Question

Calculate the gravitational field intensity and potential at the centre of the base of a solid
hemisphere of mass m, radius R.​

Answer 1

Given:

• A solid  hemisphere of mass m, radius R

To Find:

• Calculate the gravitational field intensity and potential at the centre of the base of a solid  hemisphere

Solution:

We consider the shaded elemental disc of radius $Rsin \theta$ and thickness $Rd\theta$

It mass,

      $dM = (\frac{M}{\frac{2}{3} \pi R^3}) \pi(Rsin \theta)^2 (Rd \theta sin \theta)$

     $dM= \frac{3M}{2} sin^3 \theta d \theta$

Field due to this plate at O,

 $dE= \frac{2GdM(1 - cos \theta)}{(Rsin \theta )^2}$

$dE= \frac{3GMsin \theta(1 -cos \theta)d \theta}{R^2}$    (Field due to a uniform disc)

   $E = \int\limits^\frac{\pi}{2} _0 {dE}$

 $E= \int\limits^\frac{\pi}{2} _0 \frac{3GMsin \theta(1-cos \theta)d \theta}{R^2}$

  $E = \frac{3GM}{R^2}$

Now potential due to the element under consideration at the centre of the base of the hemisphere,

     $dV= -2GdM r(cosec \theta-cot \theta)$         (Potential due to a circular plate)

      $dV= \frac{3GMsin^3 \theta(casec \theta-cot \theta)d \theta}{R sin \theta}$

      $V = \int\limits^\frac{\pi}{2} _0 \frac{ -3GM (sin \theta - cos \theta sin \theta)d \theta}{R}$

      $V = \frac{-3GM}{R}$

Diagram is attached below

Answer 2

Given:

Mass of solid hemisphere, m

Radius of solid hemisphere, R

Formula Used:

Gravitational field intensity can be obtained by:

$E=\int \frac{Gdm}{r^2}$

Gravitational potential can be obtained by:

$V=\int \frac{-Gdm}{r}$

where, dm is the mass element , r is the distance of the element from the point and G is the universal gravitational constant.

Calculations:

Mass of small element of thickness dr is:

$dm=\frac{M}{\frac{2}{3} \pi R^3}(2\pi r^2)dr$

Gravitational field intensity can be obtained by:

$E=\int \frac{Gdm}{r^2}$

$E=G\int\frac{M}{\frac{2}{3}\pi R^3}(2\pi)dr$

Integrate from 0 to R:

$E=\frac{3GM}{R^2}$

Gravitational potential can be obtained by:

$V=\int \frac{-Gdm}{r}$

Integrate from 0 to R:

$V=-G\int\frac{M}{\frac{2}{3}\pi R^3}(2\pi r)dr\\=-\frac{3GM}{2R}$

Learn more about: Gravitational potential and gravitational field intensity

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