Question

From a solid sphere of mass M and radius R , a spherical portion of radius R/2 is removed. Taking gravitational potential v=0 at r=~ the potential at the centre of the cavity thus formed is: whare G=gravitational constant​

Answer 1

The potential at the centre of the cavity is $-\frac{GM}{R}$

Explanation:

Due to full solid sphere, potential at point P,

$V_{sp}=-\frac{GM}{2R^3}[(3R^2 - ( \frac {R}{2})^2)]$

$-\frac{GM}{2R^3}(\frac{11R^2}{4})$

$= -11 \frac{kGM}{8R}$

Due to cavity part potential at point P,

$V_{ca}= - \frac{3}{2} \frac{8}{R} /(2) = -\frac{3GM}{8R}$

Thus potential at the centre of cavity,

$= V_{sp} - V_{ca} = -\frac{11GM}{8R} - (- \frac{3}{8} \frac{GM}{R}) = -\frac{GM}{R}$

Answer 2

The potential at the center of the cavity is $-\dfrac{GM}{R}$.

Explanation:

Given that,

Mass of sphere = M

Radius of sphere = R

Radius of spherical portion $r'=\dfrac{R}{2}$

We need to calculate the potential

Due to complete solid sphere

$V_{s}=-\dfrac{GM}{2R^3}((3R^2-(\dfrac{R}{2})^2))$

$V_{s}=\dfrac{GM}{2R^3}(\dfrac{11R^2}{4})=-11\dfrac{GM}{8R}$

We need to calculate the potential

Due to cavity part

$V_{c}=-\dfrac{3GM}{8R}$

We need to calculate the potential at the center of cavity

$V=V_{s}-V_{c}$

$V=-11\dfrac{GM}{8R}-(-\dfrac{3GM}{8R})$

$V=-\dfrac{GM}{R}$

Hence, The potential at the center of the cavity is $-\dfrac{GM}{R}$

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Topic : Potential

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