A mass M is divided into two parts xM and (1 | vedclass

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(A) The gravitational force $F$ between two masses $m_1 = xM$ and $m_2 = (1 - x)M$ separated by a distance $r$ is given by:$F = \frac{G m_1 m_2}{r^2}

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$1/2$

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(A) The gravitational force $F$ between two masses $m_1 = xM$ and $m_2 = (1 - x)M$ separated by a distance $r$ is given by:$F = \frac{G m_1 m_2}{r^2} = \frac{G (xM) ((1 - x)M)}{r^2}$$F = \frac{G M^2}{r^2} (x - x^2)$To find the value of $x$ for which $F$ is maximum,we differentiate $F$ with respect to $x$ and set it to zero:$\frac{dF}{dx} = \frac{G M^2}{r^2} \frac{d}{dx} (x - x^2) = 0$$\frac{G M^2}{r^2} (1 - 2x) = 0$Since $\frac{G M^2}{r^2} 
eq 0$,we have:$1 - 2x = 0$$2x = 1$$x = 1/2$Thus,the gravitational force is maximum when $x = 1/2$.

$A$ mass $M$ is divided into two parts $xM$ and $(1 - x)M$. For what value of $x$ is the gravitational force between them maximum for a given distance?

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