Question

Estimate by what factor a person can jump farther on the Moon as compared to the Earth if the take-off speed and angle are the same. The acceleration due to gravity on the Moon is one-sixth what it is on Earth.

Short answer

The factor by which a person can jump farther on the Moon compared to the Earth is 6.

Step-by-step solution

Step 1. Significance of acceleration due to gravity

A force that acts on any object because of the Earth's gravitational field is known as gravitational force.

In such cases, the object gains some acceleration in the downward direction toward the Earth’s surface. This acceleration due to which it moves downward can be termed as acceleration due to gravity (g), whose value is$\mathbf{9}\mathbf{.}\mathbf{81} \mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$.

For example, acceleration due to gravity can be observed in the case of a free falling object.

Step 2. List of assumptions

It is given that the acceleration due to gravity on the Moon is $g_M=\frac{1}{6}{g}_E$.

The assumptions can be listed below as:

• The person jumps with the initial velocity u.
• The angle made by the person while jumping from the surface of the Earth is $\theta$.
• The vertical direction is taken along the y-axis, and the horizontal direction is taken along the x-axis.
• The velocity of the person at the maximum height is $v=0$.
• The take-off speed and angle are the same on the surface of the Earth and the Moon.

Step 3. Determination of the total time taken by the person to achieve the maximum height

The horizontal velocity of the person can be expressed as:

$v_x=u\cos \theta$

The horizontal velocity of the person can be expressed as:

$v_y=u\sin \theta$

The person reaches the maximum height from the surface of the Earth. The time taken ($t_y$) by the person can be calculated with the help of the first equation of motion. This can be expressed as:

$v=v_y+a{t}_y$

Here, acceleration (a) is equal to the acceleration due to gravity (g) in the downward direction, which is equal to $\left(-g\right)$. At the maximum height, the person’s velocity can be considered as zero.

This can be further solved as:

$\begin{array}{c}v=v_y-g{t}_y\\ g{t}_y=v_y-v\end{array}$

Substitute the value 0 for v in the above equation.

$\begin{array}{c}g{t}_y=v_y-0\\ t_y=\frac{v_y}{g}\end{array}$

The total time (t) taken by the man in the air is twice the time calculated above. This can be given as:

$\begin{array}{c}t=2t_y\\ =\frac{2v_y}{g}\end{array}$

Step 4. Determination of the total distance covered by the person while jumping

The maximum distance (x) covered by the person can be given as:

$\begin{array}{c}{v}_x=\frac{x}{t}\\ x=v_{x}t\end{array}$

Substitute the values as $\frac{2v_y}{g}$for t, $u\cos \theta$for $v_x$, and $u\sin \theta$for $v_y$in the above equation.

$\begin{array}{c}x=u\cos \theta \times \frac{2v_y}{g}\\ =u\cos \theta \times \frac{2u\sin \theta }{g}\\ =\frac{u^2\times 2\sin \theta \cos \theta }{g}\\ =\frac{u^2\sin 2\theta }{g}\end{array}$

From the above relation, it can be concluded that distance is inversely proportional to g. This can be expressed as:

$x\propto \frac{1}{g}$

Step 5. Determination of the distance between the Earth and the Moon

The distance moved by the person while jumping on the surface of the Earth can be expressed as:

$x_E=\frac{u^2\sin 2\theta }{g_E}\dots \left(i\right)$

The take-off speeds and angles are the same on the Earth and the Moon. Therefore, the distance moved by the person while jumping on the surface of the Moon can be expressed as:

$x_M=\frac{u^2\sin 2\theta }{g_M}\dots \left(ii\right)$

Step 6. Determination of the ratio of the distance on the Moon to the distance on the Earth

Take the ratio of equation (i) to equation (ii).

$\begin{array}{c}\frac{x_E}{x_M}=\frac{\frac{u^2\sin 2\theta }{g_E}}{\frac{u^2\sin 2\theta }{g_M}}\\ \frac{x_E}{x_M}=\frac{g_M}{g_E}\end{array}$

Substitute the value $\frac{1}{6}{g}_E$for $g_M$in the above equation.

$\begin{array}{c}\frac{x_E}{x_M}=\frac{\frac{1}{6}{g}_E}{g_E}\\ \frac{x_M}{x_E}=\frac{6g_E}{g_E}\\ \frac{x_M}{x_E}=6\end{array}$

The maximum distance on the surface of the Moon is six times the maximum distance on the surface of the Earth. This gets applied when the person jumps with the same take-off speed and angle.

Thus, the factor by which a person can jump farther on the Moon compared to the Earth is 6.