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PhysicsMediumClass 11

Kepler's Laws — T²∝r³, elliptical orbits

Gravitation

10

JEE Qs

8%

Hard

60

min

Understand the conditions for applying each law, especially how the constant in T²∝a³ depends on the central mass and how 'a' relates to elliptical orbit parameters.

🧮 Key Formulas

dA/dt = L / (2m) = constant (Kepler's Second Law, where L is angular momentum)
T^2 = (4π^2 / (GM)) * a^3 (Kepler's Third Law, where 'a' is semi-major axis and M is mass of central body)
For circular orbits, T^2 = (4π^2 / (GM)) * R^3 (where R is orbital radius)
Semi-major axis 'a' = (r_perihelion + r_aphelion) / 2
Total Mechanical Energy E = -GMm / (2a) (for elliptical orbits)

✅ Key Points for JEE

  • 1Kepler's First Law states that all planets move in elliptical orbits with the Sun at one of the foci. This implies that the distance from the central body is not constant.
  • 2Kepler's Second Law (Law of Areas) states that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time. This is a direct consequence of the conservation of angular momentum about the central body.
  • 3Kepler's Third Law (Law of Periods) states that the square of the orbital period (T) of any planet is proportional to the cube of the semi-major axis (a) of its elliptical orbit (T² ∝ a³). The constant of proportionality (4π²/GM) depends only on the mass (M) of the central body.
  • 4Planets move faster when closer to the Sun (at perihelion) and slower when farther away (at aphelion) to maintain constant areal velocity.
  • 5For circular orbits, the semi-major axis 'a' is simply the radius 'R' of the circle.

⚠️ Common Mistakes

  • ✕Confusing the semi-major axis 'a' in Kepler's Third Law with the instantaneous radial distance for an elliptical orbit.
  • ✕Applying the constant (4π²/GM) derived for a specific central body (e.g., Sun) to systems with a different central body (e.g., Earth) without changing M.
  • ✕Assuming linear velocity is constant throughout an elliptical orbit, instead of understanding that it's the areal velocity that is constant.
  • ✕Incorrectly calculating the semi-major axis 'a' when given perihelion and aphelion distances.

📝 Practice Questions

See all

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