\; v_A = v_P\,\frac{r_P}{r_A} = \frac{v_P}{5}$

Conservation of mechanical energy (the comet’s mass $m$ cancels out): $\tfrac{1}{2}v_P^2 - \frac{GM_S}{r_P} = \tfrac{1}{2}v_A^2 - \frac{GM_S}{r_A}$

Substituting $v_A = v_P/5$: $\tfrac{1}{2}v_P^2\left(1 - \frac{1}{25}\right) = GM_S\left(\frac{1}{r_P} - \frac{1}{r_A}\right)$ $\tfrac{1}{2}v_P^2\cdot\frac{24}{25} = GM_S\left(\frac{1}{r_P} - \frac{1}{r_A}\right)$

Numerical values ($r_P = 1{,}5\cdot 10^{11}\ \text{m}$, $r_A = 7{,}5\cdot 10^{11}\ \text{m}$): $\frac{1}{r_P} - \frac{1}{r_A} = \frac{1}{1{,}5\cdot 10^{11}} - \frac{1}{7{,}5\cdot 10^{11}} = 5{,}33\cdot 10^{-12}\ \text{m}^{-1}$ $GM_S\left(\frac{1}{r_P} - \frac{1}{r_A}\right) = 1{,}33\cdot 10^{20}\cdot 5{,}33\cdot 10^{-12} \approx 7{,}09\cdot 10^{8}$

Solving for $v_P$: $v_P^2 = \frac{2\cdot 7{,}09\cdot 10^{8}}{24/25} = \frac{1{,}42\cdot 10^{9}}{0{,}96} \approx 1{,}48\cdot 10^{9}$

$\ev{v_P \approx 3{,}8\cdot 10^4\ \text{m/s}}$

At perihelion the comet “hurtles along” ($\approx 38\ \text{km/s}$); at aphelion, five times further away, it moves five times slower ($v_A = v_P/5 \approx 7{,}6\ \text{km/s}$).

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Topics: Gravitazione Concepts: Conservazione dell’energia meccanica · Conservazione del momento angolare · Energia potenziale gravitazionale universale · Orbite Skills: Conservazione dell’energia · Conservazione del momento angolare