The mass of the earth is 81 times that of the | vedclass

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

Question

(A) 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

$0.15$

Vedclass · Answer

(A) 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,and $R$ is the radius.For the moon,$g_m = \frac{GM_m}{R_m^2}$.For the earth,$g_e = \frac{GM_e}{R_e^2}$.Given that $M_e = 81M_m$ and $R_e = 3.5R_m$.The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is $\frac{g_m}{g_e} = \frac{GM_m}{R_m^2} 	imes \frac{R_e^2}{GM_e} = \frac{M_m}{M_e} 	imes \left(\frac{R_e}{R_m}ight)^2$.Substituting the given values: $\frac{g_m}{g_e} = \frac{M_m}{81M_m} 	imes (3.5)^2 = \frac{1}{81} 	imes 12.25 = \frac{12.25}{81} \approx 0.1512$.Rounding to two decimal places,the ratio is $0.15$.

The mass of the earth is $81$ times that of the moon and the radius of the earth is $3.5$ times that of the moon. The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is

Similar Questions

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:

If the radius of the Earth shrinks by $1.5 \%$ (mass remaining same),then the value of gravitational acceleration changes by ......... $\%$

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (both $x$ and $h$ being much smaller than the radius of the earth)

$A$ simple pendulum is oscillating with frequency $F$ on the surface of the earth. It is taken to a depth $R/3$ below the surface of the earth ($R =$ radius of earth). The frequency of oscillation at depth $R/3$ is

What is the value of acceleration due to gravity $(g)$ at the centre of the Earth? What will be the variation of $g$ below and above the surface of the Earth?

Vedclass Test Series

Exam Paper Generator

Online Exam Module