Physics and Mathematics
Gravitational Potential Energy
1. Concept Overview
When two masses attract each other due to gravity, work is done either by or against the gravitational field.
This work is stored as Gravitational Potential Energy (GPE) in the system.
Definition:
The gravitational potential energy of a body at a point in a gravitational field is the work done in bringing it from infinity to that point.
Mathematically,
[U = -\dfrac{G M m}{r}]
The negative sign shows that the energy is bound energy — work must be done to separate the two bodies infinitely apart.
2. Explanation and Mathematical Derivation
Let a mass [M] be fixed, and a smaller body of mass [m] is brought from infinity to a distance [r] from [M].
Work done in moving the small mass through a small distance [dr] is:
[dW = Fdr \cos\theta]
Since the gravitational force is attractive, displacement is opposite to the force, hence [\theta = 180^\circ] and [\cos\theta = -1].
Therefore,
[dW = -Fdr = -G \dfrac{M m}{r^2} dr]
Total work done in bringing the mass from infinity to [r]:
$W = -\int_{\infty}^{r} G \dfrac{M m}{r^2} dr$ $= -G M m \left[-\dfrac{1}{r}\right]_{\infty}^{r}$
[
W = -\dfrac{G M m}{r}
]
This work done is stored as potential energy:
[
U = -\dfrac{G M m}{r}
]
3. Dimensions and Units
Dimensions:
[U] = [M^1 L^2 T^{-2}]
SI Unit: Joule (J)
CGS Unit: erg
4. Key Features
- Gravitational potential energy is negative, showing attraction.
- [U \to 0] as [r \to \infty].
- The potential energy depends on two masses and their separation.
- It is a scalar quantity.
- The work done to move the body between two points equals the change in potential energy.
- Near Earth’s surface, [U = mgh] is a special case of [U = -\dfrac{G M m}{r}].
- Potential energy is additive — for a system of masses, total energy is the sum of pairwise energies.
5. Important Formulas to Remember
| Quantity | Symbol | Formula | Unit |
|---|---|---|---|
| Gravitational Potential Energy | [U] | [U = -\dfrac{G M m}{r}] | [J] |
| Change in Potential Energy | [\Delta U] | [U_2 – U_1] [= G M m \left(\dfrac{1}{r_1} – \dfrac{1}{r_2}\right)] | [J] |
| Near Earth’s Surface | [U] | [U = m g h] | [J] |
| Relation with Potential | [V] | [U = m V] | [J] |
| Work-Energy Relation | [W] | [W = -\Delta U] | [J] |
6. Conceptual Questions with Solutions
1. What is gravitational potential energy?
It is the energy possessed by a body due to its position in a gravitational field, [U = -\dfrac{G M m}{r}].
2. Why is gravitational potential energy negative?
Because work must be done against the attractive gravitational force to separate the two masses.
3. What happens to GPE as two bodies move apart?
It increases (becomes less negative) and approaches zero at infinite separation.
4. What is the GPE at infinity?
At infinite separation, [U = 0].
5. How is GPE related to gravitational potential?
[U = mV], where [V = -\dfrac{G M}{r}].
6. Why do we use negative potential energy instead of positive?
It represents that the system is bound — energy is needed to overcome attraction.
7. How is GPE connected to mechanical energy?
Total energy = kinetic energy + potential energy. ([E = K + U])
8. When does gravitational potential energy increase?
When the body moves away from the mass (work done against gravity).
9. Why does U = mgh work near Earth’s surface?
Because for small heights, [g] is constant and [\Delta U = mgh] approximates [U = -\dfrac{G M m}{r}].
10. Can GPE ever be positive?
No, gravitational potential energy is always negative for real (attractive) systems.
11. How does GPE vary with distance?
It varies inversely with distance, [U \propto -\dfrac{1}{r}].
12. What is the reference level for zero potential energy?
Infinity is taken as the point where [U = 0].
13. If U is negative, does that mean energy is lost?
Not exactly — it means energy is required to separate the masses to infinity.
14. What does negative potential energy signify physically?
It signifies a bound system where gravitational attraction holds the masses together.
15. What is the GPE of an object on Earth’s surface compared to one at height h?
[GPE_{surface}] [= -\dfrac{G M_E m}{R_E}], [GPE_{height} = -\dfrac{G M_E m}{R_E + h}]
7. FAQ / Common Misconceptions
1. Gravitational potential energy can be positive.
❌ False. It is always negative because gravity is attractive.
2. GPE depends only on the mass of one body.
❌ It depends on both masses and their separation.
3. GPE and potential are the same.
❌ No. Potential energy = potential × mass ([U = mV]).
4. GPE is a vector quantity.
❌ It’s a scalar quantity.
5. GPE at the Earth’s surface is zero.
❌ It is negative; zero is at infinity.
6. The negative sign in GPE is meaningless.
❌ It shows that work must be done to overcome attraction.
7. The formula U = mgh applies everywhere.
❌ Only valid near Earth’s surface for small heights.
8. Increasing height always increases total energy.
❌ Only potential energy increases; total energy depends on kinetic changes too.
9. Potential energy is stored in the body alone.
❌ It’s stored in the system of two interacting masses.
10. GPE is constant everywhere around Earth.
❌ It varies with distance from Earth’s center.
8. Practice Questions (With Step-by-Step Solutions)
Q1. Calculate the gravitational potential energy of a 1 kg mass at Earth’s surface.
[ M_E = 6 \times 10^{24}kg]; [R_E = 6.4 \times 10^6m]; [G = 6.67 \times 10^{-11}Nm^2/kg^2 ]
[U = -\dfrac{G M_E m}{R_E}] [= -\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{6.4 \times 10^6}] [= -6.25 \times 10^7J]
Answer: [U = -6.25 \times 10^7J]
Q2. Find the change in potential energy of a 5 kg body raised through 100 m above Earth’s surface.
[\Delta U = m g h] [= 5 \times 9.8 \times 100 = 4900J]
Answer: [\Delta U = 4.9 \times 10^3J]
Q3. Two masses, 5 kg and 10 kg, are separated by 2 m. Find their potential energy.
[U = -G \dfrac{M m}{r}] [= -6.67 \times 10^{-11} \dfrac{5 \times 10}{2}] [= -1.67 \times 10^{-9}J]
Answer: [U = -1.67 \times 10^{-9}J]
Q4. A satellite of 200 kg is at a height of [6.4 \times 10^6,m] above Earth. Find its GPE relative to infinity.
[U = -\dfrac{G M_E m}{R_E + h}] [= -\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{6.4 \times 10^6 + 6.4 \times 10^6}]
[U = -6.25 \times 10^9J]
Answer: [U = -6.25 \times 10^9J]
Q5. Compare the GPE at the surface and at double the radius from Earth’s center.
[\dfrac{U_2}{U_1}] [= \dfrac{R_E}{2R_E}] [= \dfrac{1}{2}]
Answer: Potential energy becomes half (less negative) when distance is doubled.
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