Irodov 1.233: Add Along Earth's Motion to Escape – Amit & Deepali Sinha Foundation

Problem Irodov 1.233

A body is launched from the Earth’s surface with some initial speed relative to the Earth. Determine, approximately, the third cosmic velocity $v_3$ : the smallest such launch speed that lets the body eventually leave the Solar System altogether, i.e. reach an arbitrarily large distance from the Sun. Neglect the Earth’s own axial rotation.

The Earth, mass M_E, orbiting the Sun, mass M_S, on a circle of radius r (taken circular) with orbital speed V_1, tangent to the orbit. The body is launched from the Earth's surface. — The Earth, mass $M_E$ , orbiting the Sun, mass $M_S$ , on a circle of radius $r$ (taken circular) with orbital speed $\vect V_1$ , tangent to the orbit. The body is launched from the Earth’s surface.

Given

$M_E$ , $R$ (Earth’s mass and radius)
$M_S$ , $r$ (Sun’s mass and Earth’s orbital radius)

Find

v_3

(approx.)

Solution

Step 1Escape budget relative to Earth

A body leaves the surface with speed $v$ (relative to Earth) and, once clear of Earth’s own field but still essentially at the Earth’s distance from the Sun, retains some residual speed $v_\infty$ relative to Earth; energy conservation in Earth’s field alone connects the two (this is the patched-conic approximation: the body is carried at the Earth’s heliocentric position throughout, so only the differential, tidal solar field across this short stretch is neglected, not the Sun’s potential itself, which is much larger than Earth’s but common to both ends and drops out).

\begin{align*}\frac{v^2}{2} - \frac{\gamma M_E}{R} &= \frac{v_\infty^2}{2} \\ v^2 &= v_\infty^2 + \frac{2\gamma M_E}{R}\end{align*}

Writing $v_1^2 = \gamma M_E/R$ for the speed of a circular orbit grazing the Earth’s surface (the first cosmic velocity),

v^2 = v_\infty^2 + 2v_1^2

(1)

Step 2Escaping the Sun from Earth’s orbital radius

A body at distance $r$ from the Sun needs at least the heliocentric speed $V_2$ to escape the Sun’s field entirely, i.e. to still just reach infinity, left at rest there, from the same zero-total-energy condition.

\begin{align*}\frac{V_2^2}{2} - \frac{\gamma M_S}{r} &= 0 \\ V_2^2 &= \frac{2\gamma M_S}{r}\end{align*}

Writing $V_1^2 = \gamma M_S/r$ for the Earth’s own circular orbital speed around the Sun,

V_2 = \sqrt2\,V_1

(2)

Step 3Minimum residual speed, added along Earth’s motion

Velocities added along one line: the Earth's orbital velocity V_1 and, head to tail, the body's residual velocity v_∞ relative to Earth, aimed parallel to V_1. Their sum is the heliocentric speed V_2. — Velocities added along one line: the Earth’s orbital velocity $\vect V_1$ and, head to tail, the body’s residual velocity $\vect v_\infty$ relative to Earth, aimed parallel to $\vect V_1$ . Their sum is the heliocentric speed $V_2$ .

Why

Relative to the Sun, the body’s velocity once clear of Earth is the vector sum of the Earth’s orbital velocity $\vect V_1$ and the residual velocity $\vect v_\infty$ it carries relative to Earth. The energy budget (1) fixes only the magnitude $v_\infty$ , not its direction: the body’s velocity turns through the Earth escape hyperbola between the surface and the point where Earth’s field is spent, so the asymptotic direction of $\vect v_\infty$ is free to choose at no extra energy cost. For a fixed magnitude $v_\infty$ , the triangle inequality makes the resultant largest when that asymptotic $\vect v_\infty$ , not the surface launch velocity, points parallel to $\vect V_1$ , i.e. along the Earth’s own revolution; that alignment is what makes $v_3$ a minimum.

Setting that largest possible resultant equal to the escape speed (2) isolates the smallest sufficient $v_\infty$ .

V_1 + v_\infty = V_2 = \sqrt2\,V_1 \tag{3}

v_\infty = (\sqrt2-1)\,V_1

(4)

Step 4Assemble and evaluate

v_3

Substituting the minimum residual speed (4) into the surface budget (1) gives the third cosmic velocity in closed form; numbers for $M_E$ , $R$ , $M_S$ and $r$ then give $v_1$ , $V_1$ , and the value itself.

\begin{align*}v_3^2 &= 2v_1^2 + v_\infty^2 \\ &= 2v_1^2 + (\sqrt2-1)^2 V_1^2\end{align*}

Substituting the gravitational constant $\gamma$ , the Earth’s mass and radius, the Sun’s mass, and the Earth’s orbital radius,

\begin{align*}v_1 &= \sqrt{\gamma M_E/R} = 7.90\ \text{km/s}, \\ V_1 &= \sqrt{\gamma M_S/r} = 29.6\ \text{km/s}, \\ v_3^2 &= 2(7.90\ \text{km/s})^2 + (\sqrt2-1)^2(29.6\ \text{km/s})^2 = 275\ \text{km}^2/\text{s}^2\end{align*}

v_3 = \sqrt{2v_1^2+(\sqrt2-1)^2V_1^2} \approx 16.6\ \text{km/s}

(5)

Check

Two limits confirm the formula. Letting $V_1\to0$ (drop the Sun) gives $v_3\to\sqrt2\,v_1$ , exactly the Earth’s own escape velocity, as it must once there is nothing beyond Earth’s field to escape from. Letting $v_1\to0$ (drop the Earth’s own gravity) gives $v_3\to(\sqrt2-1)V_1$ , the bare residual speed (4), as it must once the surface launch costs nothing. Numerically $16.6\ \text{km/s}$ rounds, at the book’s coarser precision, to its quoted $\approx17\ \text{km/s}$ .

Remark

The parallel direction of the asymptotic residual velocity is not a detail but the whole saving: aimed the other way once clear of Earth, against the Earth’s revolution, the vector sum (3) becomes $|v_\infty - V_1|$ instead of $v_\infty+V_1$ , so escaping needs $v_\infty=(\sqrt2+1)V_1\approx71.5\ \text{km/s}$ and $v_3\approx72\ \text{km/s}$ , more than four times the prograde answer. The same triangle-inequality argument, that a fixed relative speed buys the most absolute speed when added rather than opposed, is why real interplanetary launches are timed and aimed to ride a planet’s own orbital motion.