The acceleration due to gravity on the moon i | vedclass

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

Question

(C) We know that the acceleration due to gravity is given by $g = \frac{GM}{R^2}$.Since the mass of a planet is $M = \rho \left( \frac{4}{3} \pi R^3 \

Vedclass · Accepted Answer

$\left(\frac{5}{18}\right) R_{e}$

Vedclass · Answer

(C) We know that the acceleration due to gravity is given by $g = \frac{GM}{R^2}$.Since the mass of a planet is $M = ho \left( \frac{4}{3} \pi R^3 ight)$,we can substitute this into the gravity formula:$g = \frac{G}{R^2} \left( ho \frac{4}{3} \pi R^3 ight) = \frac{4}{3} \pi G ho R$.This shows that $g \propto ho R$.Given that $g_m = \frac{1}{6} g_e$ and the ratio of densities $\frac{ho_e}{ho_m} = \frac{5}{3}$,which implies $\frac{ho_m}{ho_e} = \frac{3}{5}$.Using the proportionality $g \propto ho R$,we have:$\frac{g_m}{g_e} = \frac{ho_m R_m}{ho_e R_e}$$\frac{1}{6} = \left( \frac{3}{5} ight) \left( \frac{R_m}{R_e} ight)$$\frac{R_m}{R_e} = \frac{1}{6} 	imes \frac{5}{3} = \frac{5}{18}$Therefore,$R_m = \frac{5}{18} R_e$.

The acceleration due to gravity on the moon is $\frac{1}{6}$ times the acceleration due to gravity on the earth. If the ratio of the density of the earth $\rho_e$ to the density of the moon $\rho_m$ is $\frac{5}{3}$,then the radius of the moon $R_m$ in terms of the radius of the earth $R_e$ is:

Similar Questions

If the radius of the Earth were to shrink by half while its density remains constant,what would be the weight of an object on the surface of the Earth?

At a certain depth $d$ below the surface of the earth,the value of acceleration due to gravity becomes four times its value at a height $3R$ above the earth's surface. Where $R$ is the radius of the earth (Take $R = 6400 \ km$). The depth $d$ is equal to $............ \ km$.

The lengths of seconds pendulums on the surface of the earth and at an altitude '$h$' from the surface of the earth are $l_s$ and $l_h$ respectively. The radius of the earth is

The mass of a planet is $\frac{1}{10}$ that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is: (in $m \ s^{-2}$)

$A$ body weighs $W$ newton on the surface of the earth. Its weight at a height equal to half the radius of the earth will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module