Irodov 1.210: Aiming Parameter Fixes the Closest Approach

Problem Irodov 1.210

A small body approaches the Sun (mass $m_S$ ) from a great distance, where it moves essentially freely with speed $v_0$ along a straight line. The perpendicular distance from the Sun’s center to that straight line, the aiming parameter, is $l$ . Under the Sun’s gravitational pull the body is deflected onto a curved path that swings around the Sun and back out to infinity. Find the minimum distance between the body and the Sun’s center reached during the encounter.

Body A arrives along the lower dotted asymptote at velocity v_0, is deflected by the Sun m_S, and leaves along the upper one. The aiming parameter l runs perpendicular from the Sun to the incoming asymptote. — Body $A$ arrives along the lower dotted asymptote at velocity $\vect v_0$ , is deflected by the Sun $m_S$ , and leaves along the upper one. The aiming parameter $l$ runs perpendicular from the Sun to the *incoming* asymptote.

Given

$v_0$
$l$
$m_S$

Find

r_{\min}

, the least distance from the Sun’s center

Solution

Step 1Angular momentum from the straight-line approach

Far from the Sun the body moves in a straight line at speed $v_0$ , so its angular momentum per unit mass about the Sun’s center, evaluated there, is $v_0$ times the perpendicular distance from the center to that line, which is exactly the aiming parameter $l$ . Angular momentum per unit mass about the center is conserved throughout the encounter, so the same value holds at every point of the path, including closest approach, where the velocity is purely tangential (the radial speed vanishes at a turning point of $r$ ).

\begin{align*}L &= v_0\, l \\ L &= r_{\min}\, v_{\min}\end{align*}

r_{\min} v_{\min} = v_0 l

(1)

Step 2Energy conservation from infinity to closest approach

The total mechanical energy per unit mass is conserved between the far away state (speed $v_0$ , potential energy negligible) and closest approach (speed $v_{\min}$ , potential energy $-\gamma m_S/r_{\min}$ ).

\frac{v_{\min}^2}{2} - \frac{\gamma m_S}{r_{\min}} = \frac{v_0^2}{2} \tag{2}

Step 3Eliminate the closest-approach speed

Solving (1) for $v_{\min}$ and substituting into the energy relation (2) leaves a single equation in $r_{\min}$ .

\begin{align*}v_{\min} &= \frac{v_0 l}{r_{\min}} \\ \frac{1}{2}\left(\frac{v_0 l}{r_{\min}}\right)^2 - \frac{\gamma m_S}{r_{\min}} &= \frac{v_0^2}{2}\end{align*}

Multiplying through by $2r_{\min}^2/v_0^2$ clears the fractions and collects everything into one quadratic in $r_{\min}$ .

r_{\min}^2 + \frac{2\gamma m_S}{v_0^2}\,r_{\min} - l^2 = 0

Step 4Solve the quadratic and keep the physical root

\begin{align*}r_{\min} &= \frac{1}{2}\left[-\frac{2\gamma m_S}{v_0^2} \pm\sqrt{\left(\frac{2\gamma m_S}{v_0^2}\right)^2+4l^2}\,\right] \\ &= -\frac{\gamma m_S}{v_0^2} \pm\frac{\gamma m_S}{v_0^2}\sqrt{1+\left(\frac{l v_0^2}{\gamma m_S}\right)^2}\end{align*}

The minus sign inside the outer $\pm$ gives a negative distance, which is unphysical; only the plus sign survives.

r_{\min} = \frac{\gamma m_S}{v_0^2}\left[\sqrt{1+\left(\frac{l v_0^2}{\gamma m_S}\right)^2}-1\right]

(3)

Check

At $l=0$ (a purely radial, head on approach) the boxed formula gives $r_{\min}=0$ : with no angular momentum the body falls straight to the center, exactly as it must. For $l\gg\gamma m_S/v_0^2$ (a distant, weakly deflected pass), expanding the square root gives $r_{\min}\to l$ , so a large aiming parameter is barely bent and passes almost as close as it would without the Sun there at all.

Remark

The result interpolates smoothly between these two limits: gravity always pulls $r_{\min}$ below the straight-line value $l$ , by an amount set by the single length scale $\gamma m_S/v_0^2$ , which compares the Sun’s pull to the body’s incoming kinetic energy. A slower body (smaller $v_0$ ) has a larger such scale and is deflected closer to the Sun for the same $l$ , which is exactly the gravitational focusing that scatters a slow-moving body far more strongly, and makes it far more likely to hit the Sun at all, than a fast one on the same straight-line course.