First slide

Gravitational potential and potential energy

Question

The work done in bringing three particles each of mass 10gm from large distance to the vertices of an equilateral triangle of side 10cm is (approximately)

Easy

A

$10^{- 3} J$
B

$2 \times 10^{- 13} J$
C

$4 \times 10^{- 3} J$
D

$10^{- 11} J$

Solution

$m = 10 g = 10 \times 10^{- 3} k g = 10^{- 2} k g r = 10 c m = 10 \times 10^{- 2} m = 10^{- 1} m W = - Δ U = - (U_{f} - U_{i}) = - (U_{f} - 0) (∵ P o t e n t i a l e n e r g y a t i n i f i n i t y = 0 f o r a l l t h e t h r e e p a r t i c l e s) W_{n e t} = - (U_{f 1} + U_{f 2} + U_{f 3}) = - 2 U = - (- \frac{3 G m_{1} m_{2}}{r}) = \frac{3 G m^{2}}{r} = \frac{3 \times 6.67 \times 10^{- 11} \times {(10^{- 2})}^{2}}{10^{- 1}} = 20.01 \times 10^{- 14} J = 2 \times 10^{- 13} J$

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