Why this chapter is the best value in JEE mechanics
Compare Gravitation with Rotational Motion. Rotational questions are long, combine with energy conservation, and require an axis decision before you can start. Gravitation questions are typically one relation applied once - find g at a height, compute an escape velocity, apply Kepler's third law. The concepts do not stack on each other the way mechanics chapters usually do.
The practical consequence is that this chapter is worth securing completely rather than partially. Unlike Rotational Motion, where diminishing returns set in, a few focused hours here can reasonably take you from unreliable to consistent - which is why it is a sensible chapter to prioritise when time is short.
How JEE actually asks Gravitation
NTA does not publish chapter-wise weightage, so any figure circulating online is a coaching estimate derived from past papers. Confirm the syllabus and pattern for your session in the official NTA information bulletin.
What past papers show is a consistent split. Roughly half the questions are direct applications - variation of g with height or depth, escape and orbital velocity, Kepler's third law. The other half test the sign and meaning of gravitational potential energy: binding energy, the energy needed to move a satellite to a higher orbit, or why the total energy of a bound satellite must be negative. Students who prepare only the first half find the second half unfamiliar.
Key concepts, compressed
- ✓Newton's law of gravitation applies between point masses, and a uniform spherical shell behaves as though all its mass were at its centre for points outside it.
- ✓Inside a uniform spherical shell the gravitational field is zero everywhere, though the potential is not - it equals its surface value throughout.
- ✓Gravitational potential V = −GM/r is a scalar and adds arithmetically. Gravitational field is a vector and must be added component-wise.
- ✓Potential energy is measured against a reference of zero at infinity, which is why every bound configuration has negative U.
- ✓A body is bound when its total energy is negative and escapes when it is zero or positive. Setting total energy to zero is what defines escape velocity.
- ✓Kepler's laws: elliptical orbits with the Sun at a focus, equal areas in equal times, and T² proportional to the cube of the semi-major axis.
Formulas you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Gravitational force | F = Gm₁m₂/r² | G = 6.674 × 10⁻¹¹ N·m²/kg² |
| Surface gravity | g = GM/R² | |
| g at height h | g' = g/(1 + h/R)², ≈ g(1 − 2h/R) for h ≪ R | |
| g at depth d | g' = g(1 − d/R) | zero at the centre |
| Gravitational potential | V = −GM/r | scalar |
| Potential energy | U = −GMm/r | zero at infinity |
| Orbital velocity | v(o) = √(GM/r) | |
| Escape velocity | v(e) = √(2GM/R) = √(2gR) | = √2 × v(o) |
| Satellite total energy | E = −GMm/2r | negative because bound |
| Satellite kinetic energy | KE = GMm/2r = −E | |
| Kepler's third law | T² ∝ r³ | r is the semi-major axis |
The five mistakes that cost the most marks
- ✓Dropping the minus sign on U = −GMm/r. Every question about binding energy, escape, or moving a satellite between orbits depends on it, and a positive U reverses the conclusion.
- ✓Confusing escape velocity with orbital velocity. Escape is √2 times orbital at the same radius - about 11.2 km/s versus 7.9 km/s near the Earth's surface.
- ✓Using g = GM/R² below the surface. Inside the Earth g decreases with depth and is zero at the centre, so the formula for height cannot be reused with a negative h.
- ✓Believing astronauts are weightless because gravity is absent. Gravity in low orbit is close to its surface value; they float because they are in free fall along with the spacecraft.
- ✓Applying Kepler's third law with an altitude rather than an orbital radius. The r in T² ∝ r³ is measured from the centre of the body, not from its surface.
Practice set 1: Newton's law and the variation of g
1. Two 10 kg masses are placed 2 m apart. What is the gravitational force between them?
About 1.67 × 10⁻⁹ N. F = Gm₁m₂/r² = 6.674 × 10⁻¹¹ × 10 × 10 / 4 = 6.674 × 10⁻¹¹ × 25 ≈ 1.67 × 10⁻⁹ N. The point of the question is the magnitude: gravity between everyday objects is utterly negligible, which is why it only matters when at least one mass is astronomical.
2. If the distance between two masses is doubled, what happens to the gravitational force?
It falls to one quarter. The force goes as 1/r², so doubling r divides the force by four. Tripling the distance divides it by nine. This inverse-square scaling appears again in Electrostatics, where Coulomb's law has exactly the same form.
3. A planet has the same mass as the Earth but twice its radius. What is its surface gravity?
One quarter of the Earth's, so 2.5 m/s². From g = GM/R², doubling R while holding M constant divides g by four. Surface gravity is far more sensitive to radius than to mass, which is why a dense small body can have surprisingly strong surface gravity.
4. What is the value of g at a height equal to the Earth's radius above the surface?
2.5 m/s². At that height the distance from the centre is 2R, so g' = GM/(2R)² = GM/4R² = g/4 = 2.5 m/s². Note the approximation g' ≈ g(1 − 2h/R) must not be used here - it is valid only for h much smaller than R, and would wrongly give a negative value.
5. What is g at a depth equal to half the Earth's radius?
5 m/s². Below the surface, g' = g(1 − d/R) = 10 × (1 − 0.5) = 5 m/s². The depth relation is linear while the height relation is inverse-square, which is why the two must be kept separate. Only the mass within your radius contributes; the shell above you exerts no net field.
6. What is the value of g at the centre of the Earth?
Zero. Using g' = g(1 − d/R) with d = R gives zero, and physically there is no mass enclosed below you at the centre, so no net field. The mass surrounding you forms a shell, and the field inside a uniform shell is zero everywhere. Note that the potential at the centre is not zero - it is at its most negative there.
7. A planet has twice the Earth's mass and twice its radius. What is the weight of a 10 kg body on its surface?
50 N. g' = G(2M)/(2R)² = 2GM/4R² = g/2 = 5 m/s², so the weight is 10 × 5 = 50 N. The mass doubling raises g while the radius doubling lowers it by four, and the radius wins. Reasoning about which factor dominates is faster than substituting numbers.
Practice this now
Practice set 2: potential and potential energy
8. Why is gravitational potential energy negative?
Because the zero of potential energy is defined at infinite separation, and gravity is attractive. Bringing two masses from infinity to a finite separation releases energy, so the system ends up below zero: U = −GMm/r. The negative value means the configuration is bound - work must be supplied to separate the masses again.
9. How much work is needed to move a mass m from a distance r to infinity, in the field of mass M?
+GMm/r. The initial energy is −GMm/r and the final is zero, so the work required is the difference, GMm/r, and it is positive because you are working against attraction. This quantity is the binding energy of the configuration, and it is what escape velocity is derived from.
10. Is gravitational potential a scalar or a vector, and how do you combine contributions from several masses?
A scalar, so contributions add arithmetically with their signs, and no direction is involved. Gravitational field, by contrast, is a vector and must be added component-wise. This distinction matters at points where the fields cancel but the potentials do not - a standard question is to identify a point where the field is zero and note that the potential there is still non-zero.
11. What are the field and the potential at the centre of a uniform spherical shell of mass M and radius R?
The field is zero and the potential is −GM/R, the same as at the surface. The field vanishes because the contributions from all parts of the shell cancel by symmetry. The potential does not vanish, because potential is the work done bringing a unit mass in from infinity, and that work is done on the way to the shell. Zero field never implies zero potential.
12. Two equal masses are placed a distance d apart. Where is the gravitational field zero, and is the potential zero there?
At the midpoint the field is zero by symmetry, but the potential is not - it is −4GM/d, twice the contribution of each mass at distance d/2. Since potential is a scalar and both contributions are negative, they reinforce rather than cancel. This is the cleanest illustration of the previous question's point.
13. How much energy is needed to remove a 1 kg mass from the Earth's surface to infinity?
6.4 × 10⁷ J. The binding energy is GMm/R, and since g = GM/R² this equals gRm = 10 × 6.4 × 10⁶ × 1 = 6.4 × 10⁷ J. Rewriting GM as gR² is the standard trick in this chapter - it lets you work with g and R, which questions give you, rather than G and M, which they usually do not.
Practice set 3: escape and orbital velocity
14. Calculate the escape velocity from the Earth's surface.
About 11.3 km/s. v(e) = √(2gR) = √(2 × 10 × 6.4 × 10⁶) = √(1.28 × 10⁸) ≈ 11314 m/s. The standard quoted value is 11.2 km/s, and the small difference comes from using g = 10 rather than 9.8 m/s². Deriving it from √(2gR) rather than √(2GM/R) avoids needing G and the Earth's mass.
15. What is the orbital velocity for a satellite orbiting just above the Earth's surface?
8 km/s. v(o) = √(GM/R) = √(gR) = √(10 × 6.4 × 10⁶) = √(6.4 × 10⁷) = 8000 m/s. With g = 9.8 this gives the familiar 7.9 km/s. Note this is the maximum orbital speed for an Earth satellite - higher orbits are slower, not faster.
16. What is the relationship between escape velocity and orbital velocity at the same radius?
v(e) = √2 × v(o), so escape velocity is about 1.41 times the orbital velocity. This follows immediately from the formulas: √(2GM/r) divided by √(GM/r) is √2. A satellite in a circular orbit therefore needs only about 41% more speed to break free entirely.
17. Does escape velocity depend on the mass of the escaping body?
No. It comes from setting total energy to zero: ½mv² − GMm/r = 0, and the mass m cancels from both terms. Escape velocity depends only on the mass and radius of the body being escaped from. A pebble and a rocket need the same launch speed - the rocket simply needs far more energy to reach it.
18. Does escape velocity depend on the direction of projection?
No, provided there is no atmosphere and no obstruction. Escape is an energy condition, and kinetic energy ½mv² depends on speed alone, not direction. A body launched at 45° escapes just as surely as one launched vertically, given the same speed. In practice, launches are angled for orbital-insertion reasons, not escape ones.
19. A planet has twice the Earth's mass and twice its radius. How does its escape velocity compare?
It is the same. v(e) = √(2GM/R), and doubling both M and R leaves the ratio M/R unchanged, so the escape velocity is unchanged at about 11.2 km/s. Contrast with question 7, where the surface gravity halved - escape velocity and surface gravity scale differently, which is exactly what this pairing tests.
Practice this now
Practice set 4: satellites and Kepler's laws
20. Find the orbital speed of a satellite at a height equal to the Earth's radius.
About 5.66 km/s. The orbital radius is 2R, so v = √(GM/2R). Using GM = gR², this is √(gR²/2R) = √(gR/2) = √(10 × 6.4 × 10⁶/2) = √(3.2 × 10⁷) ≈ 5657 m/s. Compare with 8 km/s at the surface: doubling the orbital radius divides the speed by √2, so higher orbits are slower.
21. What is the orbital period of a geostationary satellite, and why?
24 hours - strictly, one sidereal day of about 23 h 56 min. A geostationary satellite must remain above the same point on the equator, so its period must match the Earth's rotation. It must also orbit in the equatorial plane and travel west to east; an orbit at the right altitude but inclined would not stay above one point.
22. A satellite's orbital radius is increased fourfold. By what factor does its period increase?
By a factor of 8. Kepler's third law gives T² ∝ r³, so T ∝ r^(3/2). With r multiplied by 4, T is multiplied by 4^(3/2) = 8. This is why distant planets have such long years - Neptune's orbital radius is about 30 times the Earth's, giving a period of roughly 30^(3/2) ≈ 164 years.
23. What is the total energy of a satellite in a circular orbit of radius r?
E = −GMm/2r, which is negative. The kinetic energy is +GMm/2r and the potential energy is −GMm/r, and their sum is −GMm/2r. The negative sign means the satellite is bound; making the total energy zero would require exactly doubling its kinetic energy, which corresponds to reaching escape velocity.
24. What is the relationship between a satellite's kinetic energy and its total energy?
KE = −E, so the kinetic energy equals the magnitude of the total energy. Also, U = 2E and KE = −U/2. These relations hold for any circular gravitational orbit and turn many energy questions into one substitution rather than a full calculation.
25. Why do astronauts float inside an orbiting space station?
Because they are in free fall together with the station, not because gravity is absent. At the altitude of a typical low orbit, gravity is around 90% of its surface value. Both astronaut and station accelerate toward the Earth at the same rate, so there is no contact force between them - the same condition as the free-falling lift in the Laws of Motion chapter.
26. According to Kepler's second law, where in its orbit does a planet move fastest?
At perihelion, its closest approach to the Sun. The second law says the line joining planet to Sun sweeps equal areas in equal times, so when the radius is short the planet must travel further along its path to sweep the same area. It is a statement of angular momentum conservation, since gravity exerts no torque about the Sun.
Practice set 5: mixed and conceptual
27. If the Earth stopped rotating, what would happen to the value of g at the equator?
It would increase. On a rotating Earth, part of the gravitational attraction at the equator is spent providing the centripetal acceleration needed to keep you moving in a circle, so the measured g is reduced. Remove the rotation and that requirement disappears. The effect is largest at the equator and zero at the poles.
28. Where on the Earth's surface is g greatest?
At the poles. Two effects work together: the Earth is slightly flattened, so the poles are nearer the centre, and there is no rotational reduction at the poles because the radius of the circular path there is zero. Both make polar g larger than equatorial g.
29. Why must a geostationary satellite orbit in the equatorial plane?
Because the centre of any orbit must coincide with the centre of the Earth. An inclined orbit would carry the satellite north and south of the equator over each period, so it would trace a figure-of-eight relative to the ground rather than staying above a fixed point. Only an equatorial orbit with a 24-hour period stays put.
30. A satellite is moved from a lower orbit to a higher one. What happens to its speed and its total energy?
Its speed decreases and its total energy increases. Speed goes as √(GM/r), so a larger r means slower motion. Total energy is −GMm/2r, which becomes less negative - that is, larger - as r grows. This apparent paradox is a favourite exam question: raising a satellite requires energy input even though it ends up moving more slowly.
Practice this now
How to study this chapter efficiently
- ✓Learn GM = gR² as a substitution and use it constantly. Questions give you g and R, not G and M, and this identity converts between them in one step.
- ✓Write the sign of every potential energy term before combining them. Most errors in this chapter are sign errors, not conceptual gaps.
- ✓Keep the height and depth relations separate in your notes. They have different functional forms - inverse-square above the surface, linear below - and merging them is a common failure.
- ✓Memorise the energy relations for circular orbits: KE = −E and U = 2E. They shortcut a whole class of satellite question.
- ✓Practise the paired comparison questions - a planet with twice the mass and twice the radius - because they test scaling rather than substitution and appear regularly.
- ✓This is a short chapter. Finish it completely rather than partially; the marginal return here is higher than in almost any other mechanics chapter.
Turn this into active practice
Because the formula set is small, the temptation is to read it once and assume the chapter is done. That reading feels productive and does not survive contact with a question that asks for the energy needed to raise a satellite rather than its orbital speed.
The JEE Gravitation quiz on QUFF generates fresh questions across variation of g, potential energy, escape and orbital velocity and Kepler's laws, marks them instantly and explains each answer. Because this chapter is short, a single focused session can realistically cover all of it - and any question you get wrong is almost certainly a sign error worth fixing on the spot.