Mass of moon is 7.34 x10²²kg. If the acceleration due to gravity on the moon is 1.4m/s², the radius of the moon is (G = 6.667 x10⁻¹¹ Nm² / kg²)

The correct answer is 1.87 x 10⁶ m.

Step-by-step Explanation:

To find the radius of the moon, we can use the formula for the acceleration due to gravity ($g$) on the surface of a spherical body:
$g=\frac{GM}{R^2}$

Where:
• $g$ is the acceleration due to gravity on the moon ($1.4{\text{m/s}}^2$)
• $G$ is the universal gravitational constant ($6.667\times {10}^{-11}{\text{N m}}^2/{\text{kg}}^2$)
• $M$ is the mass of the moon ($7.34\times {10}^{22}\text{kg}$)
• $R$ is the radius of the moon

Rearranging the formula to solve for the radius $R$ gives:
$R^2=\frac{GM}{g}$
$R=\sqrt{\frac{GM}{g}}$

Now, let's substitute the given values into the equation:
$R=\sqrt{\frac{(6.667\times {10}^{-11})\times (7.34\times {10}^{22})}{1.4}}$

First, calculate the product in the numerator:
$6.667\times 7.34\approx 48.93578$
${10}^{-11}\times {10}^{22}={10}^{11}$
So, the numerator is approximately $48.936\times {10}^{11}$, which is equal to $4.8936\times {10}^{12}$.

Next, divide this product by $g$ ($1.4$):
$R^2\approx \frac{4.8936\times {10}^{12}}{1.4}\approx 3.4954\times {10}^{12}{\text{m}}^2$

Finally, take the square root of both sides to find $R$:
$R\approx \sqrt{3.4954\times {10}^{12}}$
$R\approx 1.8696\times {10}^6\text{m}$

Rounding to three significant figures, we get:
$R\approx 1.87\times {10}^6\text{m}$