Gravitational potential due to a spherical shell:
1) At a point outside the shell:
Consider a uniform spherical shell of radius R.
Let P be a point outside the spherical shell at a distance ‘r ' from the centre O of the shell.
Let ϱ = Mass per unit area of the surface.
As shown in figure, two planes AD and BC cut the shell vertically.
The element between the two planes is a slice ABCD in the form of a ring of small angular width dϴ with the OP as axis.
Each element of the ring is at a distance , AP = x from the outside point P.
.•. Thickness of the shell, AB = R dϴ
Radius of the shell, CK = R sinϴ
.•. Surface area of the slice= (2πR sinϴ) (R dϴ ) = 2πR² sinϴ dϴ
.•. Mass of the slice= Surface area of slice× Mass per unit area (ϱ)
Figure: Gravitational potential at a point outside the spherical shell. |
.•. Mass of the slice, Mring = 2πR²sinϴ dϴ ϱ
.•. potential at P due to the ring is :
dV = - GMring
X
X
In ∆ APO : x² = r² + R² - 2rR cosϴ
Differentiating above equation, we get :
2x dx = 2rR sinϴ dϴ
( since,R and r are constant )
( since,R and r are constant )
.•. x = rR sinϴ dϴ
dx
dx
Substituting this value of ‘x' in eq. (1) , we get :
dV = - G ( 2πR²ϱsinϴ dϴ ) dx
rR sinϴ dϴ
rR sinϴ dϴ
= - 2πRϱG dx
r
Integrating above equation for the whole shell between the limits,
But, 4πR²ϱ = M ( Mass of the whole shell )
.•. V = - GM
r ...... (2)
r ...... (2)
Thus, for a point outside the shell, the shell behabes as if the whole mass is concentrated at the centre of the shell.
Now, the intensity of gravitational field at a point outside the shell is given by:
E = - dV / dr
= - d / dr ( - GM / r )
.•. E = - GM / r² ...... (3)
Negative sign shows that intensity of gravitational field is directed towards the centre O of the shell.
(2) Potential on the surface of the shell :
For a point on the surface of the shell, r= R.
.•. Potential , V = - GM / R
Intensity of the field, E = - GM / R²
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