Question 13: Gravitation

Question

From a solid sphere of mass M and radius R, a spherical portion of radius $${R \over 2}$$ is removed as shown in the figure.
Taking gravitational potential V = 0 at r = $$\infty$$, the potential at the centre of the cavity thus formed is

Accepted Answer

Potential at point P due to the solid sphere,
$${V_1} =  - {{GM} \over {2{R^3}}}\left[ {3{R^2} - {{\left( {{R \over 2}} \right)}^2}} \right]$$
$$ =  - {{11GM} \over {8R}}$$
Potential at point P due to the cavity part,
$${V_2} =  - {3 \over 2}{{G\left( {{M \over 8}} \right)} \over {{R \over 2}}} =  - {{3GM} \over {8R}}$$
Potential due to the remaining part at point P,
$$V = {V_1} - {V_2} = {{ - 11GM} \over {8R}} + {{3GM} \over {8R}}$$
$$ \Rightarrow V =  - {{GM} \over R}$$

Answer

$$ - {{GM} \over {2R}}$$

Answer

$$ - {{GM} \over {R}}$$

Answer

$$ - {{2GM} \over {3R}}$$

Answer

$$ - {{2GM} \over {R}}$$

Gravitation

From a solid sphere of mass M and radius R, a spherical portion of radius ${R \over 2} is removed as shown in the figure. Taking gravitational potential V = 0 at r = \infty$, the potential at the centre of the cavity thus formed is

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