Irodov 1.208: Only the Semi-Major Axis Survives – Amit & Deepali Sinha Foundation

Problem Irodov 1.208

A planet of mass $m$ moves along an elliptical orbit about the Sun, of mass $m_S$ , under gravity alone. Using the conservation laws, show that the planet’s total mechanical energy $E$ depends only on the orbit’s semi-major axis $a$ , and not on its eccentricity. Find $E$ as a function of $a$ .

The planet's elliptical orbit about the Sun at a focus. At perihelion (distance r_p) the velocity v_p is purely tangential, and the same is true of v_a at aphelion (distance r_a); these are the two points the derivation compares. — The planet’s elliptical orbit about the Sun at a focus. At perihelion (distance $r_p$ ) the velocity $\vect v_p$ is purely tangential, and the same is true of $\vect v_a$ at aphelion (distance $r_a$ ); these are the two points the derivation compares.

Given

$m$
$m_S$
$a$

Find

E(a)

Solution

Step 1The turning points of the orbit

At perihelion and aphelion the radial velocity vanishes, so the velocity there is purely tangential; both points lie on the major axis, on opposite sides of the occupied focus, at distances $r_p$ and $r_a$ from it. By the definition of the semi-major axis $a$ , the major axis has length $2a$ , so

r_p+r_a = 2a

(1)

Step 2Angular momentum ties the two speeds together

Angular momentum about the Sun is conserved along the orbit, and at the two apsides the velocity is purely tangential, so evaluating it there needs no projection; equating the two apsidal values relates the two speeds.

m v_p r_p = m v_a r_a \tag{2}

so that

v_a = v_p\,\frac{r_p}{r_a}

(3)

Step 3Energy conservation between the two points

Only gravity does work on the planet, so its mechanical energy is the same at perihelion and at aphelion.

\frac12 m v_p^2 - \frac{\gamma m m_S}{r_p} = \frac12 m v_a^2 - \frac{\gamma m m_S}{r_a} \tag{4}

Step 4Eliminate the speeds

Substituting (3) for $v_a$ in (4) and dividing through by $m$ leaves an equation in $v_p$ , $r_p$ and $r_a$ alone.

\begin{align*} \frac12 v_p^2 - \frac{\gamma m_S}{r_p} &= \frac12 v_p^2\frac{r_p^2}{r_a^2} - \frac{\gamma m_S}{r_a} \\ \frac12 v_p^2\left(1-\frac{r_p^2}{r_a^2}\right) &= \gamma m_S\left(\frac{1}{r_p}-\frac{1}{r_a}\right) \\ \frac12 v_p^2\,\frac{(r_a-r_p)(r_a+r_p)}{r_a^2} &= \gamma m_S\,\frac{r_a-r_p}{r_p r_a} \tag{5}\end{align*}

Since $r_a\neq r_p$ on a genuine ellipse, the common factor $(r_a-r_p)$ on both sides of (5) cancels; using (1) for $r_a+r_p$ then isolates $v_p^2$ .

v_p^2 = \frac{2\gamma m_S}{r_p}\cdot\frac{r_a}{r_a+r_p} = \frac{\gamma m_S\,r_a}{a\,r_p} \tag{6}

Step 5The energy collapses to a function of

a

alone

Substituting (6) into the perihelion energy per unit mass, then rearranging (1) to trade $r_a-2a$ for $r_p$ , eliminates $r_p$ as well.

\begin{align*}\frac{E}{m} &= \frac12 v_p^2 - \frac{\gamma m_S}{r_p} \\ &= \frac{\gamma m_S\,r_a}{2a\,r_p} - \frac{\gamma m_S}{r_p} \\ &= \frac{\gamma m_S}{r_p}\left(\frac{r_a}{2a}-1\right) \\ &= \frac{\gamma m_S}{r_p}\cdot\frac{r_a-2a}{2a} \\ r_a - 2a &= r_a - (r_p + r_a) = -r_p \\ \frac{E}{m} &= \frac{\gamma m_S}{r_p}\cdot\frac{-r_p}{2a}\end{align*}

E = -\dfrac{\gamma m\, m_S}{2a}

(7)

Check

The distance $r_p$ canceled completely: the same algebra started from the aphelion side would have left $r_a$ instead, and both must and do give the identical result, which is exactly the demonstration asked for. As an independent check, a circular orbit ( $e\to0$ , $r_p\to r_a\to a$ ) satisfies Newton’s second law directly, $mv^2/a=\gamma m m_S/a^2$ , so $v^2\to\gamma m_S/a$ without appeal to the elliptical algebra above, and $E/m=v^2/2-\gamma m_S/a\to-\gamma m_S/(2a)$ , matching (7).

Remark

Two orbits sharing the same $a$ but very different eccentricities, one nearly circular and one a thin sliver of an ellipse, carry exactly the same total energy; only their angular momentum differs, since the quantity conserved in (2) depends on $r_p$ and $r_a$ individually, not merely on their sum. So $a$ alone fixes the energy, and by Kepler’s third law the period too, while the eccentricity is free to redistribute the same energy between kinetic and potential along the orbit however it likes.