The gravitational, field due to a mass distribution is E=(k/x3) in the x - direction, where k is a constant. The value of gravitational potential at a distance x is [Taking gravitational potential to be zero at infinity]

Question

The gravitational, field due to a mass distribution is E=(k/x3) in the x - direction, where k is a constant. The value of gravitational potential at a distance x is [Taking gravitational potential to be zero at infinity] (A)  (k/x)  (B)  (k/x3)  (C)  (k/2x2)  (D)  (k/3x3)

Tardigrade Admin · Accepted Answer

Gravitation potential =∫ 1 dx Given, 1= E =( k / x 3) ∴ ∫ x ∞ ( k / x 3) dx = k [( x -3+1/-3+1)] x ∞= k [(-1/2 x 2)] x ∞=( k /2 x 2)

Q. The gravitational, field due to a mass distribution is $E = \frac{k}{x ^{3}}$ in the $x$ - direction, where k is a constant. The value of gravitational potential at a distance $x$ is [Taking gravitational potential to be zero at infinity]

$\frac{k}{x}$

$\frac{k}{x ^{3}}$

$\frac{k}{2 x ^{2}}$

$\frac{k}{3 x ^{3}}$

Gravitation potential $= \int 1 d x$
Given, $1 = E = \frac{k}{x ^{3}}$
$∴ \int_{x}^{\infty} \frac{k}{x ^{3}} d x$
$= k [\frac{x ^{- 3 + 1}}{- 3 + 1}]_{x}^{\infty} = k [\frac{- 1}{2 x ^{2}}]_{x}^{\infty} = \frac{k}{2 x ^{2}}$

Q. The gravitational, field due to a mass distribution is E=x3k​ in the x - direction, where k is a constant. The value of gravitational potential at a distance x is [Taking gravitational potential to be zero at infinity]

xk​

x3k​

2x2k​

3x3k​