Irodov 1.203: Radial Fall as Degenerate Ellipse – Amit & Deepali Sinha Foundation

Problem Irodov 1.203

A small body is released from rest, in the heliocentric (Sun-centred) reference frame, at a distance $R$ from the Sun equal to the radius of the Earth’s orbit (taken circular), and falls straight toward the Sun under its gravity alone. Using Kepler’s laws, and taking $T$ as the period of the Earth’s revolution about the Sun, find the time $\tau$ the body takes to fall in.

The body m is released from rest a distance R from the Sun M_S and falls straight in along the line joining them; at every point of that path the gravitational pull F stays exactly radial and collinear with the velocity v. — The body $m$ is released from rest a distance $R$ from the Sun $M_S$ and falls straight in along the line joining them; at every point of that path the gravitational pull $\vect F$ stays exactly radial and collinear with the velocity $\vect v$ .

Given

$R$ (Earth’s orbital radius)
$T$ (Earth’s orbital period)

Find

the falling time

\tau

Solution

Step 1Model the fall as a degenerate ellipse

The straight-line fall (solid) read as the eccentricity-to-one limit of a highly elongated ellipse (dashed) sharing the Sun's focus. The far turning point of that ellipse is the release point, a distance R out, and the near one has collapsed onto the Sun itself. — The straight-line fall (solid) read as the eccentricity-to-one limit of a highly elongated ellipse (dashed) sharing the Sun’s focus. The far turning point of that ellipse is the release point, a distance $R$ out, and the near one has collapsed onto the Sun itself.

A body released from rest and pulled straight in traces the limiting case of an ellipse whose eccentricity has gone to 1: the near turning point (perihelion) has collapsed onto the Sun itself, and the far turning point (aphelion) is the release point, a distance $R$ away. The semi-major axis is half the sum of the two turning distances.

a = \frac{R+0}{2}

a = \dfrac{R}{2}

(1)

Step 2Scale the period by Kepler’s third law

Applying Kepler’s third law to the degenerate ellipse and to the Earth’s own orbit (semi-major axis $R$ , period $T$ ), and substituting (1),

\begin{align*}\left(\frac{T_{\mathrm{fall}}}{T}\right)^2 &= \left(\frac{a}{R}\right)^3 \\ &= \left(\frac{R/2}{R}\right)^3 \\ &= \frac{1}{8}\end{align*}

T_{\mathrm{fall}} = \dfrac{T}{2\sqrt2}

(2)

Step 3Half the period is the one-way fall time

$T_{\mathrm{fall}}$ is the period of a full trip around the degenerate ellipse, out to the release point and back; falling in from rest at aphelion to the Sun at perihelion is exactly half of that, the same way half of any orbital period carries a body from one turning point to the other. Substituting (2) and $T = 365.25$ d,

\begin{align*}\tau &= \frac{T_{\mathrm{fall}}}{2} \\ &= \frac{T}{4\sqrt2} \\ &= \frac{365.25\ \text{d}}{4\sqrt2}\end{align*}

\tau = \dfrac{T}{4\sqrt2} = 65\ \text{d}

(3)

Check

Bound the fall with no orbit theory at all: the pull per unit mass at release is $a_0=\gamma M_S/R^2=4\pi^2R/T^2$ , the Earth’s own centripetal acceleration, and held fixed at that value it would carry the body through $R$ in $\sqrt{2R/a_0}=T/(\pi\sqrt2)=0.225\,T$ . Gravity only strengthens as the body falls, so the true time must come in shorter than that, and (3) does: $T/(4\sqrt2)=0.177\,T$ , exactly $\pi/4=0.79$ of the constant-pull estimate.

Remark

The trick generalizes to any straight-line infall released from rest: model it as the zero-angular-momentum limit of an ellipse sharing the real orbit’s focus, read off $a$ from the two turning distances, and Kepler’s third law converts the geometry into a time without ever touching $\gamma$ or a central mass. What makes this problem particularly clean is that the near turning point is the force center itself, so only one distance, $R$ , is needed to fix $a$ .