Question

A long rod of mass $m$ and length $R$ is revolving around earth of radius $R$. Rod remains always in radial position such that one end is close to the surface of earth. If time period of rod is $2\pi\sqrt{\frac{x.R}{g}}$. Find $x$.

Answer 1

$\displaystyle \frac{27}{8}$

Answer 2

We note that the rod (length R, uniform mass m) always points along the radius. One end touches Earth’s surface (at r = R) and the other is at r = 2R so that the center‐of‐mass lies at

$$r_{\rm cm}=\frac{R+2R}{2}=\frac{3R}{2}\,.$$

For a free rigid body in a central gravitational field (with gravitational parameter GM) to rotate “synchronously” (i.e. keeping a fixed orientation relative to the radius) the orbital angular velocity ω must be such that

$$\omega^2=\frac{GM}{r^3_{\rm cm}}\,.$$

Since the usual acceleration due to gravity is defined by

$$g=\frac{GM}{R^2}\,,$$

we have

$$\omega^2=\frac{GM}{\left(\frac{3R}{2}\right)^3}=\frac{GM}{\frac{27}{8}R^3}=\frac{8\,GM}{27R^3}=\frac{8gR^2}{27R^3}=\frac{8g}{27R}\,.$$

Thus, the orbital time period is

$$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{1}{\omega^2}}=2\pi\sqrt{\frac{27R}{8g}}\,.$$

The given expression is

$$T=2\pi\sqrt{\frac{xR}{g}}\,.$$

Equate the two:

$$\sqrt{\frac{xR}{g}}=\sqrt{\frac{27R}{8g}}\quad\Longrightarrow\quad x=\frac{27}{8}\,.$$

Explanation (core steps):

1. The rod’s center-of-mass is at $\frac{3R}{2}$.

2. For synchronous rotation about Earth’s center, use $\omega^2 = \frac{GM}{r_{\rm cm}^3}$.

3. Replace $GM$ with $gR^2$ to get $\omega^2=\frac{8g}{27R}$.

4. Hence, $T = 2\pi\sqrt{\frac{27R}{8g}}$ which implies $x=\frac{27}{8}$.