The mass of the moon is 7.34 times 10^{22} k | vedclass

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

Question

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

Vedclass · Accepted Answer

$1.87 	imes 10^6 \ m$

Vedclass · Answer

(B) The formula for acceleration due to gravity on the surface of a celestial body is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass,and $R$ is the radius.Rearranging the formula to solve for the radius $R$,we get $R = \sqrt{\frac{GM}{g}}$.Substituting the given values: $G = 6.667 	imes 10^{-11} \ N \cdot m^2/kg^2$,$M = 7.34 	imes 10^{22} \ kg$,and $g = 1.4 \ m/s^2$.$R = \sqrt{\frac{6.667 	imes 10^{-11} 	imes 7.34 	imes 10^{22}}{1.4}}$$R = \sqrt{\frac{48.93578 	imes 10^{11}}{1.4}}$$R = \sqrt{34.954 	imes 10^{11}} = \sqrt{3.4954 	imes 10^{12}}$$R \approx 1.87 	imes 10^6 \ m$.

The mass of the moon is $7.34 \times 10^{22} \ kg$. If the acceleration due to gravity on the moon is $1.4 \ m/s^2$,calculate the radius of the moon. (Given: $G = 6.667 \times 10^{-11} \ N \cdot m^2/kg^2$)

Similar Questions

The weights of an object are measured in a coal mine of depth $h_1$,then at sea level of height $h_2=0$,and lastly at the top of a mountain of height $h_3$ as $W_1, W_2$,and $W_3$ respectively. Which one of the following relations is correct? [$h_1 \ll R, h_3 \ll R, R=$ radius of the earth]

Considering Earth to be a sphere of radius $R$ having uniform density $\rho$,the value of acceleration due to gravity $g$ in terms of $R$,$\rho$,and $G$ is:

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:

$A$ body weighs $45 \ N$ on the surface of the earth. The gravitational force on a body due to the earth at a height equal to half the radius of the earth will be: (in $N$)

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module