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 masses is zero

Answer 1

the gravitational force between M and the third mass must equal the force between the third mass and 3M. First we need to set up some facts:

M = the first mass, and the third small mass
3M = the larger mass

*Note that the masses are largely arbitrary as they will cancel out, the important thing to realize is that 3M is 3 times M.

x = the distance between first M and third M.

if x is the distance between the two small masses then:

1.0m-x is the distance between the third small mass, M and 3M.

Using our let statements we can set up our force of gravity equations:

for the gravitational force between M and M:

G(M)(M)/x^2

for the gravitational force between M and 3M:

G(M)(3M)/(1.0-x)^2

As we said these forces have to be equal so:

G(M)(M)/x^2 = G(M)(3M)/(1.0-x)^2

This might look intimidating, but G and all the M's will divide out if you are doing your alegabra correctly leaving you with:

-2x^2-2x+1 = 0

which you use the quadratic equation to solve, producing one positive and one negative value. Obviously, the negative distance is invalid leaving the positive distance of:

approximately .367m or 36.7cm