I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
[tex]F = G\frac{Mm}{r^{2} }[/tex]
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = [tex]\frac{1}{4}[/tex] G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = [tex]\frac{1}{4}[/tex] F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
