Right answer: 64 units
According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:
[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 a gravitation force [tex]F_{1}=16units[/tex], given by the formula written at the beginning. Let’s rename the distance [tex]r[/tex] as [tex]d[/tex]:
[tex]F_{1}=G\frac{m_{1}m_{2}}{d^2}[/tex] (1)
And we are asked to find the gravitation force [tex]F_{2}[/tex] with a given distance of [tex]\frac{d}{2}[/tex]:
[tex]F_{2}=G\frac{m_{1}m_{2}}{({\frac{d}{2})}^{2}}[/tex]
[tex]F_{2}=G\frac{m_{1}m_{2}}{{\frac{d^{2}}{4}}}[/tex] (2)
The gravity constant is the same for both equations, and we are assuming both masses are constants, as well. So, let’s isolate [tex]G m_{1}m_{2}[/tex] in both equations:
From (1):
[tex]Gm_{1}m_{2}=F_{1}{d}^{2}[/tex] (3)
From (2):
[tex]Gm_{1}m_{2}=F_{2}\frac{{d}^{2}}{4}[/tex] (4)
If (3)=(4):
[tex]F_{1}{d}^{2}=F_{2}\frac{{d}^{2}}{4}[/tex] (5)
Now we have to find [tex]F_{2}[/tex]:
[tex]F_{2}=F_{1}{d}^{2}\frac{4}{{d}^{2}}[/tex]
[tex]F_{2}=4F_{1}[/tex] (6)
If [tex]F_{1}=16 units[/tex]:
[tex]F_{2}=(4)(16 units)[/tex]
[tex]F_{2}=64 units[/tex]>>>>This is the new force of attraction
