do it this way
M(x)=3M(100-x)
x=300-3x
4x=300
x=75cm
or 0.75m
A small mass M and a small mass 3M are 1.00 m apart. Where should you put a third small mass so that the net gravitational force on it due to the other two masses is zero?
2 answers
Gravitational force
=GMm/r²
So for the mass m between masses M and 3M, one metre apart, equate the gravitational forces:
GMm/x² = G(3M)m/(1-x)²
Solve for x.
[(√3-1)/2]
=GMm/r²
So for the mass m between masses M and 3M, one metre apart, equate the gravitational forces:
GMm/x² = G(3M)m/(1-x)²
Solve for x.
[(√3-1)/2]