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

Question

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

ScanSolve AI Tutor · Accepted Answer

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

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, $G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$
Substitute the values into the formula.
$g = \frac{(6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (7.40 \times 10^{22} \text{ kg})}{(1.76 \times 10^6 \text{ m})^2}$
Calculate the denominator.
$(1.76 \times 10^6 \text{ m})^2 = (1.76)^2 \times (10^6)^2 \text{ m}^2$ $= 3.0976 \times 10^{12} \text{ m}^2$
Perform the multiplication in the numerator.
$(6.674 \times 10^{-11}) \times (7.40 \times 10^{22}) = (6.674 \times 7.40) \times (10^{-11} \times 10^{22})$ $= 49.3876 \times 10^{11} \text{ N m}^2/\text{kg}$
Divide the numerator by the denominator.
$g = \frac{49.3876 \times 10^{11} \text{ N m}^2/\text{kg}}{3.0976 \times 10^{12} \text{ m}^2}$ $g = \frac{49.3876}{3.0976} \times \frac{10^{11}}{10^{12}} \text{ m/s}^2$ $g \approx 15.9439 \times 10^{-1} \text{ m/s}^2$ $g \approx 1.59439 \text{ m/s}^2$
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 $1.59 \text{ m/s}^2$ .

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