Gravitation & Planetary Orbits
Practice JEE Main Gravitation chapter-wise PYQs online. Orbital velocity, escape speed v_e = √(2) v_o, Kepler’s 3rd Law, gravitational potential & solutions.
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in Gravitation & Planetary Orbits:
- Newton’s Universal Law of Gravitation & Acceleration due to gravity g
- Variation of g with Altitude g_h = g(1 - 2h/R) and Depth g_d = g(1 - d/R)
- Gravitational Potential & Potential Energy U = -G M m / r
- Orbital Velocity v_o = √(GM/R) and Escape Velocity v_e = √(2GM/R) = √(2) v_o
- Kepler’s Laws of Planetary Motion (T² ∝ r³)
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve Gravitation questions in under 45 seconds:
Escape speed to orbital speed ratio: v_e / v_o = √2 ≈ 1.414.
At depth d = R/2, acceleration due to gravity is g/2.
At height h = R, acceleration due to gravity drops to g/4.
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: 2 v_e
- Step 1: Formula for escape velocity: v_e = √(2GM / R).
- Step 2: For the new planet: M' = 8M and R' = 2R.
- Step 3: v_e' = √(2G(8M) / (2R)) = √(4 · (2GM/R)) = 2 · √(2GM / R).
- Step 4: Therefore, v_e' = 2 v_e.
⚡ Speed Tip: Scaling Rule: v_e ∝ √(M / R). √(8 / 2) = √4 = 2 ⇒ v_e' = 2 v_e.
💡 View Step-by-Step Solution & Answer
Correct Answer: 2R
- Step 1: Acceleration due to gravity at height h is g_h = g · [R / (R + h)]².
- Step 2: Given g_h = g / 9 ⇒ [R / (R + h)]² = 1/9.
- Step 3: Take the square root: R / (R + h) = 1/3.
- Step 4: Cross multiply: 3R = R + h ⇒ h = 2R.
⚡ Speed Tip: Inverse Square: g becomes 1/9 at distance 3R from center. Height above surface = 3R - R = 2R.
💡 View Step-by-Step Solution & Answer
Correct Answer: 129 days
- Step 1: By Kepler's Third Law: T² ∝ R³ ⇒ T₂ / T₁ = (R₂ / R₁)^(3/2).
- Step 2: Given R₂ = R₁ / 2 ⇒ R₂ / R₁ = 1/2.
- Step 3: T₂ = T₁ · (1/2)^(3/2) = 365 / (2√2) = 365 / 2.828.
- Step 4: T₂ ≈ 129 days.
⚡ Speed Tip: Kepler's Law: T₂ = 365 / 2^(1.5) = 365 / 2.828 ≈ 129 days.
❓ Frequently Asked Questions (FAQs)
What is the ratio of Escape Velocity to Orbital Velocity near Earth surface?
v_e = √(2) · v_o ≈ 1.414 · v_o (approximately 11.2 km/s vs 7.9 km/s).