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PhysicsMediumClass 11

Kepler's Laws + Orbital Velocity

Gravitation

40

JEE Qs

10%

Hard

75

min

Master the derivations of orbital velocity and energy from Newton's Law of Gravitation and conservation laws, as this provides a robust understanding to tackle diverse problems.

🧮 Key Formulas

dA/dt = L / (2m) = constant (Kepler's Second Law)
T^2 = (4 * pi^2 / (G * M)) * a^3 (Kepler's Third Law, 'a' is semi-major axis or orbital radius for circular)
v_circular = sqrt(G * M / r)
v_elliptical = sqrt(G * M * (2/r - 1/a)) (Vis-viva equation)
U = -G * M * m / r
K_circular = (1/2) * (G * M * m / r)
E_circular = K + U = -G * M * m / (2r)
E_elliptical = -G * M * m / (2a)

✅ Key Points for JEE

  • 1Kepler's Second Law (equal areas in equal times) is a direct consequence of the conservation of angular momentum for a central force.
  • 2Kepler's Third Law (T^2 proportional to a^3) applies to all bodies orbiting the same central mass, with the constant of proportionality depending only on the central mass.
  • 3The total mechanical energy of a satellite in a bound (elliptical or circular) orbit is always negative and constant, determined by the semi-major axis 'a'.
  • 4For circular orbits, kinetic energy is half the magnitude of potential energy (K = -U/2), and total energy is half the potential energy (E = U/2 = -K).
  • 5Orbital velocity depends on the mass of the central body and the orbital radius/semi-major axis, not on the mass of the orbiting satellite.

⚠️ Common Mistakes

  • Confusing the mass of the central body (M) with the mass of the orbiting body (m) in formulas, or using the wrong mass in Kepler's Third Law.
  • Incorrectly applying Kepler's Third Law, especially for elliptical orbits where 'a' is the semi-major axis, not simply the instantaneous radius.
  • Misinterpreting the sign of total mechanical energy; a negative energy indicates a bound orbit, while zero or positive energy implies escape or an unbound trajectory.
  • Failing to recognize that Kepler's Second Law implies conservation of angular momentum, leading to errors in problems involving varying speeds in elliptical orbits.

📝 Practice Questions

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Q38.Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. Reason (R): The mass 2025 (22 Jan Shift 2) JEE Main Previous Year Paper of the pendulum remains unchanged at Earth and the other planet. In the light of the above statements, choose the correct answer from the options given below : (1) (A) is false but (R) is true (2) (A) is true but (R) is false (3) Both (A) and (R) are true and (R) is the correct (4) Both (A) and (R) are true but (R) is NOT the explanation of (A) correct explanation of (A)

2025·Assertion ReasoningMedium

Q42.A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that θ(t) = 5t2 −8t, where θ(t) is the angular position of the rotating disc as a function of time t. How much power is delivered by the applied torque, when t = 2 s ? (1) 72MR2 (2) 8MR2 (3) 108MR2 (4) 60MR2

2025·MCQMedium

Q37.Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be : (1) 2.8 (2) 11.2 (3) 5.6 (4) 8.4 → sin [ω (t −zc )] (S.I. Units). The

2025·MCQEasy

Q35.A small point of mass m is placed at a distance 2R from the centre ' O′ of a big uniform solid sphere of mass M and radius R . The gravitational force on ' m ' due to M is F1 . A spherical part of radius R/3 is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of M is found to be F2 . The value of ratio F1 : F2 is (1) 12 : 11 (2) 11 : 10 (3) 12 : 9 (4) 16 : 9

2025·MCQHard

Q41.If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon = 27 days and gravitational attraction between the satellite and the moon is neglected. (1) 27 days (2) 1 day (3) 81 days (4) 3 days

2025·NumericalMedium

Q49.A satellite of mass M is revolving around earth in a circular orbit at a height of R from earth surface. The 2 3 angular momentum of the satellite is . The value of x is ______ , where M and R are the mass and M√GMRx radius of earth, respectively. ( G is the gravitational constant)

2025·NumericalMedium

NCERT Chapters

  • Class 11 Physics Ch 8: Gravitation