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)

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