A uniform solid sphere of mass M and radius R | vedclass

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(D) The total gravitational potential $V$ at a distance $r = \frac{3}{2}R$ is the sum of the potential due to the solid sphere $(V_1)$ and the potenti

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$ - \frac{7}{6} \frac{GM}{R} $

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(D) The total gravitational potential $V$ at a distance $r = \frac{3}{2}R$ is the sum of the potential due to the solid sphere $(V_1)$ and the potential due to the spherical shell $(V_2)$.$1$. For the solid sphere of mass $M$ and radius $R$,the point $r = \frac{3}{2}R$ is outside the sphere. The potential is given by $V_1 = -\frac{GM}{r} = -\frac{GM}{(3/2)R} = -\frac{2GM}{3R}$.$2$. For the spherical shell of mass $M$ and radius $2R$,the point $r = \frac{3}{2}R$ is inside the shell. The potential inside a shell is constant and equal to the potential at its surface,given by $V_2 = -\frac{GM}{R_{shell}} = -\frac{GM}{2R}$.$3$. The net gravitational potential is $V = V_1 + V_2 = -\frac{2GM}{3R} - \frac{GM}{2R} = -\frac{4GM + 3GM}{6R} = -\frac{7GM}{6R}$.

Column-$I$	Column-$II$
$(1)$ Can never be positive.	$(a)$ Escape velocity
$(2)$ Reason for negative potential energy of galaxies.	$(b)$ Gravitational potential energy
	$(c)$ The nature of force between different galaxies is attractive.

$A$ uniform solid sphere of mass $M$ and radius $R$ is surrounded symmetrically by a uniform thin spherical shell of equal mass and radius $2R$. The value of gravitational potential at a distance $\frac{3}{2}R$ from the center is:

Similar Questions

Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$
$(1)$ Can never be positive. $(a)$ Escape velocity
$(2)$ Reason for negative potential energy of galaxies. $(b)$ Gravitational potential energy
$(c)$ The nature of force between different galaxies is attractive.

$A$ star is modeled as a uniform spherical distribution of matter,keeping the mass of the star constant. How does the gravitational pressure on the surface depend on the volume of the star?

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