Kvant Physics Problem 182

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Problem

Published in the journal under No. 184 by mistake.

Estimate how high you could jump on the Moon.

I. Sh. Slobodetsky

Setup and Assumptions

A person performs a vertical jump from the surface of the Moon, reaching a maximum height $h_{\mathrm{M}}$. The same person can perform a vertical jump on Earth reaching height $h_{\mathrm{E}}$. The characteristic quantity determining both jumps is the initial takeoff speed $v_0$, assumed identical in both cases, since it is set by muscular power and biomechanics rather than the gravitational field.

Air resistance is neglected both on Earth and on the Moon. The motion is treated as one-dimensional motion under constant gravitational acceleration. The gravitational accelerations are denoted $g_{\mathrm{E}}$ for Earth and $g_{\mathrm{M}}$ for the Moon, with numerical values $g_{\mathrm{E}} = 9.8 ,\mathrm{m/s^2}$ and $g_{\mathrm{M}} = 1.6 ,\mathrm{m/s^2}$.

The unknown is the maximum jump height on the Moon, $h_{\mathrm{M}}$, expressed in terms of known Earth jump height $h_{\mathrm{E}}$.

Physical Principles

The vertical motion after takeoff obeys Newton’s second law in a uniform gravitational field, giving constant acceleration $-g$. The kinematic relation between velocity and displacement under constant acceleration is written as $v^2 = v_0^2 - 2 g h$.

At the highest point of the jump, the velocity becomes zero. The relation between initial speed and maximum height is therefore expressed by $v_0^2 = 2 g h$.

Derivation

For the Earth jump, the maximum height satisfies

$$v_0^2 = 2 g_{\mathrm{E}} h_{\mathrm{E}}.$$

For the Moon jump, the same initial speed $v_0$ leads to

$$v_0^2 = 2 g_{\mathrm{M}} h_{\mathrm{M}}.$$

Eliminating $v_0^2$ between these two expressions yields

$$2 g_{\mathrm{E}} h_{\mathrm{E}} = 2 g_{\mathrm{M}} h_{\mathrm{M}}.$$

Dividing both sides by $2 g_{\mathrm{M}}$ gives the symbolic relation for the lunar jump height,

$$h_{\mathrm{M}} = \frac{g_{\mathrm{E}}}{g_{\mathrm{M}}} h_{\mathrm{E}}.$$

Result

The maximum jump height on the Moon is

$$h_{\mathrm{M}} = \frac{g_{\mathrm{E}}}{g_{\mathrm{M}}} h_{\mathrm{E}}.$$

Substituting $g_{\mathrm{E}} = 9.8 ,\mathrm{m/s^2}$ and $g_{\mathrm{M}} = 1.6 ,\mathrm{m/s^2}$ gives

$$\frac{g_{\mathrm{E}}}{g_{\mathrm{M}}} = \frac{9.8}{1.6} \approx 6.125.$$

Assuming a representative Earth jump height $h_{\mathrm{E}} = 0.50 ,\mathrm{m}$, the lunar height becomes

$$h_{\mathrm{M}} \approx 6.125 \times 0.50 ,\mathrm{m} = 3.06 ,\mathrm{m}.$$

$$\boxed{h_{\mathrm{M}} \approx 3.1 ,\mathrm{m}}$$

Sanity Checks

The expression $h_{\mathrm{M}} = (g_{\mathrm{E}}/g_{\mathrm{M}}) h_{\mathrm{E}}$ has correct dimensional consistency since the ratio of gravitational accelerations is dimensionless and multiplies a length.

The dependence $h \propto 1/g$ matches the kinematic relation $h = v_0^2/(2g)$ for fixed takeoff speed, so reducing gravity increases height inversely.

The numerical factor $g_{\mathrm{E}}/g_{\mathrm{M}} \approx 6$ produces a lunar jump height of a few meters when starting from a terrestrial human jump height of order half a meter, which remains within a physically reasonable range given that the takeoff speed is unchanged. A sign error in the kinematic equation $v^2 = v_0^2 - 2gh$ would incorrectly predict a decrease of jump height with decreasing gravity, contradicting the inverse proportionality.