Irodov 1.202: The Ellipse's Hidden Equivalent Circle

Problem Irodov 1.202

A planet follows an elliptical orbit around the Sun, coming no closer than a distance $r$ from the Sun and going no farther than a distance $R$ . Using Kepler’s laws, find its orbital period $T$ in terms of $r$ , $R$ and the mass of the Sun.

The elliptical orbit with the Sun M at one focus and the planet m moving with velocity v: perihelion distance r and aphelion distance R measured from that focus, semi-major axis a from the center C. — The elliptical orbit with the Sun $M$ at one focus and the planet $m$ moving with velocity $\vect v$ : perihelion distance $r$ and aphelion distance $R$ measured from that focus, semi-major axis $a$ from the center $C$ .

Given

Find

T

Solution

Step 1The semi-major axis from the two extreme distances

The perihelion and aphelion distances are the ellipse’s semi-major axis $a$ shortened and lengthened by the same focus-to-centre offset, so their sum is exactly the major axis.

r = a(1-e), \qquad R = a(1+e)

r + R = 2a

a = \dfrac{r+R}{2}

(1)

Step 2Kepler’s third law fixes the period from the semi-major axis

Kepler’s third law states that the square of the period is proportional to the cube of the orbit’s semi-major axis, with the same constant that governs a circular orbit of that same radius; applying the circular-orbit balance, gravity supplying the centripetal force, to a circular orbit of radius $a$ fixes that constant.

\frac{\gamma M m}{a^2} = \frac{m}{a}\left(\frac{2\pi a}{T}\right)^2 \;\Longrightarrow\; \gamma M = \frac{4\pi^2 a^3}{T^2}

T^2 = \dfrac{4\pi^2}{\gamma M}a^3

(2)

Step 3Substitute the semi-major axis

Substituting (1) into (2) and solving for $T$ gives the period in terms of the two given extreme distances and the Sun’s mass.

T^2 = \frac{4\pi^2}{\gamma M}\left(\frac{r+R}{2}\right)^3

T = \frac{2\pi}{\sqrt{\gamma M}}\left(\frac{r+R}{2}\right)^{3/2}

T = \pi\sqrt{\dfrac{(r+R)^3}{2\gamma M}}

(3)

Check

At $r = R$ (a circular orbit) the boxed result gives $T = \pi\sqrt{(2r)^3/(2\gamma M)} = \pi\sqrt{4r^3/\gamma M} = 2\pi\sqrt{r^3/\gamma M}$ , exactly the circular-orbit period (2) with $a = r$ , as it must.

Remark

The period depends on the orbit only through $a = (r+R)/2$ , never through how elongated the ellipse is: a needle-thin ellipse and a near circle sharing the same semi-major axis complete a revolution in exactly the same time, even though the thin one spends most of that time crawling near aphelion and only a moment whipping through perihelion. Kepler’s second law makes the areal sweep rate constant, and that rate itself carries the same factor of $\sqrt{1-e^2}$ as the orbit’s own area, so the elongation cancels out of the ratio and leaves that trade between speed and time exact.