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  3. Kepler's Laws of Planetary Motion

Gravitation · JEE & NEET Physics

Kepler's Laws of Planetary Motion: notes and previous year questions

Orbits are ellipses, equal areas in equal times, and T² ∝ a³: Kepler's three laws, why they follow from gravity, and how to weigh a planet.

Kepler's Laws of Planetary Motion in short

  • Kepler found three laws from Tycho's observations; Newton explained them with gravity.
  • First law: orbits are ellipses with the Sun at one focus; r₁ + r₂ = 2a.
  • Eccentricity says how stretched an orbit is: e = 0 is a circle.
  • Perihelion a(1 − e), aphelion a(1 + e).

1Kepler's puzzle

Tycho Brahe measured the positions of the planets very carefully for many years, without a telescope. Johannes Kepler tried to fit Mars onto a circle round the Sun; it missed by a tiny angle. Trusting the data, he found three laws (1609 and 1619). About seventy years later Newton showed that all three follow from his law of gravitation.

2The first law: orbits are ellipses

Every planet moves on an ellipse, with the Sun at one focus. The other focus is empty.

Draw an ellipse with two pins (the foci), a loop of string and a pencil: for every point, the distances to the two foci add up to the same length:

r1+r2=2ar_1 + r_2 = 2a
  • aa: the semi-major axis, half the longest width; bb: the semi-minor axis.
  • Eccentricity e=c/a=1−b2/a2e = c/a = \sqrt{1 - b^2/a^2}, where cc is the centre-to-focus distance.
  • e=0e = 0: a circle; 0<e<10 < e < 1: an ellipse (Earth: 0.017); e=1e = 1: a parabola; e>1e > 1: a hyperbola (open paths of some comets).
rp=a(1−e),ra=a(1+e)r_p = a(1 - e),\qquad r_a = a(1 + e)Perihelion (nearest) and aphelion (farthest); they add up to 2a.

Example: a=2×1011a = 2\times10^{11} m, e=0.5e = 0.5: rp=1×1011r_p = 1\times10^{11} m, ra=3×1011r_a = 3\times10^{11} m. The aphelion is three times the perihelion when 1+e=3(1−e)1 + e = 3(1 - e), i.e. e=0.5e = 0.5.

3The second law: equal areas in equal times

The line from the Sun to the planet sweeps out equal areas in equal times. Near the Sun the line is short, so the planet moves fast; far away it moves slowly.

dAdt=12r2ω=L2m\frac{dA}{dt} = \tfrac12 r^2\omega = \frac{L}{2m}In a time dt the line sweeps a thin triangle of area ½r²dθ.

Gravity points straight at the Sun, so it has no torque about the Sun: the angular momentum LL is constant, and so is dA/dtdA/dt. At the nearest and farthest points vv is square to rr, so:

vprp=varav_p r_p = v_a r_av_p / v_a = r_a / r_p = (1 + e)/(1 − e)

Example: 30 km/s at 1.5×10111.5\times10^{11} m gives 30×1.5÷2.5=1830 \times 1.5 \div 2.5 = 18 km/s at 2.5×10112.5\times10^{11} m. The Earth moves fastest in early January.

4The third law: T² ∝ a³

The square of the period is proportional to the cube of the semi-major axis. In years and astronomical units (AU, the Earth–Sun distance), T2=a3T^2 = a^3 for every planet:

Planeta (AU)T (years)T²/a³
Mercury0.3870.2411.00
Venus0.7230.6151.00
Earth111.00
Mars1.5241.8811.00
Jupiter5.20311.861.00

Why (circular orbit): gravity gives the centripetal force, GMm/r2=mv2/rGMm/r^2 = mv^2/r, so v=GM/rv = \sqrt{GM/r}. One lap: T=2πr/v=2πr3/GMT = 2\pi r/v = 2\pi\sqrt{r^3/GM}. Squaring:

T2=4π2GM a3T^2 = \frac{4\pi^2}{GM}\,a^3For an ellipse, use the semi-major axis a. The constant depends only on the central mass M.

5Weighing a planet

M=4π2a3GT2M = \frac{4\pi^2 a^3}{G T^2}

Jupiter's moon Io: r=4.22×108r = 4.22\times10^8 m, T=1.77T = 1.77 days ≈1.53×105\approx 1.53\times10^5 s. M=39.5×7.5×1025÷(6.67×10−11×2.34×1010)≈1.9×1027M = 39.5 \times 7.5\times10^{25} \div (6.67\times10^{-11} \times 2.34\times10^{10}) \approx 1.9\times10^{27} kg, over 300 Earths. For the same period at twice the radius, the central mass is 23=82^3 = 8 times bigger.

6Speeds of orbits

QuantityFormulaGoes as
Speedv=GM/rv = \sqrt{GM/r}1/r1/\sqrt r
PeriodT=2πr3/GMT = 2\pi\sqrt{r^3/GM}rrr\sqrt r
Angular speedω=2π/T\omega = 2\pi/T1/(rr)1/(r\sqrt r)

Periods 1 : 8 → radii 1 : 4, speeds 2 : 1, angular speeds 8 : 1. A satellite skimming the Earth takes 84 min; at a height of 3R3R (r=4Rr = 4R), 84×8=67284 \times 8 = 672 min.

Summary

Key ideas

  • Kepler found three laws from Tycho's observations; Newton explained them with gravity.
  • First law: orbits are ellipses with the Sun at one focus; r₁ + r₂ = 2a.
  • Eccentricity says how stretched an orbit is: e = 0 is a circle.
  • Perihelion a(1 − e), aphelion a(1 + e).
  • Second law: equal areas in equal times, because angular momentum is conserved.
  • A planet is fastest nearest the Sun: v_p r_p = v_a r_a.
  • Third law: T² ∝ a³, with T² = (4π²/GM)a³ and M the central mass.
  • An orbit's period and size give the mass of the central body.
  • Closer orbits are faster: v ∝ 1/√r, T ∝ r√r.

Every equation

Ellipse
r1+r2=2ar_1 + r_2 = 2a
Eccentricity
e=c/a=1−b2/a2e = c/a = \sqrt{1 - b^2/a^2}
Perihelion
rp=a(1−e)r_p = a(1 - e)
Aphelion
ra=a(1+e)r_a = a(1 + e)
Areal speed
dA/dt=L/2mdA/dt = L/2m
Ends of the orbit
vprp=varav_pr_p = v_ar_a
Third law
T2=(4π2/GM) a3T^2 = (4\pi^2/GM)\,a^3
Same central body
T12/a13=T22/a23T_1^2/a_1^3 = T_2^2/a_2^3
Central mass
M=4π2a3/(GT2)M = 4\pi^2a^3/(GT^2)
Orbital speed
v=GM/rv = \sqrt{GM/r}
Angular speed
ω∝r−3/2\omega \propto r^{-3/2}

Previous year questions with solutions

Real JEE and NEET questions on kepler's laws of planetary motion. Try each one before you open the solution.

Q1NEET 2026One correct option

In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius RR is proportional to:

  1. AR3R^{3}
  2. BR1/2R^{1/2}
  3. CR3/2R^{3/2}
  4. DR2R^{2}
Show answer and solution

Answer: Option C

From the last question, T=2πr3GMT = 2\pi\sqrt{\dfrac{r^{3}}{GM}}, with MM now the Sun's mass. Everything but the radius is the same for every planet of the system, so T∝R3=R3/2T \propto \sqrt{R^{3}} = R^{3/2}; squared, T2∝R3T^{2} \propto R^{3}.

The trap is A, R3R^{3}, which is how T2T^{2} goes, not TT. B, R1/2R^{1/2}, is how the orbital speed goes, turned upside down: vo∝R−1/2v_{o} \propto R^{-1/2}. D, R2R^{2}, comes from no correct working. Check: a planet four times as far out takes 43/2=84^{3/2} = 8 times as long.

Q2JEE Main 2026One correct option

A planet (P1P_{1}) is moving around the star of mass 2M2M in the orbit of radius RR. Another planet (P2P_{2}) is moving around another star of mass 4M4M in a orbit of radius 2R2R. Ratio of time periods of revolution of P2P_{2} and P1P_{1} is ________.

  1. A12\frac{1}{2}
  2. B22
  3. C44
  4. D14\frac{1}{4}
Show answer and solution

Answer: Option B

T∝r3MT \propto \sqrt{\dfrac{r^{3}}{M}}, so T2T1=(2R)34M×2MR3=8×24=4=2\dfrac{T_{2}}{T_{1}} = \sqrt{\dfrac{(2R)^{3}}{4M} \times \dfrac{2M}{R^{3}}} = \sqrt{\dfrac{8 \times 2}{4}} = \sqrt{4} = 2.

The trap is A, 12\dfrac{1}{2}, which is T1T2\dfrac{T_{1}}{T_{2}}, the ratio upside down. C, 44, forgets the square root: it is the ratio of the squares of the periods. D, 14\dfrac{1}{4}, makes both slips. Check in two steps: the orbit twice as wide alone multiplies the period by 222\sqrt{2}, and the star twice as heavy divides it by 2\sqrt{2}, leaving 22.

Q3JEE Advanced 2021Numerical answer

The distance between two stars of masses 3M_{S} and 6M_{S} is 9R. Here R is the mean distance between the centers of the Earth and the Sun, and M_{S} is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period nT, where T is the period of Earth's revolution around the Sun. The value of n is __________.

Show answer and solution

Answer: 9

For the Earth round the Sun, T2=4π2R3GMST^{2} = \dfrac{4\pi^{2}R^{3}}{GM_{S}} (the Earth's mass is tiny beside the Sun's). For the binary, the separation is 9R9R and the total mass 3MS+6MS=9MS3M_{S} + 6M_{S} = 9M_{S}:

(nT)2=4π2(9R)3G⋅9MS=7299⋅4π2R3GMS=81T2(nT)^{2} = \dfrac{4\pi^{2}(9R)^{3}}{G \cdot 9M_{S}} = \dfrac{729}{9} \cdot \dfrac{4\pi^{2}R^{3}}{GM_{S}} = 81T^{2}

so n=9n = 9.

The trap is using one star's mass instead of the sum: with 6MS6M_{S} alone, n2=7296n^{2} = \dfrac{729}{6} and n≈11n \approx 11; with 3MS3M_{S}, n≈15.6n \approx 15.6. Another is using the radius of one star's circle instead of the separation: the law has the full distance between the stars.

Practice questions, easy to hard

Three questions from the kepler's laws of planetary motion practice ladder: one easy, one medium, one hard.

Q4One correct option

You met the escape speed earlier: the least launch speed that carries a body away from a planet's surface for good is ve=2GMRv_{e} = \sqrt{\dfrac{2GM}{R}}. Put it beside the speed of a satellite skimming the same surface, vo=GMRv_{o} = \sqrt{\dfrac{GM}{R}}. The only difference is the 22 under the root, so

ve=2 vov_{e} = \sqrt{2}\, v_{o}

The Earth's two speeds fit: 11.2÷1.414≈7.9 km/s11.2 \div 1.414 \approx 7.9\ \mathrm{km/s}.

The escape speed from the surface of a certain planet is 20 km/s20\ \mathrm{km/s}. How fast does a satellite skimming its surface move?

  1. A10 km/s10\ \mathrm{km/s}
  2. B28.3 km/s28.3\ \mathrm{km/s}
  3. C20 km/s20\ \mathrm{km/s}
  4. D14.1 km/s14.1\ \mathrm{km/s}
Show answer and solution

Answer: Option D

vo=ve2=201.414≈14.1 km/sv_{o} = \dfrac{v_{e}}{\sqrt{2}} = \dfrac{20}{1.414} \approx 14.1\ \mathrm{km/s}.

The trap is B, 28.3 km/s28.3\ \mathrm{km/s}, which multiplies by 2\sqrt{2} instead of dividing: staying in orbit needs less speed than escaping, never more. A halves, as if ve=2vov_{e} = 2v_{o}, forgetting that the 22 sits under a square root. C takes orbiting and escaping to need the same speed; a satellite skimming the surface at vev_{e} would not stay in orbit, it would leave.

Q5One or more correct options

The energy of an elliptical orbit is the circular result with the radius replaced by the semi-major axis:

E=−GMm2aE = -\dfrac{GMm}{2a}

It stays the same all the way round, while kinetic and potential energy trade places: fast and deep near perigee, slow and high near apogee. The angular momentum also stays the same all the way round.

A satellite's orbit round a planet of mass MM has perigee distance r0r_{0} and apogee distance 3r03r_{0}. Which statements are correct?

  1. AIts total energy is −GMm4r0-\dfrac{GMm}{4r_{0}}
  2. BIt moves 33 times as fast at perigee as at apogee
  3. CIts kinetic energy is the same at perigee and apogee, since its total energy is constant
  4. DIts angular momentum is larger at perigee, where it moves faster
Show answer and solution

Answer: Options A, B

A: a=r0+3r02=2r0a = \dfrac{r_{0} + 3r_{0}}{2} = 2r_{0}, so E=−GMm2×2r0=−GMm4r0E = -\dfrac{GMm}{2 \times 2r_{0}} = -\dfrac{GMm}{4r_{0}}. B: vpr0=va⋅3r0v_{p}r_{0} = v_{a} \cdot 3r_{0}, so vp=3vav_{p} = 3v_{a}.

C is the trap: the total is constant, but it is shared differently. At perigee the potential energy is lower, so the kinetic energy is higher — nine times higher here, since the speed is three times. D: the speed is larger at perigee but the distance is smaller, and mvrmvr comes out the same; that is what constant angular momentum means.

Q6One or more correct options

The orbit rules work round any planet; only its mass and radius change — or, equivalently, its density ρ\rho and radius RR. With M=43πR3ρM = \dfrac{4}{3}\pi R^{3}\rho, a satellite skimming the surface has

vo=GMR=R43πGρv_{o} = \sqrt{\dfrac{GM}{R}} = R\sqrt{\dfrac{4}{3}\pi G\rho}

and you have seen that its period depends on ρ\rho alone.

A planet has the same density as the Earth but twice its radius. Which statements are correct?

  1. AA satellite skimming its surface moves twice as fast as one skimming the Earth
  2. BA satellite skimming its surface takes twice as long per orbit as one skimming the Earth
  3. CThe escape speed from its surface is twice the Earth's
  4. DIts mass is 88 times the Earth's
Show answer and solution

Answer: Options A, C, D

A: vo∝Rρv_{o} \propto R\sqrt{\rho}, so twice the radius at the same density doubles it. C: ve=2 vov_{e} = \sqrt{2}\,v_{o} at the surface, so it doubles too. D: M∝R3ρM \propto R^{3}\rho, and 23=82^{3} = 8.

B is the trap: the path round the big planet is twice as long, but the satellite goes twice as fast, so the period is unchanged — T=3πGρT = \sqrt{\dfrac{3\pi}{G\rho}} depends on the density alone.