The mass density of a solid sphere is rho. Th | vedclass

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

Question

(A) The gravitational field $g$ inside a solid sphere of radius $R$ and uniform density $\rho$ at a distance $r$ $(r < R)$ from the centre is given by

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The gravitational field $g$ inside a solid sphere of radius $R$ and uniform density $ho$ at a distance $r$ $(r Here,$M_{in}$ is the mass of the sphere enclosed within the radius $r$.The mass $M_{in}$ can be expressed as $M_{in} = 	ext{density} 	imes 	ext{volume} = ho 	imes \frac{4}{3} \pi r^3$.Substituting this into the formula for $g$:$g = \frac{G (ho \cdot \frac{4}{3} \pi r^3)}{r^2} = \frac{4}{3} \pi G ho r$.

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

Similar Questions

Gauss's law for gravitation is given by:

Two concentric hollow spheres of radius $R$ and $2R$ have the same mass $M$,as shown in the figure. The gravitational field intensity at point $P$ (where $R < r < 2R$) is:

If gravitational potential in a region is given by $V = 4x^2$,then the gravitational field is:

$3$ particles each of mass $m$ are kept at the vertices of an equilateral triangle of side $L$. The gravitational field at the centre due to these particles is

An infinite number of masses,each of $1 \ kg$,are placed on the $+ve \ X$-axis at $1 \ m, 2 \ m, 4 \ m, \dots$ from the origin. The magnitude of the gravitational field at the origin due to this distribution of masses is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module