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: $F_{2}=\frac{3}{4}F_{1}$

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

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:

$F_{1}=G\frac{(4M)(4M)}{r^2}$   (1)

$F_{1}=G\frac{16M^2}{r^2}$   (2)

$F_{1}=16\frac{GM^2}{r^2}$   (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):

$F_{2}=G\frac{(2M)(6M)}{r^2}$   (4)

$F_{2}=G\frac{12M^2}{r^2}$   (5)

$F_{2}=12\frac{GM^2}{r^2}$   (6)

Now, if we isolate $\frac{GM^2}{r^2}$ from (3):

$\frac{F_{1}}{16}=\frac{GM^2}{r^2}$   (7)

Substituting $\frac{GM^2}{r^2}$  found in (7) in (6):

$F_{2}=12(\frac{F_{1}}{16})$   (8)

$F_{2}=\frac{12}{16}F_{1}$   (9)

Simplifying, we finally get the expression for $F_{2}$  in terms of $F_{1}$ :

$F_{2}=\frac{3}{4}F_{1}$