Consider a point mass `M` at `O` and let us calculate gravitational intensity at `A` due to this point mass. Suppose a test mass is placed at `A` .

By Newton's law of gravitation, force on test mass `F=(GMm)/r^2` along `vec(AO)`

`E=F/m= -(Gm)/r^2 hat e_r`.............(2)

Figure shows a ring or mass `M` and radius `R` . Let `P` is the point at a distance `r` from the centre of the ring. By symmetry the field must be towards the centre that is along `vec (PO)`

Let us assume that a particle of mass `dm` on the ring say, at point `A` Now the distance `A P` is `sqrt(R^+r^2)`. Again the gravitational field at `P` due to dm is along `vec (PA)` and its magnitude is

`dE=(Gdm)/Z^2`

`therefore` `dEcos theta=(Gdm)/Z^2 costheta`

Net gravitational field `E=(Gcostheta)/Z^2 intdm`

`=(GM)/Z^2 * r/Z`

`=(GMr)/((r^2+R^2)^(3//2))` along `vec (PO)`...................(3)

IMPORTANT POINTS :

(i) If `r > > R, r^2 +R^2approxr^2`
`: . E=-(GMr)/r^3=-(GM)/r^2` (where negative sign in because of attraction)
Thus, for a distant point, a ring behaves as a point mass placed at the center of the ring.

(ii) If `r < < R, r^2+R^2approxR^2`
`: . E=-(GMr)/R^3` i.e., `Epropr`

Let the mass of disc be `M` and its radius is `R` and `P` is the point on its axis where gravitational field is to be calculated. Let us draw a circle of radius `x` and centred at `O` . We draw another concentric circle of radius `x + dx` . The part of disc enclosed between two circle can be treated as a uniform ring of radius `x` . The area of this ring is `2 pi x dx`

Therefore mass `dm` of the ring `=M/(pi R^2) 2 pi x dx =(2Mxdx)/R^2`

gravitational field due to the ring is ,

`dE=(G((2Mxdx)/R^2)r)/((r^2+x^2)^(3//2))`

So,the gravitational field at `P=int dE=(2GMr)/R^2 int_0^R (xdx)/((r^+x^2)^(3/2))`

`=(2GMr)/R^2 [-1/(sqrt(r^2+x^2))]_0^R=(2GMr)/R^2[1/r-1/(sqrt(r^2+R^2))]`

So , `E=(2GM)/R^2(1-cos theta)`..................(4)

Case I : `text(When point lies inside the spherical shell)`

`E=int dE=G/r^2 int dm_(e nc l os e d)=0`..............(5)

Case II : When point P lies outside the spherical shell

`E=int dE=G/r^2 int dm_(e nc l os e d)=(GM)/r^2`..................(6)

Gravitational field due to thin spherical shell is both discontinuous and non-differentiable function.

Case l : `text(Field at an external point)`

Let the mass of sphere is M and its radius is R , we have to calculate the gravitational field at P.
`E=intdE=int(Gdm)/r^2`
`E=(GM)/r^2`..........(7)

Thus, a uniform sphere may be treated as a single particle of equal mass placed at its centre for calculating the gravitational field at an external point.

Case II : `text(Field at an internal point)`

Suppose the point P is inside the solid sphere, in this case r < R the sphere may be divided into thin spherical shells all centered at O. Suppose the mass of such a shell is `dm`. Then

`dE=(Gdm)/r^2` along `PO`
so, `E=G/r^2 intdm`
where `intdm=M/((4/3)piR^3)quad4/3 pir^3=(Mr^3)/R^3`
`therefore` `E=(GM)/R^3 r`.....(8)

Therefore gravitational field due to a uniform sphere at an internal point is proportional to the distance of the point from the centre of the sphere. At the centre `r = 0` the field is zero. At the surface of the sphere `r=R`
so, `E=(GM)/R^2`.....(9)

Gravitational field due to solid sphere is continuous but it is not differentiable function.