To find the free fall acceleration on the surface of the moon, we use the formula for acceleration due to gravity:

Question

To find the free fall acceleration on the surface of the moon, we use the formula for acceleration due to gravity:

Asked on March 29, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-03-29T09:28:10.764Z

To find the free fall acceleration on the surface of the moon, we use the formula for acceleration due to gravity: $$ g = \frac{GM}{R^2} $$ where: • $G$ is the universal gravitational constant ($6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$) • $M$ is the mass of the moon • $R$ is the radius of the moon Step 1: List the given values. Mass of the moon, $M = 7.40 \times 10^{22} \text{ kg}$ Radius of the moon, $R = 1.76 \times 10^6 \text{ m}$ Universal gravitational constant, $G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$ Step 2: Substitute the values into the formula. $$ g = \frac{(6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (7.40 \times 10^{22} \text{ kg})}{(1.76 \times 10^6 \text{ m})^2} $$ Step 3: Calculate the denominator. $$ (1.76 \times 10^6 \text{ m})^2 = (1.76)^2 \times (10^6)^2 \text{ m}^2 $$ $$ = 3.0976 \times 10^{12} \text{ m}^2 $$ Step 4: Perform the multiplication in the numerator. $$ (6.674 \times 10^{-11}) \times (7.40 \times 10^{22}) = (6.674 \times 7.40) \times (10^{-11} \times 10^{22}) $$ $$ = 49.3876 \times 10^{11} \text{ N m}^2/\text{kg} $$ Step 5: Divide the numerator by the denominator. $$ g = \frac{49.3876 \times 10^{11} \text{ N m}^2/\text{kg}}{3.0976 \times 10^{12} \text{ m}^2} $$ $$ g = \frac{49.3876}{3.0976} \times \frac{10^{11}}{10^{12}} \text{ m/s}^2 $$ $$ g \approx 15.9439 \times 10^{-1} \text{ m/s}^2 $$ $$ g \approx 1.59439 \text{ m/s}^2 $$ Step 6: Round to an appropriate number of significant figures (3 significant figures, based on the given data). $$ g \approx 1.59 \text{ m/s}^2 $$ The free fall acceleration on the surface of the moon is $\boxed{1.59 \text{ m/s}^2}$.

Accepted Answer · 2026-03-29T09:28:10.764Z

To find the free fall acceleration on the surface of the moon, we use the formula for acceleration due to gravity: $$ g = \frac{GM}{R^2} $$ where: • $G$ is the universal gravitational constant ($6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$) • $M$ is the mass of the moon • $R$ is the radius of the moon Step 1: List the given values. Mass of the moon, $M = 7.40 \times 10^{22} \text{ kg}$ Radius of the moon, $R = 1.76 \times 10^6 \text{ m}$ Universal gravitational constant, $G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$ Step 2: Substitute the values into the formula. $$ g = \frac{(6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (7.40 \times 10^{22} \text{ kg})}{(1.76 \times 10^6 \text{ m})^2} $$ Step 3: Calculate the denominator. $$ (1.76 \times 10^6 \text{ m})^2 = (1.76)^2 \times (10^6)^2 \text{ m}^2 $$ $$ = 3.0976 \times 10^{12} \text{ m}^2 $$ Step 4: Perform the multiplication in the numerator. $$ (6.674 \times 10^{-11}) \times (7.40 \times 10^{22}) = (6.674 \times 7.40) \times (10^{-11} \times 10^{22}) $$ $$ = 49.3876 \times 10^{11} \text{ N m}^2/\text{kg} $$ Step 5: Divide the numerator by the denominator. $$ g = \frac{49.3876 \times 10^{11} \text{ N m}^2/\text{kg}}{3.0976 \times 10^{12} \text{ m}^2} $$ $$ g = \frac{49.3876}{3.0976} \times \frac{10^{11}}{10^{12}} \text{ m/s}^2 $$ $$ g \approx 15.9439 \times 10^{-1} \text{ m/s}^2 $$ $$ g \approx 1.59439 \text{ m/s}^2 $$ Step 6: Round to an appropriate number of significant figures (3 significant figures, based on the given data). $$ g \approx 1.59 \text{ m/s}^2 $$ The free fall acceleration on the surface of the moon is $\boxed{1.59 \text{ m/s}^2}$.