Question

Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags mutually exert a gravitational attraction F1 on each other. You now take two bricks from one bag and add them to the other bag, causing the bags to attract each other with a force F2. What is the closest expression for F2 in terms of F1?

Answer 1

Answer:
`F_(2)=(3)/(4)F_(1)`

Step-by-step explanation:

According to Newton's law of universal gravitation:

`F=G(m_(1)m_(2))/(r^2)`

Where:

`F` is the module of the force exerted between both bodies

`G` is the universal gravitation constant.

`m_(1)` and
`m_(2)` are the masses of both bodies.

`r` is the distance between both bodies

In this case we have two situations:

1) Two bags with masses
`4M` and
`4M` mutually exerting a gravitational attraction
`F_(1)` on each other:

`F_(1)=G((4M)(4M))/(r^2)` (1)

`F_(1)=G(16M^2)/(r^2)` (2)

`F_(1)=16(GM^2)/(r^2)` (3)

2) Two bags with masses
`2M` and
`6M` mutually exerting a gravitational attraction
`F_(2)` on each other (assuming the distance between both bags is the same as situation 1):

`F_(2)=G((2M)(6M))/(r^2)` (4)

`F_(2)=G(12M^2)/(r^2)` (5)

`F_(2)=12(GM^2)/(r^2)` (6)

Now, if we isolate
`(GM^2)/(r^2)` from (3):

`(F_(1))/(16)=(GM^2)/(r^2)` (7)

Substituting
`(GM^2)/(r^2)` found in (7) in (6):

`F_(2)=12((F_(1))/(16))` (8)

`F_(2)=(12)/(16)F_(1)` (9)

Simplifying, we finally get the expression for
`F_(2)` in terms of
`F_(1)` :

`F_(2)=(3)/(4)F_(1)`