A 4.9 kg particle and a 2.1 kg particle have a gravitational attraction with a m

Question

A 4.9 kg particle and a 2.1 kg particle have a gravitational attraction with a magnitude of 2.3 Times 10^-12 N. What is the gravitational potential energy of the two-particle system? If you triple the separation between the particles, how much work is done by the gravitational force between the particles? How much work is done by you? The work done by the gravitational force between the particles is equal to the negative of the change in the potential energy of the two-particle system. The work done by you is equal to the change in that potential energy.

Explanation / Answer

a) GM1*M2/r^2 = 2.3*10^-12

=> r = sqrt(GM1M2/(2.3*10^-12)) = sqrt(6.674*(10^-11)*4.9*2.1/(2.3*10^-12)) = 17.279 m

=> Potential energy = GM1M2/r = 17.279*2.3*10^-12 = 3.974*10^-11 Joule Answer

b) work done by gravitation force = potential_final - potential_initial = 2*3.974*(10^-11)/3 = -2.6493*10^-11 Joule Answer

c) work done by me = -work done by gravitation force = 2.6493*10^-11 Joule Answer

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