First slide

Gravitational potential and potential energy

Question

The gravitational field due to a mass distribution is $E = K / x^{3}$ in the x-direction. (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is

Moderate

A

$K / x$
B

$K / 2 x$
C

$K / x^{2}$
D

$K / 2 x^{2}$

Solution

$G r a v i t a t i o n a l f i e l d = E = - \frac{d V}{d x} \Rightarrow d V = - E d x \Rightarrow \int d V = - \int E d x$
Gravitational potential $= \int_{x}^{\infty} E d x = \int_{x}^{\infty} \frac{K}{x^{3}} d x = K \int_{x}^{\infty} x^{- 3} d x$
$= K {(\frac{x^{- 3 + 1}}{- 3 + 1})}_{x}^{\infty} = {|\frac{- K}{2 x^{2}}|}_{x}^{\infty} = |\frac{- K}{2 {(\infty)}^{2}} - \frac{- K}{2 {(x)}^{2}}| = 0 - \frac{K}{2 x^{2}} = \frac{K}{2 x^{2}} (∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C)$

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