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Standard XII

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:

K / x

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K / 2x

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K / x 2

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K / 2x²

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Solution

The correct option is D
K / 2x²

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

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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:

The correct option is D K / 2x2 V=−∫Edx=−∫Kx3dx=K2x2.

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
K / 2x²

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