Mass M is distributed uniformly over a rod of length $L$ . Find the gravitational field at P at a distance a on perpendicular bisector.

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Answer 1

$d m = \frac{M}{L} d x$

$d E = \frac{G \frac{M}{L} d x}{( x ^{2} + a ^{2} )} d E_{n e t} = d E c o s θ = \frac{G M a d x}{L ( x ^{2} + a ^{2} ) ^{3 / 2}}$

$E = \int d E c o s θ = \frac{G M a}{L} \int_{- L / 2}^{L / 2} \frac{d x}{( x ^{2} + a ^{2} ) ^{3 / 2}}$

Put $x = a t a n θ$ then $d x = a s e c^{2} θ d θ$

$E = \frac{G M a}{L} \int \frac{a s e c ^{2} θ d θ}{a ^{3} s e c ^{3} θ} = \frac{G M}{L a} \int c o s θ d θ$

$= \frac{G M}{L a} s i n θ = \frac{G M x}{L a x ^{2} + a ^{2}} \int_{- L / 2}^{L / 2}$

$= \frac{G M ( 2 L )}{a L ( L ^{2} + 4 a ^{2} ) ^{1 / 2}} = \frac{2 G M}{a ( L ^{2} + 4 a ^{2} ) ^{1 / 2}}$

Answer 2

$d m = \frac{M}{L} d x$

$d E = \frac{G \frac{M}{L} d x}{( x ^{2} + a ^{2} )} d E_{n e t} = d E c o s θ = \frac{G M a d x}{L ( x ^{2} + a ^{2} ) ^{3 / 2}}$

$E = \int d E c o s θ = \frac{G M a}{L} \int_{- L / 2}^{L / 2} \frac{d x}{( x ^{2} + a ^{2} ) ^{3 / 2}}$

Put $x = a t a n θ$ then $d x = a s e c^{2} θ d θ$

$E = \frac{G M a}{L} \int \frac{a s e c ^{2} θ d θ}{a ^{3} s e c ^{3} θ} = \frac{G M}{L a} \int c o s θ d θ$

$= \frac{G M}{L a} s i n θ = \frac{G M x}{L a x ^{2} + a ^{2}} \int_{- L / 2}^{L / 2}$

$= \frac{G M ( 2 L )}{a L ( L ^{2} + 4 a ^{2} ) ^{1 / 2}} = \frac{2 G M}{a ( L ^{2} + 4 a ^{2} ) ^{1 / 2}}$