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Question: Inside a uniform sphere of mass $M$ and radius $R$, a cavity of Radius $\frac{R}{3}$ is made as show...

Inside a uniform sphere of mass MM and radius RR, a cavity of Radius R3\frac{R}{3} is made as shown in figure. Gravitational potential at the centre of sphere is:

A

−14GM9R\frac{-14GM}{9R}

B

−3GM2R\frac{-3GM}{2R}

C

−3GMR\frac{-3GM}{R}

D

−13GM9R\frac{-13GM}{9R}

Answer

−13GM9R\frac{-13GM}{9R}

Explanation

Solution

Solution:

  1. Step 1. Treat the problem by superposition. A sphere of mass MM and radius RR has gravitational potential at its center

    ϕfull=−3GM2R.\phi_{\text{full}}=-\frac{3GM}{2R}.

  2. Step 2. The cavity is removed. Removing a region is equivalent to adding a “negative mass” distribution. Its density is the same as the full sphere:

    ρ=3M4πR3.\rho=\frac{3M}{4\pi R^3}.

    The mass removed (if the cavity were filled) is

    mcav=ρ (43π(R3)3)=3M4πR3⋅4πR381=M27.m_{\rm cav}=\rho\,\left(\frac{4}{3}\pi\left(\frac{R}{3}\right)^3\right)=\frac{3M}{4\pi R^3}\cdot\frac{4\pi R^3}{81}=\frac{M}{27}.

    Thus we have a negative mass −mcav=−M27-m_{\rm cav}=-\frac{M}{27}.

  3. Step 3. It is given (from the figure) that the cavity is such that its surface touches the outer sphere. Hence the distance from the center of the full sphere (point O) to the center of the cavity (point B) is

    OB=R−R3=2R3.OB=R-\frac{R}{3}=\frac{2R}{3}.

    Since OO lies outside the cavity, the potential due to the negative mass is the same as that of a point mass:

    ϕcav=−G(−mcav)OB=−G(−M27)2R3=GM27⋅32R=GM18R.\phi_{\rm cav}=-\frac{G(-m_{\rm cav})}{OB}=-\frac{G\left(-\frac{M}{27}\right)}{\frac{2R}{3}}=\frac{GM}{27}\cdot\frac{3}{2R}=\frac{GM}{18R}.

  4. Step 4. The net potential at OO is the sum:

    ϕ(O)=ϕfull+ϕcav=−3GM2R+GM18R\phi(O)=\phi_{\text{full}}+\phi_{\rm cav}=-\frac{3GM}{2R}+\frac{GM}{18R}

    Writing with a common denominator:

    −3GM2R=−27GM18R⟹ϕ(O)=−27GM18R+GM18R=−26GM18R=−13GM9R.-\frac{3GM}{2R}=-\frac{27GM}{18R}\quad \Longrightarrow\quad \phi(O)=-\frac{27GM}{18R}+\frac{GM}{18R}=-\frac{26GM}{18R}=-\frac{13GM}{9R}.

Thus, the gravitational potential at the center of the sphere is

−13GM9R.\frac{-13GM}{9R}.

Core Explanation:

  • Full sphere potential at center: −3GM2R-\frac{3GM}{2R}
  • Cavity mass removed: M/27M/27 with its center at OB=2R/3OB=2R/3
  • Potential due to cavity at center: +GM18R+\frac{GM}{18R}
  • Total potential: −3GM2R+GM18R=−13GM9R-\frac{3GM}{2R}+\frac{GM}{18R}=-\frac{13GM}{9R}