Orbital Velocity & Escape Velocity
Gravitation
10
JEE Qs
8%
Hard
75
min
Master the energy conservation approach for escape velocity and force balance for orbital velocity, paying close attention to the definition of 'r' in each case.
🧮 Key Formulas
✅ Key Points for JEE
- 1Orbital velocity is the precise horizontal velocity required for an object to maintain a stable circular orbit around a central body, where gravitational force provides the necessary centripetal force.
- 2Escape velocity is the minimum initial velocity an object needs to completely escape the gravitational pull of a planet and reach an infinite distance with zero kinetic energy.
- 3Both orbital velocity and escape velocity are independent of the mass of the orbiting/escaping body, depending only on the mass of the central body and the radial distance 'r' from its center.
- 4The total mechanical energy of a satellite in a stable circular orbit is negative, signifying it is gravitationally bound. For a body to escape, its total mechanical energy must become zero or positive.
- 5For geostationary satellites, the orbital radius is fixed (approx. 42,164 km from Earth's center), the orbital period is 24 hours, and the orbital plane is equatorial, matching Earth's rotation.
⚠️ Common Mistakes
- ✕Confusing 'r' (distance from the center of the central body) with 'R' (radius of the central body) or 'h' (height above the surface). Remember r = R + h.
- ✕Incorrectly using the sign convention for gravitational potential energy; it is always negative (U = -GMm/r) for bound systems.
- ✕Assuming escape velocity is always calculated from the surface (r=R), instead of from a given distance 'r' from the center of the planet.
- ✕Not understanding that the mass 'm' in kinetic energy and potential energy terms cancels out when deriving orbital and escape velocities.
📝 Practice Questions
See allQ38.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)
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
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
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
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
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)
NCERT Chapters
- Class 11 Physics Ch 8: Gravitation