Gravitational PE — U = -GMm/r
Gravitation
10
JEE Qs
8%
Hard
75
min
Master the physical interpretation of the negative sign in U = -GMm/r and its application in work-energy problems and conservation of mechanical energy.
🧮 Key Formulas
✅ Key Points for JEE
- 1The reference point for zero gravitational potential energy (U=0) is conventionally taken at infinity (r → ∞).
- 2Gravitational potential energy (U) is always negative for a system of two masses at a finite separation, implying an attractive force and a bound system.
- 3Gravitational potential energy is a scalar quantity, so for a system of multiple particles, the total PE is the algebraic sum of the potential energies of all unique pairs of particles.
- 4Work done by the conservative gravitational force (W_g) is equal to the negative of the change in gravitational potential energy (W_g = -ΔU = U_initial - U_final).
- 5The negative sign in U = -GMm/r fundamentally indicates that gravity is an attractive force.
⚠️ Common Mistakes
- ✕Forgetting the negative sign in U = -GMm/r or misinterpreting its physical significance.
- ✕Confusing gravitational potential energy (U) with gravitational potential (V) or gravitational force (F).
- ✕Incorrectly calculating the total potential energy for a system of multiple point masses by not summing the potential energies of *all unique pairs* of masses.
- ✕Not understanding the implications of the U=0 reference point at infinity when solving problems involving work or energy changes.
📝 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