Three co-linear masses?

Question

One mass is 102 kg, one mass is 407 kg, and they are separated horizontally by 0.475 m. Midway between them is an object that is 21.7 kg.

Find the magnitude of the net gravitational force exerted by the larger objects on the 21.7 kg object in units of N. The universal gravitational constant is 6.672 x 10^-11 N m^2/kg^2.

Also, If the distance bettween the 102 kg and the 407 kg masses remains fixed, at what distance from the 407 kg mass (other than infinitely remote ones) does the 21.7 kg mass experience a net force of zero (in units of m).

Answer 1

F=F1-F2
F=qm(M1/R1^2-M2/R2^2)
Where
q- universal gravitational constant (6.672 x 10^-11 N m^2/kg^2)
M1- mass of 407kg
M2 – mass of 102kg
M - 21.7 kg
R1-distance between M1 and the object
R1-distance between M2 and the object

In the first case R1=R2=Rso we have

F=qm(M1 -M2)/R^2
F=6.672 x 10^-11 x 21.7(407 + 102kg) / (0.475)^2=3.266e-6 N



For the second question

F=F1-F2=0

So

M1/R1^2=M2/R2^2

R1/R2 = (M2/M1)^.5

And we know that R1+R2=0.475 m
So R2=0.475 –R1

Then we have

R1 = (M2/M1)^.5 (0.475 –R1)
R1=((M2/M1)^.5 x 0.475)/(1+(M2/M1)^.5)
R1=(1.997 x.475)/(1 + 1.997)= 0.317 m
And R2 must me then 0.475-.317=0.158 m