The radius of the Moon is frac{1}{4} times th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The acceleration due to gravity on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M

Vedclass · Accepted Answer

$\frac{g}{5}$

Vedclass · Answer

(B) The acceleration due to gravity on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $R$ is its radius.For Earth: $g_e = \frac{GM_e}{R_e^2}$.For the Moon: $g_m = \frac{GM_m}{R_m^2}$.Given that $M_m = \frac{M_e}{80}$ and $R_m = \frac{R_e}{4}$.Taking the ratio of the two accelerations:$\frac{g_m}{g_e} = \frac{M_m}{M_e} 	imes \left( \frac{R_e}{R_m} ight)^2$Substituting the given values:$\frac{g_m}{g_e} = \left( \frac{1}{80} ight) 	imes \left( \frac{4}{1} ight)^2 = \frac{1}{80} 	imes 16 = \frac{16}{80} = \frac{1}{5}$.Therefore,$g_m = \frac{g_e}{5} = \frac{g}{5}$.

The radius of the Moon is $\frac{1}{4}$ times that of the Earth,and its mass is $\frac{1}{80}$ times that of the Earth. If the acceleration due to gravity on Earth is $g$,what is the acceleration due to gravity on the Moon?

Similar Questions

The height $h$ above the earth's surface at which the value of acceleration due to gravity $g$ becomes $\frac{g}{3}$ is ($R=$ radius of the earth).

The value of acceleration due to gravity at Earth's surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$,is close to: (Radius of Earth $= 6.4 \times 10^6\, m$)

Assuming the earth to be a sphere of uniform density,the acceleration due to gravity inside the earth at a distance of $r$ from the centre is proportional to

$A$ body thrown on Earth reaches a height of $90\,m$. If the same body is thrown on a planet with $\frac{1}{10}$ of the mass and $\frac{1}{3}$ of the radius of Earth,it will reach a height of ....... $m$.

$A$ mine is located at depth $\frac{R}{3}$ below the earth's surface. The acceleration due to gravity at that depth in the mine is ($R = \text{radius of earth}$,$g = \text{acceleration due to gravity at surface}$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module