Chapter 3, Problem 61GP

To determine

The acceleration due to gravity on the surface of the moon.

Answer to Problem 61GP

The acceleration due to gravity on the surface of the moon is $1.91\text{m}/{\text{s}}^2$ .

Explanation of Solution

Given:

Height of the ball on the moon, $h_m=35\text{}\text{m}$

Height of the ball on the earth, $h_e=180\text{}\text{m}$

Formula Used:

Let the gravitational acceleration on earth is $g_e$ and the gravitational acceleration on moon is $g_m$ . Since the launch angle and the velocity is same in both the cases. Consider the second law of motion to calculate the acceleration on the moon.

$u^2=v^2+2g_m{h}_m$    … (1)

Where,

  $u$ is the initial velocity.

  $v$ is the final velocity.

Write the expression to calculate the acceleration on the earth using second law of motion.

$u^2=v^2+2g_e{h}_e$    … (2)

Calculation:

Since, the initial velocity is zero. Solve equation (1).

$\begin{array}{c}0=v^2+2g_m{h}_m\\ v^2=-2g_m{h}_m\end{array}$

Substitute the value of $v^2$ in equation (2) and solve.

$\begin{array}{c}0=-2g_m{h}_m+2g_e{h}_e\\ g_m{h}_m=g_e{h}_e\\ g_m=\frac{g_e{h}_e}{h_m}\end{array}$

Substitute all the values in the above expression.

$\begin{array}{c}{g}_m=\frac{\left(9.8\text{m}/{\text{s}}^2\right)\left(35\text{m}\right)}{180\text{m}}\\ =1.91\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Thus, the acceleration due to gravity on the surface of the moon is $1.91\text{m}/{\text{s}}^2$ .