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

$A$ body weighs $200 \; N$ on the surface of the earth. How much will it weigh halfway down to the centre of the earth (in $; N$)?

The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by $40 \%$ of that at the surface of the earth,is (radius of earth is $6400 \ km$) (in $km$)

Which graph correctly represents the variation of acceleration due to gravity $(g)$ with radial distance $(r)$ from the centre of the earth (radius of the earth $= R_e$)?

For a body of mass $m$,the acceleration due to gravity at a distance $R$ from the surface of the earth is $\frac{g}{4}$. Its value at a distance $\frac{R}{2}$ from the surface of the earth is ($R = \text{radius of the earth}$,$g = \text{acceleration due to gravity at the surface}$)

What should be the angular speed of the Earth so that a body lying on the equator may appear weightless? $(g = 10\,m/s^2, R = 6400\,km)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module