How to find the equilibrium point between two masses? (Topic 6.2)

[Balti]

Homework Statement

Two masses of $9.0×10^{24}kg$ and $5.0×10^{24}kg$ respectively, are separated by $1.6×10^{11}$m. A third mass is placed between the other two masses, such that it remains stationary. How far from the $9.0×10^{24}kg$ mass is the third mass placed? Give your answer in giga-metres (Gm) but without a unit. Give your answer to two significant figures.

Relevant Equations

##g=\frac {F}{m}

Morning all

I've recently come across a problem where I get conceptually but cannot apply mathematically if that makes sense.

I understand the position of the third mass must be at the equilibrium point of $m_1$ ($9.0×10^{24}kg$), so $\Sigma F = 0$ right? And not even necessarily zero, something must equal zero for the mass to be at the neutral point or the Lagrange point.

The farthest I've come is having something like $\frac {9}{x^2} = \frac {5}{(1.6×10^11-x)^2}$ and even then I have no idea how I got to that.

Can someone please give me any tips? I don't necessarily want answers, but just hints or a clue.

 

[PeroK]

Homework Statement: Two masses of $9.0×10^{24}kg$ and $5.0×10^{24}kg$ respectively, are separated by $1.6×10^{11}$m. A third mass is placed between the other two masses, such that it remains stationary. How far from the $9.0×10^{24}kg$ mass is the third mass placed? Give your answer in giga-metres (Gm) but without a unit. Give your answer to two significant figures.
Homework Equations: ##g=\frac {F}{m}

Morning all

I've recently come across a problem where I get conceptually but cannot apply mathematically if that makes sense.

I understand the position of the third mass must be at the equilibrium point of $m_1$ ($9.0×10^{24}kg$), so $\Sigma F = 0$ right? And not even necessarily zero, something must equal zero for the mass to be at the neutral point or the Lagrange point.

The farthest I've come is having something like $\frac {9}{x^2} = \frac {5}{(1.6×10^11-x)^2}$ and even then I have no idea how I got to that.

Can someone please give me any tips? I don't necessarily want answers, but just hints or a clue.

Tip: forget the specific numbers; they just get in the way.

Hint: $g_1 = \frac{Gm_1}{x^2} = g_2 = \frac{Gm_2}{(r-x)^2}$

Where $r$ is the distance between masses $m_1$ and $m_2$.

Hint: can you turn that into a quadratic equation in $x$?

 

[RPinPA]

I understand the position of the third mass must be at the equilibrium point of $m_1$ ($9.0×10^{24}kg$), so $\Sigma F = 0$ right? And not even necessarily zero, something must equal zero for the mass to be at the neutral point or the Lagrange point.

What does equilibrium mean? You already answered that question. The total force is 0.

The farthest I've come is having something like $\frac {9}{x^2} = \frac {5}{(1.6×10^11-x)^2}$ and even then I have no idea how I got to that.

Presumably by saying the total force is 0, i.e., the magnitudes of the two forces acting in opposite directions are equal. Write the expressions for the force [magnitudes], set them equal and you get an equation like the above.

This equation can be rearranged into a quadratic in $x$ as @PeroK said. Personally I find it easier in these problems to solve for the ratio $f = r_1/r_2$ of the two distances $r_1$ and $r_2 = r - r_1$. Once you know $f$ you can use the fact that $r_1 = f r_2$ and $r_1 + r_2 = r$ to solve for the individual distances.

 

[kuruman]

What does equilibrium mean? You already answered that question. The total force is 0.

Although I understand the spirit of this problem, I don't like its unrealistic presentation at all. The word equilibrium has a loaded meaning. The middle mass is as much in equilibrium as a sphere balanced on top of another sphere: an infinitesimally small displacement from the point where the net force is zero, will result in an increasing acceleration as there is no restoring force (or torque). Even if we could accept that the middle mass "remains stationary" in non stable equilibrium, what about the masses on either side of it? We must also imagine and accept two gigantic disembodied hands (or equivalent), separated by $1.6\times 10^{11}~\mathrm{m}$, holding each mass in place against the gravitational attraction from the other two. It would have been much more relevant to the spirit of the problem to ask, "find the point in space where a third mass will experience zero force if placed there."