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 I 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

d= sqrt(Gm1m2/F)
So,
d^2 = 6.67 * 10^-11 * 4.9 x 2.1 / [2.3 x 10^-12]
So, d = 17.27 m
Gravitational potential energy is given by:
U = - G m1 m2 / r
Right from this you could get value of d and then use it in the following formula
Potential energy = - 6.67 * 10^-11 * 4.9 x 2.1 / (17.2745)
= - 3.973 * 10^-11 J

b)
Tripling the separation d, the potential energy becomes 1/3.
So, U = - 1.324 * 10^-11 J

c)
The work you do is the difference between the final and initial potential energies.

W = U(nr) - U(r)

Remember that the potential energy is negative, so you have to DO work to push the particle further away

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