Question

Two point masses M and 3M are placed at a

distance L apart. Another point mass m is placed in

between on the line joining them so that the net

gravitational force acting on it due to masses M and

3M is zero. The magnitude of gravitational force

acting due to mass M on mass m will be


$(a) \frac{GMm {(1 + \sqrt{3} )}^{2} }{ {l}^{2} } \\ \\ (b) \frac{3gmm}{ {l}^{2} (1 + \sqrt{ 3} )} \\ \\ (c) \frac{gmm {(1 - \sqrt{3} )}^{2} }{ {l}^{2} } \\ \\ (d) \frac{gmm(1 - \sqrt{3} )}{ {l}^{2} }$

Answer 1

Net force on any mass at the position between two masses is zero

Let say the distance between M and m is "x" so the net force due to M and 3M on mass m is zero

so we have

$\frac{GMm}{x^2} = \frac{G3Mm}{(L-x)^2}$

now by solving above equation

$L - x = \sqrt3 x$

$x = \frac{L}{1 + \sqrt3}$

now by the formula of force between two mass

$F = \frac{GMm}{x^2}$

plug in the value of "x" in it

$F = \frac{GMm}{(\frac{L}{\sqrt3 +1})^2}$

$F = \frac{GMm(\sqrt3 + 1)^2}{L^2}$

so the correct answer is option A

Answer 2

Net force on any mass at the position between two masses is zero

Let say the distance between M and m is "x" so the net force due to M and 3M on mass m is zero

so we have

\frac{GMm}{x^2} = \frac{G3Mm}{(L-x)^2}x2GMm​=(L−x)2G3Mm​

now by solving above equation

L - x = \sqrt3 xL−x=3​x

x = \frac{L}{1 + \sqrt3}x=1+3​L​

now by the formula of force between two mass

F = \frac{GMm}{x^2}F=x2GMm​

plug in the value of "x" in it

F = \frac{GMm}{(\frac{L}{\sqrt3 +1})^2}F=(3​+1L​)2GMm​

F = \frac{GMm(\sqrt3 + 1)^2}{L^2}F=L2GMm(3​+1)2​

so the correct answer is option A

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