First slide

Gravitational potential and potential energy

Question

On the x-axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{\frac{3}{2}}}$ in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity, is :

Difficult

A

$A {(x^{2} + a^{2})}^{\frac{3}{2}}$
B

$\frac{A}{{(x^{2} + a^{2})}^{\frac{3}{2}}}$
C

$\frac{A}{{(x^{2} + a^{2})}^{\frac{1}{2}}}$
D

$A {(x^{2} + a^{2})}^{\frac{1}{2}}$

Solution

$\begin{array}{l} Given E_{x} = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}} \\ \int_{v = \infty}^{v = v_{x}} d v = - \int_{\infty}^{x} E d x (∵ E = \frac{d v}{d r}) \\ V (x) - V (\infty) = - \int_{\infty}^{x} \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}} d x - - - - (1) \\ L e t (x^{2} + a^{2}) = z \\ d i f f e r e n t i a t e b o t h s i d e s w i t h r e s p e c t i v e x w e g e t \\ 2 x d x + 0 = d z \Rightarrow x d x = \frac{d z}{2} - - - - (2) \\ F r o m (1) a n d (2) \\ Given V (\infty) = 0 \Rightarrow V (x) - 0 = - \frac{A}{2} \int_{\infty}^{x} \frac{d z}{{(z)}^{3 / 2}} = - \frac{A}{2} \int_{\infty}^{x} {(z)}^{- \frac{3}{2}} d z = - \frac{A}{2} (\frac{z^{- 3 / 2 + 1}}{- 3 / 2 + 1}) \\ = - {\frac{A}{2} \frac{{(x^{2} + a^{2})}^{- 1 / 2}}{(- \frac{1}{2})}|}_{\infty}^{x} = A {(x^{2} + a^{2})}^{- 1 / 2} \end{array}$

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