Answer:
The force of gravity after you double the mass and the distance is half of the initial force: [tex]F_{2}=\frac{1}{2}F_{1}[/tex]
Explanation:
The initial force of gravity is:
[tex]F_{1}=\frac{Gm_{1}m_{2}}{r^2}[/tex]
where [tex]G[/tex] is the universal gravitational constant, [tex]m_{1}[/tex] is the mass of the first object, [tex]m_{2}[/tex] is the mass of the second object, and [tex]r[/tex] is the distance between the objects.
If the mass of the second object is doubled, now we have [tex]2m_{2}[/tex], and if the distance between the objects is also doubled instead of [tex]r[/tex] now we have [tex]2r[/tex].
So the force of gravity now is:
[tex]F_{2}=\frac{Gm_{1}(2m_{2})}{(2r)^2}\\ F_{2}=\frac{2Gm_{1}m_{2}}{4r^2} \\F_{2}=\frac{1}{2} \frac{Gm_{1}m_{2}}{r^2}[/tex]
and we know that [tex]F_{1}=\frac{Gm_{1}m_{2}}{r^2}[/tex]
so the new force of gravity is:
[tex]F_{2}=\frac{1}{2}F_{1}[/tex]
The force of gravity after you double the mass and the distance is half of the initial force.
