A large spherical mass M is fixed at one position and two identical masses m are kept on a line pass

Q: Why is the acceleration of both masses the same?

Because they are connected by a rigid rod, they must move as a single unit, meaning their displacement, velocity, and acceleration at any instant must be identical.

Q: Why does the gravitational force of $m_2$ on $m_1$ point to the right, while $m_1$ on $m_2$ points to the left?

Gravitational force is always attractive. Therefore, the farther mass $m_2$ pulls the closer mass $m_1$ towards itself (to the right), and $m_1$ pulls $m_2$ towards itself (to the left).

Q: What would happen if the tension in the rod was not zero?

If the tension were non-zero, we would have an additional force variable $T$ in our equations. The value of $m$ would then depend on the specific tension value.

Q: Why did we ignore the gravitational force between $M$ and the rod?

The problem states that the rod is massless, so it experiences no gravitational force and does not contribute to the equations of motion.

Q: Is this system in equilibrium?

No, the system is accelerating towards the massive sphere $M$. It is in a state of free fall, but the relative distance between the two masses remains constant due to the rigid rod.

Question

A large spherical mass $M$ is fixed at one position and two identical masses $m$ are kept on a line passing through the centre of $M$ (see figure). The point masses are connected by a rigid massless rod of length $l$ and this assembly is free to move along the line connecting them.

All three masses interact only through their mutual gravitational interaction. When the point mass nearer to $M$ is at a distance $r = 3l$ from $M$ the tension in the rod is zero for $m = k\left(\frac{M}{288}\right)$. The value of $k$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a massive, fixed sphere of mass $M$ anchored at the origin of our coordinate system.

Along a radial line extending outwards from this giant mass, we place two identical, smaller particles, each of mass $m$.

These two particles are connected by a rigid, massless rod of length $l$.

This entire rod-and-mass assembly is free to slide along the radial line, pulled inexorably by the gravity of the massive sphere $M$.

The closer mass, which we will call $m_1$, is located at a distance $r = 3l$ from the center of $M$.

Consequently, the farther mass, $m_2$, is located at a distance of $r + l = 4l$ from the center of $M$.

Our goal is to find the specific value of the mass $m$ (expressed in terms of $M$) such that the tension in the connecting rod is exactly zero.

### The Physics of Zero Tension

Because the rod is rigid, the two masses must move together as a single unit.

This means they must share a common acceleration, $\vec{a}$, directed towards the massive sphere $M$.

Normally, a connecting rod would exert a tension force $T$ to keep the two masses at a fixed distance $l$ if their natural gravitational accelerations differed.

However, we are given that the tension in the rod is **exactly zero**.

This means that the gravitational forces acting on each mass are perfectly balanced by nature to produce the exact same acceleration for both particles, without requiring any physical push or pull from the rod.

Let's write down the equations of motion for each mass individually.

### Setting Up the Equations of Motion

For the closer mass $m_1$ (at distance $3l$):

It experiences a leftward gravitational pull from the massive sphere $M$:

$$F_{M1} = \frac{GMm}{(3l)^2} = \frac{GMm}{9l^2}$$

It also experiences a rightward gravitational pull from its companion mass $m_2$:

$$F_{21} = \frac{Gm^2}{l^2}$$

Since the net acceleration $a$ is directed to the left (towards $M$), the equation of motion is:

$$\frac{GMm}{9l^2} - \frac{Gm^2}{l^2} = ma \quad 	ext{--- (Equation 1)}$$

Now, let's look at the farther mass $m_2$ (at distance $4l$):

It experiences a leftward gravitational pull from the massive sphere $M$:

$$F_{M2} = \frac{GMm}{(4l)^2} = \frac{GMm}{16l^2}$$

It also experiences a leftward gravitational pull from the closer mass $m_1$:

$$F_{12} = \frac{Gm^2}{l^2}$$

Since both forces act to the left, they add up to produce the acceleration $a$:

$$\frac{GMm}{16l^2} + \frac{Gm^2}{l^2} = ma \quad 	ext{--- (Equation 2)}$$

### Solving the System

Since both equations are equal to $ma$, we can set their left-hand sides equal to each other:

$$\frac{GMm}{9l^2} - \frac{Gm^2}{l^2} = \frac{GMm}{16l^2} + \frac{Gm^2}{l^2}$$

Notice how beautifully the universal gravitational constant $G$ cancels out from every single term:

$$\frac{Mm}{9l^2} - \frac{m^2}{l^2} = \frac{Mm}{16l^2} + \frac{m^2}{l^2}$$

Next, let's group the terms containing $M$ on the left side and the terms containing $m$ on the right side:

$$\frac{Mm}{9l^2} - \frac{Mm}{16l^2} = \frac{m^2}{l^2} + \frac{m^2}{l^2}$$

$$\frac{Mm}{l^2}\left(\frac{1}{9} - \frac{1}{16}ight) = \frac{2m^2}{l^2}$$

We can cancel out the common factor of $\frac{m}{l^2}$ from both sides:

$$M\left(\frac{16 - 9}{144}ight) = 2m$$

$$M\left(\frac{7}{144}ight) = 2m$$

Solving for $m$ gives:

$$m = \frac{7M}{288}$$

Comparing this with the given expression $m = k\left(\frac{M}{288}ight)$, we find:

$$\mathbf{k = 7}$$

This elegant result shows that when the mass of the small particles is exactly $\frac{7}{288}$ of the central mass, the system falls freely without any internal stress in the connecting rod!

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Identifying Gravitational Forces

Condition for Zero Tension

Equation of Motion for the Nearer Mass $m_{1}$

Equation of Motion for the Farther Mass $m_{2}$

Equating the Accelerations

Algebraic Simplification

Solving for $m$ and Finding $k$

The Way Forward

The Sigma Insight: Gravitational Force

Analyzing the Setup

The Physics of Zero Tension

Setting Up the Equations of Motion

Solving the System

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