Question

a small area is removed from a uniform spherical shell of mass M and radius R then the gravitational feild intensity near the hollow portion is?​

Answer 1

Answer:

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Answer 2

The gravitational field outside the small part is $\dfrac{GM}{2R^2}$.

Explanation:

Given that,

Mass of spherical shell = M

Radius = R

Let the small area is A.

$E_{exrt}$= gravitation field due to the rest of spherical shell

$E_{self}$=gravitation field due to the small part

$E_{in}$ = gravitation field inside the small part due to whole  shell

$E_{out}$ = gravitation field outside the small part due to whole  shell

$E_{out}=\dfrac{GM}{R^2}$

We know that,

$E_{in}=E_{ext}-E_{self}=0$

$E_{ext}=E_{self}$

We need to calculate the gravitational field outside the small part

Using formula of gravutationa fied

$E_{out}=E_{ext}+E_{self}$

Put the vaue into the formula

$\dfrac{GM}{R^2}=2E_{ext}$

$E_{ext}=\dfrac{GM}{2R^2}$

After the small area has been removed there is no gravitation field due to the rest of spherical shell and no gravitation field due to the small part.

Hence, The gravitational field outside the small part is $\dfrac{GM}{2R^2}$.

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Topic : gravitationa fied intensity

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