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(A) Let $m_1 = 100 \,kg$ and $m_2 = 8100 \,kg$ be the masses separated by distance $d = 1 \,m$.Let $x$ be the distance of the null point (where gravit

Vedclass · Accepted Answer

$-6.67 	imes 10^{-7}$

Vedclass · Answer

(A) Let $m_1 = 100 \,kg$ and $m_2 = 8100 \,kg$ be the masses separated by distance $d = 1 \,m$.Let $x$ be the distance of the null point (where gravitational field is zero) from $m_1$.The condition for zero gravitational field is $\frac{G m_1}{x^2} = \frac{G m_2}{(d-x)^2}$.Taking the square root on both sides: $\frac{\sqrt{m_1}}{x} = \frac{\sqrt{m_2}}{d-x}$.Substituting values: $\frac{10}{x} = \frac{90}{1-x} \implies 10 - 10x = 90x \implies 100x = 10 \implies x = 0.1 \,m$.The distance from $m_2$ is $d-x = 0.9 \,m$.The gravitational potential $V$ at this point is $V = -\frac{G m_1}{x} - \frac{G m_2}{d-x}$.$V = -G \left( \frac{100}{0.1} + \frac{8100}{0.9} ight) = -G (1000 + 9000) = -10000 G$.$V = -10000 	imes 6.67 	imes 10^{-11} = -6.67 	imes 10^{-7} \,J/kg$.

The bodies of masses $100 \,kg$ and $8100 \,kg$ are held at a distance of $1 \,m$. The gravitational field at a point on the line joining them is zero. The gravitational potential at that point in $J/kg$ is $\left(G = 6.67 \times 10^{-11} \,Nm^2/kg^2\right)$

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