Irodov 1.207: Angular Momentum from the Two Apsides – Amit & Deepali Sinha Foundation

Problem Irodov 1.207

A planet of mass $m$ travels on an elliptical orbit about the Sun (mass $m_S$ ), with the Sun at one focus. The planet’s greatest and least distances from the Sun, at the two apsides of the orbit, are $r_1$ (aphelion) and $r_2$ (perihelion) respectively. Find the magnitude $M$ of the planet’s angular momentum about the Sun’s center.

The elliptical orbit with the Sun m_S at the near focus: aphelion at the left, distance r_1, and perihelion at the right, where the planet m sits at distance r_2 with velocity v_2 tangent to the ellipse. — The elliptical orbit with the Sun $m_S$ at the near focus: aphelion at the left, distance $r_1$ , and perihelion at the right, where the planet $m$ sits at distance $r_2$ with velocity $\vect v_2$ tangent to the ellipse.

Given

$m$
$r_1$ (aphelion distance)
$r_2$ (perihelion distance)
$m_S$ , the Sun’s mass
$\gamma$ , the gravitational constant

Find

the angular momentum

M

of the planet about the Sun

Solution

Step 1Angular momentum at the two apsides

At an apsis the radial distance is momentarily extremal, so its rate of change vanishes and the velocity has no radial component: it is purely tangential, perpendicular to the radius vector. The angular momentum’s magnitude, $m$ times the perpendicular distance times the speed, therefore reduces to a plain product at each apsis, and conservation of angular momentum equates the two.

M = m v_1 r_1 = m v_2 r_2 \tag{1}

Solving for $v_2$ in terms of $v_1$ ,

v_2 = v_1\dfrac{r_1}{r_2}

(2)

Step 2Energy conservation between the apsides

Mechanical energy is conserved between the two points; the gravitational potential energy at distance $r$ is $-\gamma m m_S/r$ .

\frac{1}{2}mv_1^2 - \frac{\gamma m m_S}{r_1} = \frac{1}{2}mv_2^2 - \frac{\gamma m m_S}{r_2}

Dividing by $m$ and substituting (2) for $v_2$ ,

\begin{align*}\frac{1}{2}v_1^2 - \frac{\gamma m_S}{r_1} &= \frac{1}{2}v_1^2\frac{r_1^2}{r_2^2} - \frac{\gamma m_S}{r_2} \\ \frac{1}{2}v_1^2\left(1-\frac{r_1^2}{r_2^2}\right) &= \gamma m_S\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\end{align*}

\begin{align*}\frac{1}{2}v_1^2\frac{(r_2-r_1)(r_2+r_1)}{r_2^2} &= \gamma m_S\frac{r_2-r_1}{r_1r_2} \\ \frac{1}{2}v_1^2\frac{r_1+r_2}{r_2^2} &= \frac{\gamma m_S}{r_1r_2} \\ v_1^2 &= \frac{2\gamma m_S r_2}{r_1(r_1+r_2)}\end{align*}

v_1 = \sqrt{\dfrac{2\gamma m_S r_2}{r_1(r_1+r_2)}}

(3)

Step 3Assemble the angular momentum

Multiplying (3) by $m r_1$ , as (1) requires,

\begin{align*}M &= m v_1 r_1 \\ &= m r_1\sqrt{\frac{2\gamma m_S r_2}{r_1(r_1+r_2)}} \\ &= m\sqrt{\frac{r_1^2\cdot 2\gamma m_S r_2}{r_1(r_1+r_2)}}\end{align*}

M = m\sqrt{\dfrac{2\gamma m_S r_1 r_2}{r_1+r_2}}

(4)

Check

For a circular orbit, $r_1=r_2=r$ , the boxed result gives $M=m\sqrt{2\gamma m_S r^2/2r}=m\sqrt{\gamma m_S r}=mr\sqrt{\gamma m_S/r}$ , exactly $mrv_{\mathrm{circ}}$ with the standard circular speed $v_{\mathrm{circ}}=\sqrt{\gamma m_S/r}$ , as it must; this is the $r_1\to r_2$ limit of the boxed formula, not a case the derivation itself covers, since at $r_1=r_2$ both sides of the energy step above reduce to $0=0$ .

Remark

$M/2m$ is the constant areal velocity of Kepler’s second law, so this same computation, done once at either apsis, is really a statement about the whole orbit: the ellipse sweeps out area at the steady rate $\frac{1}{2}\sqrt{2\gamma m_S r_1r_2/(r_1+r_2)}$ , all the way round, fast near the Sun and slow far from it, in exactly the ratio that keeps $M$ unchanged.