Q39 (2026)
Given below are two statements :
Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.
Statement II: The time period of revolution of the satellite is $T = 2\pi \sqrt{\frac{R_{e}}{g}}$ (for satellite very close to the earth surface), where $R_{e}$ radius of earth and g acceleration due to gravity.
In the light of the above statements, choose the correct answer from the options given below :
- Both Statement I and Statement II are false
- Both Statement I and Statement II are true
- Statement I is true but Statement II is false
- Statement I is false but Statement II is true
### Related Formula
$$T = 2\pi \sqrt{\frac{R^3}{GM}}$$
$$M = \rho \cdot \frac{4}{3}\pi R^3$$
$$g = \frac{GM}{R^2}$$
### Core Logic
Evaluating Statement I:
Substituting total mass $M = \rho \cdot \frac{4}{3}\pi R^3$ into time period formula:
$$T = 2\pi \sqrt{\frac{R^3}{G \left(\rho \cdot \frac{4}{3}\pi R^3\right)}} = \sqrt{\frac{3\pi}{G \rho}}$$
Hence, time period depends inversely on the square root of Earth's density $\rho$. Statement I is true.
Evaluating Statement II:
Substituting $g = \frac{GM}{R_e^2}$ into $T = 2\pi \sqrt{\frac{R_e^3}{GM}}$:
$$T = 2\pi \sqrt{\frac{R_e}{g}}$$
Statement II is also true.
### Step 1: Final Conclusion
Both Statement I and Statement II are true.
### Pattern Recognition
Near-surface satellite shortcut: $T = \sqrt{3\pi / (G\rho)} \approx 84.6 \text{ min}$. Shows direct dependence on Earth's density $\rho$ and formula $T = 2\pi \sqrt{R_e / g}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Gravitation
