Irodov 1.232: Escaping One Well, Landing in Another – Amit & Deepali Sinha Foundation

Problem Irodov 1.232

A spaceship of mass $m=2.0\times10^3\ \text{kg}$ starts at rest on the Earth’s surface. Using the Earth’s mass $M_1$ and radius $R_1$ , the Moon’s mass $M_2$ and radius $R_2$ , and the Earth-Moon center-to-centre distance $D$ , find the minimum work that must be performed to bring the ship to rest on the Moon’s surface.

The Earth (mass M_1, radius R_1) and Moon (mass M_2, radius R_2), centers D apart, not to scale. The ship m starts on the Earth's surface; the dashed path runs to the Moon's surface. — The Earth (mass $M_1$ , radius $R_1$ ) and Moon (mass $M_2$ , radius $R_2$ ), centers $D$ apart, not to scale. The ship $m$ starts on the Earth’s surface; the dashed path runs to the Moon’s surface.

Given

$m=2.0\times10^3\ \text{kg}$
$M_1,R_1$ (Earth)
$M_2,R_2$ (Moon)
$D$

Find

A_{\min}

Solution

Step 1Climbing out of Earth’s own field

The Earth-Moon separation $D\approx3.84\times10^5\ \text{km}$ dwarfs either radius, and the two fields exactly balance at the null point found in Problem 1.231, about $3.8\times10^4\ \text{km}$ from the Moon; the potential energy there is a small correction to the Earth’s own surface potential (checked below), so to leading order the first leg is an escape from the Earth alone, from rest at $R_1$ to rest at an arbitrarily large distance.

0 = W_1 - \frac{\gamma M_1m}{R_1}

W_1=\frac{\gamma M_1m}{R_1}

(1)

Step 2Paying to land softly on the Moon

0 = W_2 - \frac{\gamma M_2m}{R_2}

W_2=\frac{\gamma M_2m}{R_2}

(2)

Step 3Total minimum work

The two legs are independent contributions to the propulsion budget, the first spent climbing out and the second spent braking, and because the engine cannot recover the braking energy the two costs add; adding (1) and (2) and substituting the tabulated masses and radii gives the total.

\begin{align*}\frac{\gamma M_1m}{R_1} &= \frac{(6.67\times10^{-11})(5.96\times10^{24})(2.0\times10^3)}{6.37\times10^6} = 1.25\times10^{11}\ \text{J} \\ \frac{\gamma M_2m}{R_2} &= \frac{(6.67\times10^{-11})(7.35\times10^{22})(2.0\times10^3)}{1.74\times10^6} = 5.64\times10^{9}\ \text{J} \\ A &= 1.25\times10^{11}\ \text{J} + 5.64\times10^{9}\ \text{J} = 1.30\times10^{11}\ \text{J} = 1.3\times10^{8}\ \text{kJ}\end{align*}

A=\gamma m\left(\frac{M_1}{R_1}+\frac{M_2}{R_2}\right)\approx1.3\times10^8\ \text{kJ}

(3)

Check

$\gamma M_1/R_1^2=(6.67\times10^{-11})(5.96\times10^{24})/(6.37\times10^6)^2=9.8\ \text{m/s}^2$ , reproducing the tabulated surface $g$ , a consistency check on the Earth data used. The two legs above are stitched together at infinity, but the true minimum-work route need only be given enough energy to reach the point where the Earth’s and Moon’s pulls balance, $3.8\times10^4\ \text{km}$ from the Moon (Problem 1.231), arriving there with zero speed, and then brake all the way down the far side: that exact route gives $A_{\min}=1.274\times10^{11}\ \text{J}=1.274\times10^8\ \text{kJ}$ , so the boxed two-leg formula overestimates it by only $2.4\%$ (its $1.3045\times10^8\ \text{kJ}$ against the exact $1.274\times10^8\ \text{kJ}$ ), and both values round to the printed $1.3\times10^8\ \text{kJ}$ .

Remark

The Earth leg alone accounts for about $96\%$ of the total ( $1.25\times10^{11}$ of $1.30\times10^{11}\ \text{J}$ ): climbing out of the Earth’s own gravity well is by far the larger cost, and landing gently on the much smaller, less massive Moon is comparatively cheap. This is a direct consequence of $\gamma M_1/R_1$ exceeding $\gamma M_2/R_2$ , not a general rule that the origin body always dominates; a trip the other way, launched from the Moon and landed on the Earth, would use the same two terms with the roles of “escape” and “land” swapped between them, so it is now the landing leg, $\gamma M_1m/R_1$ , that costs $96\%$ of the total, and escaping the Moon that is cheap.