JEE Main Physics Gravitation 2027: Satellites, Escape Velocity & Kepler's Laws
Variation of g, orbital and escape velocity, satellite energy and Kepler's laws, taught with worked examples and the traps that repeat every year.
Edurack
September 28, 2026

Why does an astronaut float even though gravity at the space station is still about 90% of what you feel on the ground? Gravitation answers questions like this with a small, tidy set of formulas. Learn them once and this chapter becomes some of the fastest marks in Physics.
Floating in orbit is not the absence of gravity. It is continuous free fall.
Chapter at a Glance
| Snapshot | Detail |
|---|---|
| NTA unit | Unit 6 of 20 — Gravitation |
| Priority (trend-based) | Moderate |
| Typical question style | Formula-based MCQs on g variation, satellites and energy |
| Best first step | Learn g with height and depth, then satellite energies |
Priority reflects past-paper trends, not an official NTA weightage.
What the NTA Syllabus Covers
- Universal law of gravitation
- Acceleration due to gravity and its variation with altitude and depth
- Kepler's laws of planetary motion
- Gravitational potential energy and potential, escape velocity
- Motion of a satellite: orbital velocity, time period and energy
Master These Topics
1. Variation of g: Height, Depth and the Two Formulas
At height h above Earth's surface: g' = g R² / (R + h)². For small h, this becomes g' ≈ g (1 − 2h/R).
Inside the Earth at depth d: g' = g (1 − d/R). Gravity falls linearly to zero at the centre.
Worked example: At height h = R, g' = g R² / (2R)² = g/4. At depth d = R/2, g' = g (1 − ½) = g/2.
Trap: Do not use the approximate formula 1 − 2h/R when h is comparable to R. Use the exact form.
2. Orbital and Escape Velocity
A satellite in a circular orbit of radius r has v_o = √(GM/r) = √(gR²/r). Close to the surface, v_o = √(gR).
Worked example: With g = 10 m/s² and R = 6.4 × 10⁶ m, v_o = √(10 × 6.4 × 10⁶) = 8000 m/s = 8 km/s.
Escape velocity is the minimum speed needed to leave the gravitational field: v_e = √(2GM/R) = √(2gR) ≈ 11.2 km/s. It is exactly √2 times the orbital speed near the surface, and it does not depend on the mass or direction of the projectile.
3. Satellite Energy
For a circular orbit of radius r around mass M:
- Kinetic energy:
KE = GMm / 2r - Potential energy:
PE = −GMm / r - Total energy:
E = −GMm / 2r
Total energy is negative for a bound satellite, and its magnitude equals the kinetic energy. The work needed to free the satellite completely is +GMm / 2r, its binding energy. Time period follows Kepler's third law: T² ∝ r³, precisely T = 2π √(r³ / GM).
4. Kepler's Laws in One Glance
- Orbits: planets move in ellipses with the Sun at one focus.
- Areas: the line joining planet and Sun sweeps equal areas in equal times (angular momentum conservation).
- Periods:
T² ∝ a³, where a is the semi-major axis.
Common Traps to Avoid
- Applying the small-height approximation for large heights.
- Thinking escape velocity depends on the mass of the object being launched.
- Assuming a satellite speeds up when it moves to a higher orbit. Orbital speed decreases as r increases.
- Mixing up total energy (negative) with binding energy (positive).
60-Second Revision Sheet
- Height:
g' = gR²/(R+h)²; depth:g' = g(1 − d/R) v_o = √(GM/r),v_e = √(2gR),v_e = √2 · v_o- Satellite:
KE = GMm/2r,PE = −GMm/r,E = −GMm/2r - Kepler:
T² ∝ r³, equal areas in equal times
Your Study Plan
- Day 1: g with altitude and depth, at least ten mixed problems.
- Day 2: orbital velocity, time period and geostationary reasoning.
- Day 3: energy of satellites and change of orbit.
- Day 4: timed mixed set including Kepler's-law ratios.
Practice Gravitation Questions Free → (opens in a new tab)
Continue Your Physics Journey
- Previous chapter: Rotational Motion
- Next chapter: Properties of Solids & Liquids
- All 20 JEE Main Physics chapters
- Complete JEE Main Syllabus 2027 guide
Frequently Asked Questions
What is the value of escape velocity from Earth?
About 11.2 km/s. It equals √(2gR), and is √2 times the orbital velocity close to Earth's surface (about 8 km/s).
Why is a satellite's total energy negative?
Because it is gravitationally bound. Its negative potential energy is twice as large in magnitude as its positive kinetic energy.