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Standard XIII

Physics

Newton's Law of Gravitation

The gravitati...

Question

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

$\frac{K}{x}$

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$\frac{K}{2 x}$

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$\frac{K}{x 2}$

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

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Solution

The correct option is D
$\frac{K}{2 x^{2}}$

$V = - \int E d x = - \int \frac{K}{x^{3}} d x = \frac{K}{2 x^{2}}$ .

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The gravitational field due to a mass distribution is E=Kx3 in the x - direction (K is a constant).Taking the gravitational potential to be zero at infinity, its value at a distance x is:

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

The correct option is D
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$V = - \int E d x = - \int \frac{K}{x^{3}} d x = \frac{K}{2 x^{2}}$ .