Gravitation is the chapter that connects a falling apple to an orbiting planet using one single law — and once you see that a satellite in orbit is really just continuously falling and continuously missing the Earth, half the chapter's formulae stop needing separate memorisation.
Newton's law of gravitation gives the force between any two masses. Everything else in this chapter — orbital velocity, escape velocity, time period of a satellite, energy of an orbiting body — is that same one force, set equal to whatever kind of motion or energy balance the situation demands. Kepler's three laws, discovered empirically before Newton's law existed, are shown in this chapter to be direct consequences of it, which is itself one of the most satisfying derivations in the entire syllabus.
Core gravitation formulae
| Quantity | Formula |
|---|---|
| Newton's law of gravitation | F = Gm₁m₂/r² |
| Acceleration due to gravity at Earth's surface | g = GM/R² |
| Variation of g with height h | g' = g(1 − 2h/R), h ≪ R |
| Variation of g with depth d | g' = g(1 − d/R) |
| Orbital velocity | v₀ = √(GM/r) |
| Escape velocity | ve = √(2GM/R) |
| Relation between escape and orbital velocity | ve = √2 · v₀ |
| Time period of a satellite | T = 2π√(r³/GM) |
| Gravitational potential energy | U = −GMm/r |
Topics covered
- Kepler's three laws — the law of orbits, law of areas, and law of periods, and how each follows directly from Newton's law
- Newton's universal law of gravitation — the inverse-square force law and the meaning of the gravitational constant G
- Acceleration due to gravity — its derivation from the gravitation law, and why it's called an acceleration rather than a force
- Variation of g with height, depth and latitude — three separate reasons g is not truly constant across the Earth's surface
- Gravitational potential energy — why it's negative, and what that negative sign actually means physically
- Escape velocity — the minimum speed needed to leave a gravitational field permanently, derived from energy conservation
- Orbital velocity and time period — the speed and period required to maintain a stable circular orbit at a given radius
- Geostationary and polar satellites — the specific orbital conditions that make each type useful for different applications
Why this chapter matters
In NEET and JEE, escape velocity and orbital velocity numericals are asked almost every year, and the g-variation-with-height-or-depth formulae are a reliable source of quick, formula-based marks. Questions connecting Kepler's laws to Newton's gravitation law test conceptual depth rather than calculation, and are a favourite for distinguishing well-prepared students.
For Class 11 boards, expect derivations of escape velocity or orbital velocity from first principles, and a numerical on g variation with height or depth.
Mistakes students make in this chapter
- Confusing gravitational potential energy's negative sign with a mistake. The negative sign is correct and meaningful — it reflects that gravitational potential energy is taken as zero at infinite separation, and any finite separation is therefore at lower (negative) energy.
- Mixing up the height and depth formulae for g. g decreases going up and decreases going down too, but by different rates and for different physical reasons — the two formulae are not interchangeable.
- Forgetting escape velocity doesn't depend on the mass of the escaping object. It depends only on the mass and radius of the body being escaped from, a fact many students overlook when solving numericals.
- Treating a satellite's orbit as requiring continuous propulsion. Once in stable orbit, gravity alone provides exactly the centripetal force needed — no engine thrust is required to maintain it.
- Misapplying Kepler's second law. It states equal areas are swept in equal times, which means a planet moves faster when closer to the Sun, not that its speed is constant throughout the orbit.
Worked example: orbital speed, escape speed and the link between them
- Derive the orbital velocity. For a circular orbit, gravity provides exactly the centripetal force needed:GMm/R² = mv²/R → v = √(GM/R)since GM = gR², v₀ = √(gR)
- Substitute the numbers. Convert the radius to metres first:v₀ = √(9.8 × 6.4×10⁶) = √(6.27×10⁷)v₀ ≈ 7.9 × 10³ m/s = 7.9 km/s
- Derive the escape velocity. A body escapes when its kinetic energy equals the magnitude of its gravitational potential energy:½mv² = GMm/R → v_e = √(2GM/R) = √(2gR)
- Substitute again — or simply scale the previous answer, since only a factor of 2 has appeared under the root:v_e = √2 × v₀ = 1.414 × 7.9v_e ≈ 11.2 km/s
- State the relation, which holds for any planet and is worth memorising in this form:v_e = √2 · v₀ (about 41% faster)
- Find the time period. The satellite covers one circumference at constant speed:T = 2πR/v₀ = 2π(6.4×10⁶)/7900T ≈ 5080 s ≈ 84.6 minutes
What the notes contain
The PDF derives Kepler's laws from Newton's gravitation law, works through the full escape velocity derivation using energy conservation, and includes solved numericals on satellite orbits, geostationary satellite altitude calculation, and g variation with height and depth side by side for direct comparison.
How to use this: build the concept from these notes first, then do one quick NCERT pass before the exam for exact wording and diagram labels, and spend whatever time is left on previous year questions.
Questions
Why is gravitational potential energy negative?
Because the zero reference point is chosen at infinite separation, where the gravitational force (and hence potential energy) vanishes. Any finite separation means the masses have lower potential energy than at infinity, which shows up as a negative value — the PDF explains this with a worked comparison.
What's the relationship between escape velocity and orbital velocity?
Escape velocity is always √2 (about 1.414) times the orbital velocity at the same radius. Both formulae are derived from first principles in the PDF so you can see exactly where this factor comes from.
Does escape velocity depend on the direction of launch?
No — escape velocity is a scalar minimum speed requirement and, ignoring atmospheric effects, doesn't depend on launch direction, only on the masses and radius involved.
Is this enough to derive Kepler's laws from Newton's law of gravitation in an exam?
Yes, the full derivation connecting the two is worked through step by step, which is exactly the kind of derivation-based question boards and JEE both favour.
All other chapters
Every chapter of Class 11 and Class 12 Physics, Chemistry and Biology — each with its own notes and PDF.