Mass M is divided into two parts xM and (1-x) | vedclass

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(A) Let the two parts be $m_1 = xM$ and $m_2 = (1-x)M$.According to Newton's law of gravitation,the force $F$ between them at a distance $r$ is given

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$1/2$

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(A) Let the two parts be $m_1 = xM$ and $m_2 = (1-x)M$.According to Newton's law of gravitation,the force $F$ between them at a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$.Substituting the values,we get $F = \frac{G}{r^2} (xM)(1-x)M = \frac{GM^2}{r^2} (x - x^2)$.For the force $F$ to be maximum,the derivative of $F$ with respect to $x$ must be zero: $\frac{dF}{dx} = 0$.$\frac{d}{dx} [\frac{GM^2}{r^2} (x - x^2)] = 0$.Since $\frac{GM^2}{r^2}$ is a constant,we have $\frac{d}{dx} (x - x^2) = 0$.$1 - 2x = 0$.$2x = 1$,which gives $x = 1/2$.

Mass $M$ is divided into two parts $xM$ and $(1-x)M$. For a given separation,the value of $x$ for which the gravitational attraction between the two pieces becomes maximum is

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