Kvant Physics Problem 182

Verified: yes
Verdicts: PASS + PASS
Solve time: 6m46s
Source on kvant.digital

Problem

The problem was mistakenly published in the magazine under number 184.

Estimate how high you would be able to jump on the Moon.

I. Sh. Slobodetskyi

Setup and Assumptions

A person jumps vertically upward from the surface of the Moon. The quantity to be estimated is the maximum height of the jump, denoted by $h$ and measured in meters.

Let the person's mass be $m$, the initial upward speed at takeoff be $v_0$, the acceleration due to gravity on Earth be $g_{\mathrm E} \approx 9.8,\mathrm{m/s^2}$, and the acceleration due to gravity on the Moon be $g_{\mathrm M} \approx 1.62,\mathrm{m/s^2}$.

The problem provides no numerical data about the jumper, so the estimate must be based on familiar terrestrial experience. A typical healthy person can jump vertically about $h_{\mathrm E} \approx 0.5,\mathrm m$ on Earth. The same muscular effort is assumed to produce the same takeoff speed on the Moon, because the jump is generated during the brief push against the ground before the flight begins.

Air resistance is neglected. The Moon has essentially no atmosphere, so this approximation is even better than on Earth. The lunar gravitational field is treated as uniform over the height of the jump. Rotational and terrain effects are neglected.

Physical Principles

The motion after takeoff is vertical motion under constant gravitational acceleration.

The kinetic energy at takeoff is converted into gravitational potential energy at the highest point. Thus,

$$\frac12 m v_0^2 = mgh,$$

where $g$ is the local gravitational acceleration.

Equivalently,

$$h = \frac{v_0^2}{2g}.$$

Because the jumper's muscles generate the same takeoff speed on Earth and on the Moon, the quantity $v_0$ is the same in both cases.

For Earth,

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

and for the Moon,

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

Derivation

The Earth relation gives

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

Substituting this expression into the lunar formula,

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

The jump height on the Moon exceeds the jump height on Earth by the factor

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

Using the numerical values,

$$\frac{g_{\mathrm E}}{g_{\mathrm M}} = \frac{9.8,\mathrm{m/s^2}} {1.62,\mathrm{m/s^2}} \approx 6.05.$$

Hence

$$h_{\mathrm M} \approx 6.05,h_{\mathrm E}.$$

For a typical terrestrial vertical jump of

$$h_{\mathrm E}\approx 0.5,\mathrm m,$$

the lunar jump height is

$$h_{\mathrm M} \approx 6.05 \times 0.5,\mathrm m = 3.03,\mathrm m.$$

Result

The symbolic result is

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

Substituting

$$h_{\mathrm E}=0.5,\mathrm m, \qquad g_{\mathrm E}=9.8,\mathrm{m/s^2}, \qquad g_{\mathrm M}=1.62,\mathrm{m/s^2},$$

gives

$$h_{\mathrm M} = 0.5,\mathrm m \cdot \frac{9.8,\mathrm{m/s^2}} {1.62,\mathrm{m/s^2}} = 3.03,\mathrm m.$$

$$\boxed{h_{\mathrm M}\approx 3,\mathrm m}$$

Thus a person who can jump about half a meter high on Earth could be expected to jump roughly three meters high on the Moon.

Sanity Checks

The dimensions are correct because the factor $g_{\mathrm E}/g_{\mathrm M}$ is dimensionless, so the result has units of length.

If the lunar gravitational acceleration were equal to Earth's, $g_{\mathrm M}=g_{\mathrm E}$, the formula would give $h_{\mathrm M}=h_{\mathrm E}$, which matches physical expectations.

As gravity decreases, the formula predicts larger jump heights. In the formal limit $g_{\mathrm M}\to 0$, the height tends to infinity. This reflects the fact that with no gravity the concept of a finite maximum height ceases to apply.

The estimate of about $3,\mathrm m$ is consistent with the Moon's gravity being approximately six times weaker than Earth's and with the assumption that the takeoff speed remains unchanged.

The most common source of error is to assume that the jump height scales directly with body weight rather than with gravitational acceleration in the energy relation. Another frequent mistake is to use the factor $6$ twice, once when estimating the takeoff speed and again when computing the height, which would overestimate the result by a factor of six.