1. Introduction to Gravitational Potential Energy

Gravitational potential energy (U) is the energy possessed by an object due to its position in a gravitational field. It arises because of the work done against the gravitational force to bring an object from a reference point (usually infinity or ground) to its current position.

2. Formula for Gravitational Potential Energy

The gravitational potential energy is given by:

U = -Gm1m2/r

Where:

G: Universal gravitational constant (6.674 x 10^-11 N m^2/kg^2)

m1, m2: Masses of two objects

r: Distance between the centers of the two masses

3. Explanation of the Negative Sign

The negative sign indicates that the gravitational force is attractive, and energy decreases as objects come closer. At infinity (r → ∞), U = 0. This is the reference point.

4. Gravitational Potential Energy near Earth's Surface

For small heights (h) above the Earth's surface:

U = mgh

Where:

m: Mass of the object

g: Acceleration due to gravity (9.8 m/s^2 near Earth's surface)

h: Height above the surface

Assumption: g is constant, which holds for small heights compared to Earth's radius.

Read Also: Class 11 Physics: Gravitational Constant

5. Derivation of U = -Gm1m2/r

Work is done to move an object of mass m2 from infinity to a distance r in a gravitational field of mass m1:

Work done, W = ∫ (F dr) = -∫ (Gm1m2/r^2 dr)

Solving the integral gives:

W = -Gm1m2/r

This work done becomes the gravitational potential energy.

6. Gravitational Potential

Gravitational potential (V) at a point is the gravitational potential energy per unit mass:

V = U / m2 = -Gm1/r

It is a scalar quantity.

7. Key Properties

1. Conservative Force: Gravitational force is conservative, meaning the work done depends only on initial and final positions, not the path taken.

2. Additivity: The potential energy due to multiple masses can be added algebraically.

  U_net = Σ U_i = Σ (-Gm1mi/r_i)

8. Applications

1. Escape Velocity: The minimum velocity required for an object to escape Earth’s gravity:

  v_escape = √(2GM / R)

2. Satellites: Gravitational potential energy plays a role in understanding orbits and total mechanical energy of satellites:

  E_total = K + U = -Gm1m2/2r (For a circular orbit)

9. Graphical Representation

A graph of U vs. r shows a hyperbolic curve where U → 0 as r → ∞ and U → -∞ as r → 0.

10. Problem-Solving Tips

1. Use U = mgh only for small heights where g is approximately constant.

2. For large distances, always use U = -Gm1m2/r.

3. Remember the total energy in an orbital motion is the sum of potential and kinetic energy.

11. Solved Example

Q. Calculate the gravitational potential energy between Earth and a satellite of mass 1000 kg orbiting at an altitude of 500 km.

Given:

- Mass of Earth (M) = 5.97 x 10^24 kg

- Radius of Earth (R) = 6.371 x 10^6 m

- G = 6.674 x 10^-11 N m^2/kg^2

Solution:

1. Distance from Earth's center: r = R + 500 km = 6.371 x 10^6 + 5 x 10^5 = 6.871 x 10^6 m

2. Potential energy:

  U = -GM m / r

  = - (6.674 x 10^-11 .  5.97 x 10^24 . 1000) / 6.871 x 10^6

  U ≈ -5.79 x 10^10 J

Conclusion:

Gravitational potential energy is a fundamental concept in physics that explains the interaction between masses due to gravity. It helps us understand phenomena ranging from simple objects falling near the Earth’s surface to the motion of celestial bodies in space. By using the derived formulas and understanding the principles, students can solve problems related to gravitational forces, orbits, and energy conservation in gravitational fields. This topic serves as a critical stepping stone to advanced concepts in mechanics and astrophysics, providing insights into the universal nature of gravitational interactions.