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(D) Let a unit mass $m$ be placed at a distance $x$ from the centre of the earth,where the net gravitational force is zero.At this point,the gravitati

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$\frac{9 D}{10}$

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(D) Let a unit mass $m$ be placed at a distance $x$ from the centre of the earth,where the net gravitational force is zero.At this point,the gravitational pull from the earth must be equal in magnitude to the gravitational pull from the moon.$\frac{G m M_e}{x^2} = \frac{G m M_m}{(D-x)^2}$ ... $(i)$where $M_e$ is the mass of the earth and $M_m$ is the mass of the moon.Given that $M_e = 81 M_m$.Substituting this into equation $(i)$:$\frac{G m (81 M_m)}{x^2} = \frac{G m M_m}{(D-x)^2}$$\frac{81}{x^2} = \frac{1}{(D-x)^2}$Taking the square root on both sides:$\frac{9}{x} = \frac{1}{D-x}$$9(D - x) = x$$9D - 9x = x$$9D = 10x$$x = \frac{9D}{10}$Thus,the gravitational force is zero at a distance of $\frac{9D}{10}$ from the centre of the earth.

The distance between the centre of the moon and the earth is $D$ and the mass of the earth is $81$ times the mass of the moon. At what distance from the centre of the earth will the net gravitational force on an object be zero?

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