What is the gravitational potential energy of the mass $m$ ?

- \frac{2}{\sqrt{3}} \frac{G M m}{l} (1 - 2 \sqrt{3})

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- \frac{2}{\sqrt{3}} \frac{G M m}{l} (1 + 2 \sqrt{3})

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- \frac{\sqrt{3}}{2} \frac{G M m}{l} (1 - 2 \sqrt{3})

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- \frac{\sqrt{3}}{2} \frac{G M m}{l} (1 + 2 \sqrt{3})

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Solution

The correct option is B $- \frac{2}{\sqrt{3}} \frac{G M m}{l} (1 + 2 \sqrt{3})$
Gravitational potential of mass m due to one of the nearer masses= $\frac{- G M m}{(l / 2)} = \frac{- 2 G M m}{l}$
Gravitational potential due to the other mass $= \frac{- 2 G M m}{l}$
Gravitational potential due to third mass $= \frac{- G M m}{(\sqrt{3} l / 2)} = \frac{- 2 G M m}{\sqrt{3} l}$

Hence the total gravitational potential $= - \frac{4 G M m}{l} - \frac{2 G M m}{\sqrt{3} l} = - \frac{2}{\sqrt{3}} \frac{G M m}{l} (1 + 2 \sqrt{3})$

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Similar questions

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L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}, L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

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L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}

and

L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

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L_{2} : \frac{x}{1} = \frac{y}{- 1} = \frac{z - 5}{3}

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A body of mass m kg. starts falling from a point 2R above the Earth's surface. Its kinetic energy when it has fallen to a point 'R' above the Earth's surface [R-Radius of Earth, M-Mass of Earth, G-Gravitational Constant]

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What is the gravitational potential energy of the mass m?

What is the gravitational potential energy of the mass $m$ ?