Question

The gravitational field intensity at a point 'p' which is at a distance x from the centre of the disc is

Answer 1

Explanation:

Let us consider a uniform disc of mass M. The center of mass is located at point O. We have to calculate the field due to this uniform disc at test point P. Distance between O and P is r. The resultant field is along

PO

is figure.

Let us draw a circle of radius x. We draw another concentric circle of radius n+dx. These two concentric circle of forms a disc of corresponding mass is

dM=

πa

2

m

(2πxdx)=

a

2

2mxdx

dE=

(r

2

+x

2

)

3/2

h(

a

2

2mxdx

)r

=

a

2

2amr

=

(r

2

+x

2

)

3/2

xdx

Answer 2

Answer:

The gravitational field intensity at a point 'p' which is at a distance x from the center of the disc is

$E = \frac{2 G M x}{R^{2} } [\frac{1}{x} - \frac{1}{\sqrt{x^{2} +R^{2} } }]$

Explanation:

To Find:

The gravitational field intensity at a point 'p' which is at a distance x from the center of the disc

Given:

• Mass = m
• Radius = R
• x be the distance between the point 'p' to the disc
• dE = Gravitational field intensity
• dr = radius of disc (small element)
• r = radius
• Distance between the point 'p' and center of disc = $\sqrt{x^{2} + r^{2} }$

Derivation:

Gravitational field due to elemental ring:

$dE= \frac{G dM x}{(x^{2}+r^{2}) ^{\frac{3}{2} } }$    ...........................(1)

$dM = \frac{M}{\pi R^{2} } 2\pi r dr\\dM = \frac{2 M}{R^{2} } r dr$...........................(2)

Substitute the value of dM in equation (1)

$dE = \frac{G \frac{2M}{R^{2} } r dr x }{R^{2}(x^{2} +r^{2}) ^{\frac{3}{2} } }$  ..........................(3)

Integrate the above equation:

With the limit : r --> 0 to R

$E = \int\limits^R_0 ({\frac{2 G M x}{R^{2} })\frac{r dr}{(x^{2} +r^{2} )^{2} } } \,$ .....................(4)

$E = \frac{2 G M x}{R^{2} } [\frac{1}{x} - \frac{1}{\sqrt{x^{2} +R^{2} } }]$ ..................(5)

Equation (5) gives the The gravitational field intensity at a point 'p' which is at a distance x from the center of the disc.

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