Irodov 1.209: One Point Fixes the Whole Ellipse – Amit & Deepali Sinha Foundation

Problem Irodov 1.209

A planet $A$ travels along an elliptical orbit about the Sun. At some instant its distance from the Sun is $r_0$ , its speed is $v_0$ , and the angle between the radius vector (Sun to planet) and the velocity vector at that instant is $\alpha$ . Find the maximum and minimum distances the planet ever attains from the Sun over the course of its orbit.

Given

$r_0$
$v_0$
$\alpha$

Find

r_{\max}

and

r_{\min}

Solution

Step 1Angular momentum from the given point

Angular momentum about the Sun is conserved, so its value at the given point equals its value at either turning point, a distance $r$ away with speed $v$ there.

m r_0 v_0\sin\alpha = m r v

v = \frac{r_0 v_0\sin\alpha}{r}

(1)

Step 2Energy conservation between the two points

Mechanical energy is conserved along the whole orbit, so it too is the same at the given point and at the turning point; dividing through by the planet’s mass $m$ turns this into a statement about the specific energy alone.

\frac12 v_0^2 - \frac{\gamma m_S}{r_0} = \frac12 v^2 - \frac{\gamma m_S}{r} \tag{2}

Step 3A quadratic for the turning radius

Two shorthands carry the rest of the algebra: $\rho$ for the turning radius measured in units of the given distance, and $\eta$ for the ratio of the given kinetic energy scale to the gravitational one at that same distance.

\rho = \dfrac{r}{r_0},\qquad \eta = \dfrac{r_0v_0^2}{\gamma m_S}

(3)

Substituting (1) for $v$ into the energy balance (2) and rewriting it in these two shorthands turns it into a quadratic in $\rho$ .

\begin{align*}\frac12 v_0^2 - \frac{\gamma m_S}{r_0} &= \frac12\frac{v_0^2\sin^2\alpha}{\rho^2} - \frac{\gamma m_S}{\rho r_0} \\ 1 - \frac{2}{\eta} &= \frac{\sin^2\alpha}{\rho^2} - \frac{2}{\eta\rho} \\ \rho^2\left(1-\frac{2}{\eta}\right) &= \sin^2\alpha - \frac{2\rho}{\eta}\end{align*}

Multiplying through by $\eta$ and collecting every term on one side,

(2-\eta)\rho^2 - 2\rho + \eta\sin^2\alpha = 0

(4)

Step 4Solve the quadratic

The quadratic formula applied to (4) gives both roots, one for each turning point, at once.

\rho = \frac{2\pm\sqrt{4-4(2-\eta)\eta\sin^2\alpha}}{2(2-\eta)} = \frac{1\pm\sqrt{1-(2-\eta)\eta\sin^2\alpha}}{2-\eta}

Multiplying back by $r_0$ , with the plus sign giving the farther turning point ( $r_{\max}$ ) and the minus sign the nearer one ( $r_{\min}$ ); the book itself reports both distances as the two signs of this one expression, so a single boxed formula carries both asked quantities.

\begin{gathered}r_{\max/\min} = \frac{r_0}{2-\eta}\left[1\pm\sqrt{1-(2-\eta)\eta\sin^2\alpha}\right],\\ \eta=\frac{r_0v_0^2}{\gamma m_S}\end{gathered}

(5)

Check

If the given point is itself a turning point, $\alpha=90^\circ$ and $\sin^2\alpha=1$ ; the radicand becomes $1-(2-\eta)\eta=(1-\eta)^2$ , so the square root is $|1-\eta|$ , and for $\eta<1$ the plus sign in the boxed formula gives $r=r_0(2-\eta)/(2-\eta)=r_0$ , exactly recovering the starting point as one of its own turning points, as it must.

Remark

The denominator $2-\eta$ vanishes at $\eta=2$ , i.e. at $v_0^2=2\gamma m_S/r_0$ , which is precisely the escape speed at $r_0$ : $r_{\max}$ diverges there because the trajectory is no longer a closed ellipse but a parabola, consistent with $\eta<2$ being the actual condition for a bound orbit to exist at all.