CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Consider a spherical gaseous cloud of mass density $$\rho (r)$$ in a free space where r is the radial distance from its centre. The gaseous cloud is made of particles of equal mass m moving in circular orbits about their common centre with the same kinetic energy K. The force acting on the particles is their mutual gravitational force. If $$\rho (r)$$ is constant in time. The particle number density $$n(r)=\rho(r)/m$$ is? (G$$=$$ universal gravitational constant)


A
K6πr2m2G
loader
B
Kπr2m2G
loader
C
3Kπr2m2G
loader
D
K2πr2m2G
loader

Solution

The correct option is D $$\dfrac{K}{2\pi r^2m^2G}$$
$$\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}$$
$$=\dfrac{2}{r}\left(\dfrac{1}{2}mv^2\right)$$
$$\Rightarrow \dfrac{GMm}{r^2}=\dfrac{2K}{r}$$
$$\Rightarrow M=\dfrac{2Kr}{Gm}$$
$$\Rightarrow dM=\dfrac{2K}{Gm}dr$$
$$\Rightarrow 4\pi r^2dr\rho =\dfrac{2K}{Gm}dr$$
$$\therefore \rho =\dfrac{K}{2\pi Gmr^2}$$
$$\Rightarrow \dfrac{\rho}{m}=\dfrac{K}{2\pi Gm^2r^2}$$ Alternative:
$$\dfrac{GM(r)}{r^2}=\dfrac{V^2}{r}$$
Where $$M(r)=$$ total mass upto radius(r)
$$\Rightarrow K=\dfrac{GMm}{2r}$$
$$\Rightarrow M(r)=\dfrac{2Kr}{Gm}$$
$$\Rightarrow dM(r)=\dfrac{2K}{Gm}dr=\rho dV=\rho 4\pi r^2dr$$
$$\Rightarrow \rho =\dfrac{K}{G2\pi r^2m}$$
$$\Rightarrow \dfrac{\rho}{m}=\dfrac{K}{2\pi Gm^2r^2}$$
Correct option $$4$$.

1286377_1633304_ans_b34d556303ca4edfaa01fe3a0aa9e467.png

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image