Three point masses m, m and $2$ m are placed on the vertex of an equilateral triangle of side 'a' Another point mass m is placed at the centroid of the triangle as shown in the figure If the magnitude of net force on the point mass placed at centroid is $\frac{K G m^{2}}{a^{2}}$ then find the value of K (G = universal gravitational constant)

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Solution

Force between masses $m$ and $m$ , $F_{1} = \frac{G m m}{(a / \sqrt{3})^{2}}$
Force between masses $m$ and $2 m$ , $F_{2} = \frac{G (2 m) m}{(a / \sqrt{3})^{2}}$
Hence net force on $m$ towards $2 m,$
$F_{2} - 2 F_{1} c o s 60^{\circ}$ $= \frac{3 G m^{2}}{a^{2}}$
$⟹ K = 3$

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Force between masses m and m, F1=Gmm(a/√3)2Force between masses m and 2m, F2=G(2m)m(a/√3)2Hence net force on m towards 2m,F2−2F1cos60∘=3Gm2a2⟹K=3

Force between masses $m$ and $m$ , $F_{1} = \frac{G m m}{(a / \sqrt{3})^{2}}$
Force between masses $m$ and $2 m$ , $F_{2} = \frac{G (2 m) m}{(a / \sqrt{3})^{2}}$
Hence net force on $m$ towards $2 m,$
$F_{2} - 2 F_{1} c o s 60^{\circ}$ $= \frac{3 G m^{2}}{a^{2}}$
$⟹ K = 3$