The acceleration due to gravity at the moon's surface is $1.67 \, m s^{-2}$. If the radius of the moon is $1.74 \times 10^{6} \, m$,calculate the mass of the moon. (Use $G = 6.67 \times 10^{-11} \, N m^{2} kg^{-2}$)

Vedclass pdf generator app on play store
Vedclass iOS app on app store
The formula for acceleration due to gravity $(g)$ is given by $g = \frac{GM}{R^{2}}$,where $G$ is the universal gravitational constant,$M$ is the mass of the celestial body,and $R$ is its radius.
Rearranging the formula to solve for mass $(M)$,we get $M = \frac{g R^{2}}{G}$.
Given values are:
$g = 1.67 \, m s^{-2}$
$R = 1.74 \times 10^{6} \, m$
$G = 6.67 \times 10^{-11} \, N m^{2} kg^{-2}$
Substituting these values into the formula:
$M = \frac{1.67 \times (1.74 \times 10^{6})^{2}}{6.67 \times 10^{-11}}$
$M = \frac{1.67 \times 3.0276 \times 10^{12}}{6.67 \times 10^{-11}}$
$M \approx 0.758 \times 10^{23} \, kg = 7.58 \times 10^{22} \, kg$.
Thus,the mass of the moon is approximately $7.58 \times 10^{22} \, kg$.

Explore More

Similar Questions

What is the name given to the path followed by a projectile?

$A$ ball is thrown vertically upwards with a velocity $u$. Calculate the velocity with which it falls to the earth again.

At which of the following locations is the value of $g$ maximum?

The equation $F = G \frac{m_1 m_2}{r^2}$ is valid only for:

Write True or False for the following statement:
If a spring balance holding a heavy object is released,it will read zero weight.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo