Physics and Mathematics
Binding Energy of a Satellite
1. Concept Overview
The binding energy of a satellite is the minimum energy required to remove the satellite completely from the gravitational field of the Earth — that is, to take it to infinity where gravitational influence becomes zero.
In simple terms, it is the energy needed to overcome the Earth’s gravitational pull on the satellite.
Binding energy is a measure of how tightly the satellite is bound to Earth.
2. Explanation and Derivation
When a satellite of mass ([m]) revolves around the Earth of mass ([M]) in an orbit of radius ([r]), its total energy (E) is the sum of kinetic energy (K) and potential energy (U):
[
E = K + U
]
(i) Kinetic Energy
[
K = \dfrac{1}{2} m v^2
]
From orbital velocity, ([v^2 = \dfrac{G M}{r}]), we get:
[K] [= \dfrac{1}{2} m \dfrac{G M}{r}] [= \dfrac{G M m}{2r}]
(ii) Gravitational Potential Energy
[
U = – \dfrac{G M m}{r}
]
(iii) Total Energy
[E = K + U] [= \dfrac{G M m}{2r} – \dfrac{G M m}{r}] [= -\dfrac{G M m}{2r}]
Thus, the total energy of a satellite is negative, indicating a bound system.
Now, to remove the satellite to infinity (where energy becomes zero), we must supply energy equal in magnitude to its total energy:
[\text{Binding Energy}] [= |E|] [= \dfrac{G M m}{2r}]
Hence,
[\boxed{\text{Binding Energy (B)} = \dfrac{G M m}{2r}}]
3. Dimensions and Units
| Quantity | Symbol | Dimensions | SI Unit |
|---|---|---|---|
| Binding Energy | [B] | [M L² T⁻²] | [Joule (J)] |
| Gravitational constant | [G] | [M⁻¹ L³ T⁻²] | [N·m²·kg⁻²] |
| Mass of Earth | [M] | [M] | [kg] |
| Mass of Satellite | [m] | [M] | [kg] |
| Orbital radius | [r] | [L] | [m] |
4. Key Features
- Binding energy is always positive in value (though total energy is negative).
- It represents the stability of the satellite in orbit.
- A satellite in a lower orbit has greater binding energy, i.e., it is more tightly bound.
- To remove the satellite from orbit, an external agency (like rocket thrust) must supply at least this much energy.
- Binding energy per unit mass is given by ([\dfrac{G M}{2r}]).
5. Important Formulas to Remember
| Formula | Description |
|---|---|
| [E = K + U] | Total energy of the satellite |
| [K = \dfrac{G M m}{2r}] | Kinetic energy of the satellite |
| [U = -\dfrac{G M m}{r}] | Gravitational potential energy |
| [E = -\dfrac{G M m}{2r}] | Total energy (negative for bound system) |
| [B = \dfrac{G M m}{2r}] | Binding energy of the satellite |
6. Conceptual Questions with Solutions
1. What is meant by the binding energy of a satellite?
It is the minimum energy required to remove the satellite from Earth’s gravitational field to infinity.
2. Why is the total energy of a satellite negative?
Because the satellite is bound to the Earth; its potential energy is negative and greater in magnitude than kinetic energy.
3. What does negative total energy signify?
It signifies that the satellite cannot escape Earth’s gravity without external energy — a bound system.
4. How is binding energy related to total energy?
Binding energy is the magnitude of total energy: [B = |E| = \dfrac{G M m}{2r}].
5. Which satellite is more stable: one closer or farther from Earth?
The closer one, because it has greater binding energy and is more tightly bound.
6. What happens to binding energy if orbital radius doubles?
Since [B ∝ \dfrac{1}{r}], doubling [r] halves the binding energy.
7. What happens to binding energy if satellite mass doubles?
Since [B ∝ m], binding energy doubles.
8. Does binding energy depend on Earth’s mass?
Yes, it is directly proportional to Earth’s mass [M].
9. Why is energy needed to launch a satellite?
To overcome the gravitational binding energy holding it to Earth.
10. Is binding energy the same as potential energy?
No. Potential energy is negative; binding energy is the positive energy required to escape Earth’s field.
11. What happens to binding energy for a satellite at infinity?
At infinity, [r → ∞], so [B → 0]; the satellite is free from Earth’s gravitational field.
12. Is the total energy of a satellite equal to its kinetic energy?
No. Total energy is half the potential energy and negative: [E = -K].
13. Can total energy of a satellite ever be zero?
Yes, when the satellite escapes Earth’s gravity (i.e., at infinity).
14. If the total energy is -100 J, what is its binding energy?
Binding energy = 100 J (the positive magnitude of total energy).
15. What is the relation between binding energy and orbital velocity?
Since [v = \sqrt{\dfrac{G M}{r}}], binding energy [B = \dfrac{1}{2} m v^2].
7. FAQ / Common Misconceptions
1. Binding energy is negative.
❌ False. Total energy is negative; binding energy is positive.
2. Higher orbit means higher binding energy.
❌ False. Binding energy decreases with height.
3. A free satellite has binding energy.
❌ False. A free satellite (at infinity) has zero binding energy.
4. Kinetic energy equals potential energy in orbit.
❌ False. [K = -\dfrac{1}{2}U].
5. Binding energy depends only on satellite mass.
❌ False. It depends on [M], [m], and [r].
6. Binding energy is always constant.
❌ False. It changes with orbital radius.
7. Total energy of a satellite can be positive.
❌ False. It is negative for bound orbits.
8. Potential energy is the same as binding energy.
❌ False. Binding energy is the magnitude of total energy, not potential energy.
9. When a satellite is closer to Earth, it is less stable.
❌ False. It is more stable because its binding energy is greater.
10. To make a satellite escape, we must reduce its binding energy.
❌ False. We must increase its total energy by supplying energy equal to binding energy.
8. Practice Questions (With Step-by-Step Solutions)
Q1. Derive the expression for binding energy of a satellite.
Solution:
[
E = K + U = \dfrac{G M m}{2r} – \dfrac{G M m}{r} = -\dfrac{G M m}{2r}
]
[
\therefore B = |E| = \dfrac{G M m}{2r}
]
Q2. A satellite of mass [1000 kg] is orbiting Earth at a radius of [7 × 10^6 \text{ m}].
Find its binding energy.
(Given [G = 6.67 × 10^{-11}], [M = 6 × 10^{24}])
[
B = \dfrac{G M m}{2r}
]
[
B = \dfrac{6.67 × 10^{-11} × 6 × 10^{24} × 1000}{2 × 7 × 10^6}
]
[
B ≈ 2.86 × 10^{10} \text{ J}
]
Q3. If the radius of orbit is halved, what happens to the binding energy?
Since [B ∝ \dfrac{1}{r}], halving [r] doubles [B].
Q4. What is the total energy of a satellite whose binding energy is [5 × 10^9 \text{ J}]?
[
E = -B = -5 × 10^9 \text{ J}
]
Q5. If a satellite’s total energy is [−4 × 10^9 \text{ J}], find the energy required to free it.
[
\text{Energy required} = |E| = 4 × 10^9 \text{ J}
]
Sign in with Google