Question 192: Gravitation

Question

On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is
given by $${{Ax} \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}$$ 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:

Accepted Answer

Given $${E_x} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$$
$$ 	herefore $$ $${{ - dV} \over {dx}} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$$
$$ \Rightarrow $$ $$\int\limits_0^V {dV}  =  - \int\limits_\infty ^x {{{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}dx} $$
$$ \Rightarrow $$ $$V = {A \over {{{({x^2} + {a^2})}^{1/2}}}}$$

Answer

$${A{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}$$

Answer

$${A{{\left( {{x^2} + {a^2}} \right)}^{1/2}}}$$

Answer

$${A \over {{{\left( {{x^2} + {a^2}} \right)}^{1/2}}}}$$

Answer

$${A \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}$$

Gravitation

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