Gravitational Potential

1. Concept Overview

Every mass exerts a gravitational influence around it. When another body is brought into this region, work is done against or by the gravitational field.
This work done per unit mass is called gravitational potential.

Definition:
The gravitational potential at a point in a gravitational field is the work done per unit mass in bringing a body from infinity to that point, without acceleration.

Mathematically,

[V = -\dfrac{W}{m}]
or,
[V = -\dfrac{G M}{r}]

The negative sign indicates that gravitational potential is always negative because the force of gravity is attractive, and work is done against the field to bring the mass from infinity.

2. Explanation and Mathematical Derivation

Let a point mass [M] produce a gravitational field around it.
We bring a small test mass [m] from infinity to a point at distance [r] from [M].

Work done in moving the test mass through a small distance [dr] is:
[dW = Fdr \cos\theta]

Since the force is attractive and displacement is opposite to force direction, [\theta = 180^\circ],
so [\cos\theta = -1].

Now, [F = G \dfrac{M m}{r^2}]

Thus,

[dW = -G \dfrac{M m}{r^2} dr]

Total work done from infinity to distance [r] is:

[W = \int_{\infty}^{r} dW] [= -G M m \int_{\infty}^{r} \dfrac{dr}{r^2}]

$W = -G M m \left[-\dfrac{1}{r}\right]_{\infty}^{r}$ $= -G M m \left(\dfrac{1}{r} – 0\right)$

[
W = -\dfrac{G M m}{r}
]

Hence, gravitational potential at distance r is:

[V = \dfrac{W}{m} = -\dfrac{G M}{r}]

3. Dimensions and Units

Dimensions:
[V] = [M^0 L^2 T^{-2}]

SI Unit: [J kg^{-1}] or [m^2 s^{-2}]
CGS Unit: [erg g^{-1}]

4. Key Features

1. Gravitational potential is a scalar quantity (has magnitude only).
2. It is negative everywhere (since gravity is attractive).
3. At infinity, V = 0.
4. It decreases (becomes more negative) as we move closer to the mass.
5. The potential difference between two points gives the work done per unit mass to move between them.
6. The gravitational field intensity is the negative gradient of potential:
   [
   E = -\dfrac{dV}{dr}
   ]
7. The potential obeys the principle of superposition.

5. Important Formulas to Remember

6. Conceptual Questions with Solutions

7. FAQ / Common Misconceptions

8. Practice Questions (With Step-by-Step Solutions)

Q1. Find the gravitational potential at a point 10,000 km from the center of Earth.
[ M_E = 6 \times 10^{24}kg,] [G = 6.67 \times 10^{-11}Nm^2/kg^2 ]

Solution:
[V = -\dfrac{G M_E}{r}] [= -\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{10^7}] [= -4.002 \times 10^7J/kg]
Answer: [V = -4.0 \times 10^7J/kg]

Q2. Calculate the potential at the surface of Earth.
[ M_E = 6 \times 10^{24}kg], [R_E = 6.4 \times 10^6m ]

[V = -\dfrac{G M_E}{R_E}] [= -\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 10^6}] [= -6.25 \times 10^7J/kg]
Answer: [V = -6.25 \times 10^7,J/kg]

Q3. What is the potential difference between two points 10,000 km and 20,000 km from Earth’s center?

[\Delta V = V_2 – V_1] [= -G M_E \left(\dfrac{1}{r_2} – \dfrac{1}{r_1}\right)]
Substitute:
[\Delta V = -6.67 \times 10^{-11}] [\times 6 \times 10^{24} \left(\dfrac{1}{2 \times 10^7} – \dfrac{1}{10^7}\right)]
[
\Delta V = 2.0 \times 10^7J/kg
]
Answer: [\Delta V = 2.0 \times 10^7J/kg]

Q4. A satellite is moved from Earth’s surface to a height where potential becomes half. Find the height.

Let [V_1 = -\dfrac{G M}{R}], [V_2 = \dfrac{V_1}{2} = -\dfrac{G M}{2R’}]

[\dfrac{V_1}{V_2} = \dfrac{R’}{R}] [= 2 \Rightarrow R’ = 2R]
Thus, height [h = R’ – R = R]
Answer: Height = one Earth radius = [6.4 \times 10^6m]

Q5. What is the work done in moving a 2 kg body from infinity to Earth’s surface?

[W = mV] [= 2 \times (-6.25 \times 10^7)] [= -1.25 \times 10^8J]
Answer: [W = -1.25 \times 10^8J]