Gravitation – IIT/JEE Solved Problems
Phase 1 • Page 3 • Escape Velocity & Energy Concepts
Problem 11: Escape Velocity from Earth
Q: Derive the expression for escape velocity of a body from Earth.
Solution:
Escape velocity is obtained using conservation of energy.
- Kinetic energy at surface = ½mv²
- Gravitational potential energy at surface = −GMm/R
- At infinity, total energy = 0
½mv² − GMm/R = 0
v = √(2GM/R)
Core IIT Idea: Escape velocity comes from energy balance, not force equations.
Problem 12: Escape Velocity on Moon
Q: If escape velocity on Earth is 11.2 km/s, find escape velocity on Moon. (gmoon = g/6, Rmoon ≈ R/4)
Solution:
ve ∝ √(gR)
vmoon/vearth = √[(g/6 × R/4)/(g × R)]
vmoon = vearth / √24
Shortcut: Never calculate GM again—use proportionality.
Problem 13: Energy of Satellite
Q: Find total mechanical energy of a satellite in circular orbit of radius r.
Solution:
Kinetic energy = GMm / (2r)
Potential energy = −GMm / r
Total energy = −GMm / (2r)
IIT Favourite: Negative total energy ⇒ satellite is bound to Earth.
Problem 14: Effect of Height on Escape Velocity
Q: Does escape velocity depend on height from Earth’s surface?
Answer:
Escape velocity depends on distance from centre of Earth. At a height h, escape velocity becomes:
v = √(2GM / (R + h))
Important: Escape velocity decreases with increase in height.
Problem 15: Satellite Falling Towards Earth
Q: Why does a satellite not fall straight down to Earth?
Answer:
Satellite has sufficient horizontal velocity. While it falls towards Earth, Earth curves away at the same rate. Thus, satellite remains in orbit.
Concept Insight: Orbit is continuous free fall.
Phase-1 Master Formula Set
- g = GM/R²
- v = √(GM/r)
- T = 2π √(r³/GM)
- ve = √(2GM/R)
- E = −GMm / (2r)
Phase 1 • Page 3 Completed Successfully ✅
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