The average density of the Earth is [ g is ac | vedclass

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

Question

(B) The acceleration due to gravity $g$ on the surface of the Earth is given by the formula $g = \frac{GM}{R^2}$.Substituting the mass of the Earth $M

Vedclass · Accepted Answer

directly proportional to $g$

Vedclass · Answer

(B) The acceleration due to gravity $g$ on the surface of the Earth is given by the formula $g = \frac{GM}{R^2}$.Substituting the mass of the Earth $M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3$,where $\rho$ is the average density and $R$ is the radius of the Earth:$g = \frac{G}{R^2} \cdot \left( \frac{4}{3} \pi R^3 \rho \right) = \frac{4}{3} \pi G R \rho$.Since $G$,$\pi$,and $R$ are constants for the Earth,we have $g \propto \rho$.Therefore,the average density $\rho$ is directly proportional to the acceleration due to gravity $g$.

The average density of the Earth is [ $g$ is acceleration due to gravity]:

Similar Questions

The period of a seconds pendulum on a planet,whose mass and radius are three times that of Earth,is

If the density of the Earth increases $4$ times and its radius becomes half of its present value,our weight will

If the value of $g$ (acceleration due to gravity) at the Earth's surface is $10 \ m/s^2$,what is its value in $m/s^2$ at the center of the Earth,assuming the Earth is a sphere of radius $R$ meters with uniform mass density?

What is the gravitational force on a particle at the centre of the Earth?

Vedclass Test Series

Exam Paper Generator

Online Exam Module