Mass density of a solid sphere is rho . | vedclass

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

Question

$\mathrm{g}=\frac{\mathrm{GMr}}{\mathrm{R}^{3}}=\frac{\mathrm{G} 4 \pi \mathrm{R}^{3} \mathrm{pr}}{3 \mathrm{R}^{3}}=\frac{4 \pi \mathrm{G} \rho \math

Vedclass · Accepted Answer

$\frac{{4\rho G\pi r}}{3}$

Vedclass · Answer

$\mathrm{g}=\frac{\mathrm{GMr}}{\mathrm{R}^{3}}=\frac{\mathrm{G} 4 \pi \mathrm{R}^{3} \mathrm{pr}}{3 \mathrm{R}^{3}}=\frac{4 \pi \mathrm{G} \rho \mathrm{r}}{3}$

Mass density of a solid sphere is $\rho $ . Radius of the sphere is $R$. The gravitational field at a distance $r$ from the centre of the sphere inside it is

Similar Questions

Radius of earth is around $6000\, km$. The weight of body at height of $6000 \,km$ from earth surface becomes

At what distance from the centre of the earth, the value of acceleration due to gravity $g$ will be half that on the surface ($R =$ radius of earth)

Spot the wrong statement :The acceleration due to gravity $‘g’$ decreases if

The diameters of two planets are in ratio $4:1$ . Their mean densities have ratio $1:2$ . The ratio of gravitational acceleration on the surface of planets will be

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$, the radius of the earth, is