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PhysicsMediumClass 11

Satellites — Geostationary, binding energy

Gravitation

10

JEE Qs

8%

Hard

75

min

Master the interrelationships between kinetic, potential, total, and binding energies as they are frequently tested in energy conservation problems related to orbital changes.

🧮 Key Formulas

Orbital velocity: v_o = sqrt(GM/r)
Time period of satellite: T = 2πr / v_o = 2π * sqrt(r^3 / GM)
Kinetic Energy (KE) of satellite: KE = (1/2)mv_o^2 = GMm / (2r)
Gravitational Potential Energy (PE) of satellite: PE = -GMm / r
Total Mechanical Energy (E_total) of satellite: E_total = KE + PE = -GMm / (2r)
Binding Energy (BE) of satellite: BE = -E_total = GMm / (2r)
Height of geostationary satellite: r_geo = (GMT^2 / (4π^2))^(1/3) where T = 24 hours, h_geo = r_geo - R_earth
For geostationary satellite: angular velocity ω_sat = ω_earth = 2π / T_earth

✅ Key Points for JEE

  • 1Geostationary satellites have a time period of 24 hours, orbit in the equatorial plane, and their angular velocity matches that of the Earth, making them appear stationary from Earth's surface.
  • 2Binding energy is the minimum energy required to remove a satellite from its orbit to infinity, meaning it's the negative of the satellite's total mechanical energy.
  • 3For a satellite in a stable circular orbit, the kinetic energy is positive, potential energy is negative, and the total mechanical energy is negative, with KE = |E_total| and PE = 2 * E_total.
  • 4Orbital velocity and time period depend only on the orbital radius and the mass of the central body (M), not on the mass of the satellite (m).
  • 5Any change in orbit requires a change in the satellite's total mechanical energy; understanding energy relationships is key to solving orbit transition problems.

⚠️ Common Mistakes

  • Confusing the orbital radius 'r' (R_earth + height) with the Earth's radius 'R_earth' in formulas.
  • Incorrectly calculating binding energy as just the absolute value of potential energy instead of the negative of total mechanical energy.
  • Forgetting that a geostationary satellite must orbit in the equatorial plane and in the same direction as Earth's rotation.
  • Errors in unit conversion, especially for time (hours to seconds) when calculating time period or geostationary height.

📝 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