The ratio of the weights of a body on the Ear | vedclass

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

Question

(D) Since the mass of the object remains constant,the weight of the object is proportional to the acceleration due to gravity $g$.Given: $\frac{W_{ear

Vedclass · Accepted Answer

$\frac{R}{2}$

Vedclass · Answer

(D) Since the mass of the object remains constant,the weight of the object is proportional to the acceleration due to gravity $g$.Given: $\frac{W_{earth}}{W_{planet}} = \frac{9}{4} = \frac{g_{earth}}{g_{planet}}$.We know that $g = \frac{GM}{R^2}$. Since density $\rho$ is constant,$M = \rho \cdot \frac{4}{3}\pi R^3$,so $g = \frac{G \cdot \rho \cdot \frac{4}{3}\pi R^3}{R^2} = G \cdot \rho \cdot \frac{4}{3}\pi R \propto R$.However,the problem states the mass of the planet is $\frac{1}{9}$ of the Earth's mass. Let $M_p = \frac{M_e}{9}$.Using $g = \frac{GM}{R^2}$,we have $\frac{g_e}{g_p} = \frac{M_e}{M_p} \cdot \frac{R_p^2}{R_e^2} = 9 \cdot \frac{R_p^2}{R^2}$.Given $\frac{g_e}{g_p} = \frac{9}{4}$,so $\frac{9}{4} = 9 \cdot \frac{R_p^2}{R^2}$.$\frac{1}{4} = \frac{R_p^2}{R^2} \implies \frac{R_p}{R} = \frac{1}{2}$.Therefore,$R_p = \frac{R}{2}$.

The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is $9 : 4$. The mass of the planet is $\frac{1}{9}^{th}$ of that of the Earth. If $R$ is the radius of the Earth,what is the radius of the planet? (Assume the planets have the same mass density)

Similar Questions

An astronaut on a strange planet finds that the acceleration due to gravity is twice that on the surface of the Earth. Which of the following could explain this?

How much is the radius of the Earth at the equator greater than the radius at the poles?

Weight of a body of mass $m$ decreases by $1\%$ when it is raised to height $h$ above the Earth's surface. If the body is taken to a depth $h$ in a mine,then its weight will

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

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^o$ latitude becomes zero is (Radius of earth $= 6400 \, km$. At the poles $g = 10 \, m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module