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Three masses 1 kg 2 kg and 3 kg are placed at the vertices of an equilateral triangle of side 1m. Find the gravitational potential energy of the system $(G=6.67 \times 10^{-11} Nm^2 /kg^2)$

\[(a)\;-7.33 \times 10^{-10}\;J \quad (b)\; -6.42 \times 10^{-10}J \quad (c)\;-7.33 \times 10^{-9}\;J \quad (d)\;-6.42 \times 10^{-9}J \]

1 Answer

$V=-G \bigg( \large\frac{m_3m_2}{r_{32}}+\frac{m_3m_1}{r_{31}}+\frac{m_2m_1}{r{21}}\bigg)$
$m_1=1 \;kg \quad m_2 =2\; kg \quad m_3= 3\;kg$
Substituing we get
$U= -7.33 \times 10^{-10}\;J$
Hence a is the correct answer. 
answered Aug 20, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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