Question

A satellite is orbiting extremely close to the surface of a planet of average density $\rho$. The time period of revolution of the satellite depends only on $\rho$ as:

• A. Proportional to $\sqrt{\rho}$
• B. Inversely proportional to $\sqrt{\rho}$
• C. Proportional to $\rho$
• D. Inversely proportional to $\rho$ Correct Answer: (B) Inversely proportional to $\sqrt{\rho}$

Hint:

This reveals an amazing astronomical fact: the orbital period of a surface-skimming satellite is completely independent of the planet's size or radius! Whether it is a small rock or a giant world, if they share the exact same density $\rho$, a low-flying satellite will take the exact same amount of time to complete one full lap. Always remember the shortcut: $\mathbf{T \propto \frac{1}{\sqrt{\rho}}}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When a satellite moves in a stable circular orbit around a massive celestial planet, the required centripetal force keeping it in its circular track is provided entirely by the gravitational attraction pulling inward toward the planet's center. For a satellite orbiting extraordinarily close to the surface of the planet, we can make a practical geometric approximation: the radius of the satellite's circular orbit ($r$) is effectively equal to the physical radius of the planet itself ($R$). By linking orbital mechanics with mass-volume geometry, we can express the time period of the satellite purely as a function of the planet's intrinsic material density.

Step 2: Key Formula or Approach:
1. Orbital Velocity ($v_o$): Equating gravitational force to centripetal force for an orbital radius $r \approx R$ gives: $$ \frac{G M m}{R^2} = \frac{m v_o^2}{R} \implies v_o = \sqrt{\frac{G M}{R}} $$ 2. Time Period of Revolution ($T$): The time taken to complete one full circular lap of circumference $2\pi R$ is: $$ T = \frac{2\pi R}{v_o} = \frac{2\pi R}{\sqrt{\frac{GM}{R}}} = 2\pi \sqrt{\frac{R^3}{G M}} $$ 3. Mass-Density Relationship: Assuming the planet is a uniform solid sphere, its total mass ($M$) can be written in terms of its average density ($\rho$) and volume ($V = \frac{4}{3}\pi R^3$): $$ M = \text{Volume} \times \text{Density} = \frac{4}{3}\pi R^3 \rho $$

Step 3: Detailed Explanation:
Let us substitute the mass-density expression into our time period equation to eliminate the explicit mass variable: $$ T = 2\pi \sqrt{\frac{R^3}{G \left(\frac{4}{3}\pi R^3 \rho\right)}} $$ Now, simplify the algebraic terms hidden under the square root radical: 1. Notice that the radius term $R^3$ appears in both the numerator and denominator, allowing them to cancel out completely: $$ T = 2\pi \sqrt{\frac{1}{\frac{4}{3}\pi G \rho}} $$ 2. Rearrange the fractions under the radical sign to clean up the expression: $$ T = 2\pi \sqrt{\frac{3}{4\pi G \rho}} = \sqrt{\frac{4\pi^2 \times 3}{4\pi G \rho}} = \sqrt{\frac{3\pi}{G \rho}} $$ Isolating the variables to see the proportionalities gives: $$ T = \left(\text{Constant}\right) \times \frac{1}{\sqrt{\rho}} \implies T \propto \frac{1}{\sqrt{\rho}} $$ This mathematical derivation proves that the time period of a low-altitude orbital satellite is inversely proportional to the square root of the planet's average density ($\sqrt{\rho}$). This matches option (B).

Step 4: Final Answer:
The time period of revolution of the satellite is inversely proportional to $\sqrt{\rho}$.