The magnitudes of the gravitational force at | vedclass

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

Question

(D) For a uniform sphere of mass $M$ and radius $R$,the gravitational field $g$ at a distance $r$ from the centre is given by:$1$. Inside the sphere $

Vedclass · Accepted Answer

Both $(a)$ and $(b)$

Vedclass · Answer

(D) For a uniform sphere of mass $M$ and radius $R$,the gravitational field $g$ at a distance $r$ from the centre is given by:$1$. Inside the sphere $(r $2$. Outside the sphere $(r > R)$: $g = \frac{GM}{r^2}$,which implies $g \propto \frac{1}{r^2}$.Since the gravitational force $F = mg$,the ratio of forces $\frac{F_1}{F_2}$ is equal to the ratio of gravitational fields $\frac{g_1}{g_2}$.If $r_1 If $r_1 > R$ and $r_2 > R$,then $\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$. This matches option $(b)$.Therefore,both $(a)$ and $(b)$ are correct.

The magnitudes of the gravitational force at distances $r_1$ and $r_2$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively. Then

Similar Questions

Two planets of radii in the ratio $2 : 3$ are made from the material of density in the ratio $3 : 2$. Then the ratio of acceleration due to gravity $g_1/g_2$ at the surface of the two planets will be

The highest temperature, density, and pressure on Earth are found:

In order to make the effective acceleration due to gravity equal to zero at the equator,the angular velocity of rotation of the earth about its axis should be $(g = 10\,m/s^2$ and radius of earth is $6400\,km)$.

If the radius of the earth shrinks by $1.5\%$ (mass remaining same),then the value of acceleration due to gravity changes by ....... $\%$.

The weight of a body on the surface of the earth is $63 \ N$. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth? (in $N$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module