Analysis: Both the rod and the bead are in free fall under Earth’s gravity. Since the gravitational field is non-uniform ($g \propto 1/r^2$), parts of the system closer to Earth accelerate faster than parts further away. We need to find the relative acceleration between the bead and the rod to determine the separation time.

1. Acceleration of the Rod

The rod is an extended body of length $l$ and mass $m$. The total gravitational force on it is obtained by integrating the force elements from $r = r_0$ to $r = r_0 + l$.

$$ F_{rod} = \int_{r_0}^{r_0+l} \frac{GM (\lambda dr)}{r^2} = GM\lambda \left[ -\frac{1}{r} \right]_{r_0}^{r_0+l} = GM\lambda \left( \frac{1}{r_0} – \frac{1}{r_0+l} \right) $$

Since total mass $m = \lambda l$, the acceleration of the rod $a_{rod}$ (directed towards Earth) is:

$$ a_{rod} = \frac{F_{rod}}{m} = \frac{GM}{l} \left( \frac{l}{r_0(r_0+l)} \right) = \frac{GM}{r_0(r_0+l)} $$

Using the approximation $l \ll r_0$ (since $10\text{m} \ll 4\times10^8\text{m}$):

$$ a_{rod} = \frac{GM}{r_0^2 \left(1 + \frac{l}{r_0}\right)} \approx \frac{GM}{r_0^2} \left( 1 – \frac{l}{r_0} \right) $$

2. Acceleration of the Bead

The bead is a point mass located at $r = r_0 + x_0$. Its acceleration towards Earth is:

$$ a_{bead} = \frac{GM}{(r_0+x_0)^2} = \frac{GM}{r_0^2 \left(1 + \frac{x_0}{r_0}\right)^2} \approx \frac{GM}{r_0^2} \left( 1 – \frac{2x_0}{r_0} \right) $$

3. Relative Acceleration

Comparing the accelerations, since $x_0 \ll l$, the term $(1 – 2x_0/r_0)$ is larger than $(1 – l/r_0)$. Thus, $a_{bead} > a_{rod}$. The bead accelerates towards the Earth faster than the rod does. Relative to the rod, the bead moves towards the Earth (to the left in our diagram).

The magnitude of relative acceleration is:

$$ a_{rel} = a_{bead} – a_{rod} \approx \frac{GM}{r_0^2} \left[ \left( 1 – \frac{2x_0}{r_0} \right) – \left( 1 – \frac{l}{r_0} \right) \right] $$ $$ a_{rel} \approx \frac{GM}{r_0^2} \left( \frac{l – 2x_0}{r_0} \right) $$

Since $x_0$ (2 cm) is negligible compared to $l$ (10 m), we can approximate $l – 2x_0 \approx l$.

$$ a_{rel} \approx \frac{GM l}{r_0^3} $$

4. Time to Separate

The bead starts at distance $x_0$ from the left end and moves left relative to the rod. Therefore, it must cover a relative distance $S = x_0$ to fall off.

$$ S = \frac{1}{2} a_{rel} t^2 \implies x_0 = \frac{1}{2} \left( \frac{GM l}{r_0^3} \right) t^2 $$ $$ t = \sqrt{ \frac{2 x_0 r_0^3}{GM l} } $$

We know that surface gravity $g = \frac{GM}{R^2}$, so $GM = gR^2$. Substituting this:

Numerical Value:

Given: $x_0 = 0.02$ m, $r_0 = 4 \times 10^8$ m, $R = 6.4 \times 10^6$ m, $l = 10$ m, $g = 10$ m/s².

$$ t = \sqrt{ \frac{2(0.02)(64 \times 10^{24})}{10(40.96 \times 10^{12})(10)} } = \sqrt{ \frac{2.56 \times 10^{24}}{4096 \times 10^{12}} } \approx 2.5 \times 10^4 \text{ s} $$