Irodov 1.228: Escaping a Smaller World – Amit & Deepali Sinha Foundation

Problem Irodov 1.228

Find the orbital speed, for a circular orbit skimming the surface, and the escape speed from the surface, of the Moon. Compare both results with the corresponding orbital and escape speeds for the Earth.

Two paths leaving the same surface point of a body of mass M, radius R: a closed circular orbit at speed v_1, and the open, just-escaping trajectory at speed v_2. — Two paths leaving the same surface point of a body of mass $M$ , radius $R$ : a closed circular orbit at speed $v_1$ , and the open, just-escaping trajectory at speed $v_2$ .

Given

Moon: $M_{\text{Moon}}=7.35\times10^{22}$ kg, $R_{\text{Moon}}=1.74\times10^6$ m
Earth: $M_\text{Earth}=5.96\times10^{24}$ kg, $R_\text{Earth}=6.37\times10^6$ m

Find

The orbital speed

v_1

and escape speed

v_2

at each surface, and their ratio

Solution

Step 1Find the orbital speed from the centripetal condition

A body skimming the surface in a circular orbit is held in by gravity alone, so gravity supplies exactly the centripetal force at radius $R$ ,

\frac{\gamma Mm}{R^2} = \frac{mv_1^2}{R} \;\;\Longrightarrow\;\; v_1 = \sqrt{\frac{\gamma M}{R}}. \tag{1}

Step 2Find the escape speed from energy conservation

Launched radially at the escape speed, the body just reaches infinity with zero kinetic energy left, so the total mechanical energy at the surface equals zero,

\frac12 mv_2^2 - \frac{\gamma Mm}{R} = 0 \;\;\Longrightarrow\;\; v_2 = \sqrt{\frac{2\gamma M}{R}} = \sqrt2\,v_1. \tag{2}

Step 3Evaluate both speeds at the Moon and at the Earth

\begin{align*} v_{1,\text{Moon}} &= \sqrt{\frac{\gamma M_{\text{Moon}}}{R_{\text{Moon}}}} \approx 1.68\text{ km/s}, &\qquad v_{2,\text{Moon}} &= \sqrt2\,v_{1,\text{Moon}} \approx 2.37\text{ km/s}, \end{align*}

\begin{align*} v_{1,\text{Earth}} &= \sqrt{\frac{\gamma M_\text{Earth}}{R_\text{Earth}}} \approx 7.90\text{ km/s}, &\qquad v_{2,\text{Earth}} &= \sqrt2\,v_{1,\text{Earth}} \approx 11.2\text{ km/s}. \tag{3}\end{align*}

Moon:

v_1\approx1.68

km/s,

v_2\approx2.37

km/s

Check

The ratio $v_{2,\text{Moon}}/v_{1,\text{Moon}} = 2.37/1.68 \approx 1.41$ matches $\sqrt2$ from (2) regardless of which body it is evaluated at, since that factor came from the energy equation alone and never involved $M$ or $R$ . The Moon’s two speeds are both about $4.7$ times smaller than the Earth’s, the same factor by which $\sqrt{M/R}$ is smaller for the Moon. The Moon’s orbital speed computes to $1678$ m/s from the constants above, which rounds to the printed $1.68$ km/s; the book prints $1.67$ km/s, consistent with a slightly different Moon mass, e.g. $M_{\text{Moon}}=7.3\times10^{22}$ kg gives $1673$ m/s, which rounds to $1.67$ km/s, rather than a discrepancy in the computation here.

Remark

The Moon’s escape speed is far closer to the thermal speeds of light gas molecules than the Earth’s is, which is the reason the Moon has essentially no retained atmosphere while the Earth does.