Answer: [tex]F_{2}=\frac{3}{4}F_{1}[/tex]
Explanation:
According to Newton's law of universal gravitation:
[tex]F=G\frac{m_{1}m_{2}}{r^2}[/tex]
Where:
[tex]F[/tex] is the module of the force exerted between both bodies
[tex]G[/tex] is the universal gravitation constant.
[tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of both bodies.
[tex]r[/tex] is the distance between both bodies
In this case we have two situations:
1) Two bags with masses [tex]4M[/tex] and [tex]4M[/tex] mutually exerting a gravitational attraction [tex]F_{1}[/tex] on each other:
[tex]F_{1}=G\frac{(4M)(4M)}{r^2}[/tex] (1)
[tex]F_{1}=G\frac{16M^2}{r^2}[/tex] (2)
[tex]F_{1}=16\frac{GM^2}{r^2}[/tex] (3)
2) Two bags with masses [tex]2M[/tex] and [tex]6M[/tex] mutually exerting a gravitational attraction [tex]F_{2}[/tex] on each other (assuming the distance between both bags is the same as situation 1):
[tex]F_{2}=G\frac{(2M)(6M)}{r^2}[/tex] (4)
[tex]F_{2}=G\frac{12M^2}{r^2}[/tex] (5)
[tex]F_{2}=12\frac{GM^2}{r^2}[/tex] (6)
Now, if we isolate [tex]\frac{GM^2}{r^2}[/tex] from (3):
[tex]\frac{F_{1}}{16}=\frac{GM^2}{r^2}[/tex] (7)
Substituting [tex]\frac{GM^2}{r^2}[/tex] found in (7) in (6):
[tex]F_{2}=12(\frac{F_{1}}{16})[/tex] (8)
[tex]F_{2}=\frac{12}{16}F_{1}[/tex] (9)
Simplifying, we finally get the expression for [tex]F_{2}[/tex] in terms of [tex]F_{1}[/tex] :
[tex]F_{2}=\frac{3}{4}F_{1}[/tex]
