AP Physics C Mechanics- 6.6 Motion of Orbiting Satellites- Study Notes- New Syllabus
AP Physics C Mechanics- 6.6 Motion of Orbiting Satellites – Study Notes
AP Physics C Mechanics- 6.6 Motion of Orbiting Satellites – Study Notes – per latest Syllabus.
Key Concepts:
- Negligible Motion of the Central Object
- Conservation Laws in Orbital Motion
- Total Energy of a Satellite in a Circular Orbit
- Escape Velocity
Negligible Motion of the Central Object
In a system consisting of a massive central object (e.g., Earth) and a small satellite (e.g., a moon or spacecraft), the satellite’s mass is negligible compared to that of the central object. Thus, the motion of the central object is negligible, and it can be treated as fixed in space.
The center of mass of the system lies very close to the center of the massive body, so the satellite’s motion dominates the observable dynamics. This simplification allows orbital motion to be modeled as the satellite moving under the gravitational influence of a stationary central mass.
\( \mathrm{M_{central} \gg m_{satellite} \Rightarrow \text{central body approximately stationary}} \)
Example
Earth’s mass is \( \mathrm{5.97\times10^{24}\,kg} \), while a typical satellite’s mass is \( \mathrm{500\,kg} \). Determine whether Earth’s motion due to the satellite’s gravitational pull is significant.
▶️ Answer / Explanation
The ratio of satellite to Earth mass is:
\( \mathrm{\dfrac{m_{sat}}{M_{E}} = \dfrac{500}{5.97\times10^{24}} \approx 8.4\times10^{-23}} \)
Result: The effect is extremely small — Earth’s motion due to the satellite is negligible. Hence, the satellite can be analyzed as orbiting a stationary Earth.
Conservation Laws in Orbital Motion
Satellite motion in orbits is governed by the laws of conservation — particularly, the conservation of mechanical energy and angular mombentum. The gravitational force acts as a central fobrce, always directed toward the massive object, ensuring these quantities remain constant (in the absence of external influences).
Circular Orbits
In a circular orbit, all physical quantities remain constant in time:
- Total mechanical energy (\( \mathrm{E} \))
- Gravitational potential energy (\( \mathrm{U_g} \))
- Kinetic energy (\( \mathrm{K} \))
- Angular momentum (\( \mathrm{L} \))
Key Equations:
\( \mathrm{U_g = -\dfrac{GMm}{r}} \)
\( \mathrm{K = \tfrac{1}{2}mv^2 = \dfrac{GMm}{2r}} \)
\( \mathrm{E_{total} = K + U_g = -\dfrac{GMm}{2r}} \)
\( \mathrm{L = mvr = m\sqrt{GMr}} \)
Key Idea: In circular orbits, the balance between gravitational attraction and centripetal acceleration ensures a constant speed, distance, and energy distribution.
Example
Find the orbital speed of a satellite in a circular orbit \( \mathrm{300\,km} \) above Earth’s surface. Use \( \mathrm{R_E = 6.37\times10^6\,m} \) and \( \mathrm{M_E = 5.97\times10^{24}\,kg} \).
▶️ Answer / Explanation
Step 1: From centripetal motion:
\( \mathrm{\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}} \Rightarrow v = \sqrt{\dfrac{GM}{r}} \)
Step 2: Substitute \( \mathrm{r = R_E + 300\times10^3 = 6.67\times10^6\,m} \):
\( \mathrm{v = \sqrt{\dfrac{(6.67\times10^{-11})(5.97\times10^{24})}{6.67\times10^6}} = 7.73\times10^3\,m/s} \)
Result: The satellite’s orbital speed is \( \mathrm{7.73\,km/s} \), and its total mechanical energy per unit mass is \( \mathrm{-3.0\times10^7\,J/kg} \).
Elliptical Orbits
In an elliptical orbit, the satellite’s distance from the central object changes continuously. Consequently, its kinetic and gravitational potential energies vary, but the total mechanical energy and angular momentum remain constant.
Conserved quantities:
- Total mechanical energy: \( \mathrm{E_{total} = \text{constant}} \)
- Angular momentum: \( \mathrm{L = \text{constant}} \)
Changing quantities:
- Kinetic energy (\( \mathrm{K} \)) increases as satellite approaches the central object.
- Potential energy (\( \mathrm{U_g} \)) decreases (becomes more negative) near the central body.
At Periapsis (closest point): \( \mathrm{v} \) is maximum, \( \mathrm{U_g} \) is minimum.
At Apoapsis (farthest point): \( \mathrm{v} \) is minimum, \( \mathrm{U_g} \) is maximum.
Key Relationship:
\( \mathrm{E = -\dfrac{G M m}{2a}} \)
(where \( \mathrm{a} \) is the semi-major axis of the ellipse)
Example
A \( \mathrm{500\,kg} \) satellite moves in an elliptical orbit around Earth with a periapsis of \( \mathrm{6.5\times10^6\,m} \) and an apoapsis of \( \mathrm{8.0\times10^6\,m} \). Find the semi-major axis and total mechanical energy of the system.
▶️ Answer / Explanation
Step 1: Find semi-major axis:
\( \mathrm{a = \dfrac{r_{min} + r_{max}}{2} = \dfrac{6.5\times10^6 + 8.0\times10^6}{2} = 7.25\times10^6\,m} \)
Step 2: Compute total energy:
\( \mathrm{E = -\dfrac{G M_E m}{2a}} \)
\( \mathrm{E = -\dfrac{(6.67\times10^{-11})(5.97\times10^{24})(500)}{2(7.25\times10^6)} = -1.37\times10^{10}\,J} \)
Result: Total mechanical energy \( \mathrm{E = -1.37\times10^{10}\,J} \) remains constant throughout the orbit.
Definition of Gravitational Potential Energy in Satellite Systems
The gravitational potential energy (\( \mathrm{U_g} \)) of a system consisting of a satellite and a massive central object is defined to be zero when the satellite is infinitely far away from the central object.
This establishes a natural reference point for potential energy in gravitational systems.
Relevant equation:
\( \mathrm{U_g = -\dfrac{G M m}{r}} \)
- \( \mathrm{U_g} \) is negative for all finite \( \mathrm{r} \) values, since the gravitational force is attractive.
- As \( \mathrm{r \to \infty} \), \( \mathrm{U_g \to 0} \).
A negative potential energy indicates that the satellite is bound to the central object; it would require positive work (energy input) to escape the gravitational influence completely.
Example
Calculate the gravitational potential energy of a \( \mathrm{1200\,kg} \) satellite located \( \mathrm{3.0\times10^7\,m} \) above Earth’s surface. (Earth’s radius \( \mathrm{R_E = 6.37\times10^6\,m} \))
▶️ Answer / Explanation
Step 1: Determine distance from Earth’s center:
\( \mathrm{r = R_E + h = 6.37\times10^6 + 3.0\times10^7 = 3.637\times10^7\,m} \)
Step 2: Apply gravitational potential energy formula:
\( \mathrm{U_g = -\dfrac{G M_E m}{r}} \)
\( \mathrm{U_g = -\dfrac{(6.67\times10^{-11})(5.97\times10^{24})(1200)}{3.637\times10^7} = -1.32\times10^{10}\,J} \)
Result: The satellite’s gravitational potential energy relative to infinity is \( \mathrm{-1.32\times10^{10}\,J} \).
Total Energy of a Satellite in a Circular Orbit
A satellite moving in a circular orbit around a massive central object (like Earth or the Sun) has two types of mechanical energy:
- Kinetic energy due to its motion, and
- Gravitational potential energy due to the gravitational attraction of the central body.
These two energies are connected by the conditions of uniform circular motion and Newton’s Law of Gravitation.
Derivation of Relationships
For a satellite of mass \( \mathrm{m} \) orbiting a central object of mass \( \mathrm{M} \) at a radius \( \mathrm{r} \):
\( \mathrm{F_{grav} = \dfrac{G M m}{r^2}} \)
and since this gravitational force provides the necessary centripetal force,
\( \mathrm{\dfrac{G M m}{r^2} = \dfrac{m v^2}{r}} \Rightarrow v^2 = \dfrac{G M}{r}} \)
1. Relationship Between Kinetic and Potential Energy
The kinetic energy of the satellite is:
\( \mathrm{K = \tfrac{1}{2} m v^2 = \tfrac{1}{2} m \left(\dfrac{G M}{r}\right)} \Rightarrow K = \dfrac{G M m}{2r}} \)
The gravitational potential energy of the system is:
\( \mathrm{U_g = -\dfrac{G M m}{r}} \)
Therefore,
\( \mathrm{K = -\tfrac{1}{2} U_g} \)
Key Idea: In a stable circular orbit, the kinetic energy is always half the magnitude of the potential energy, but positive meaning \( \mathrm{K} = -\tfrac{1}{2} U_g} \).
2. Total Mechanical Energy of the Orbiting System
The total energy of the system (sum of kinetic and potential) is:
\( \mathrm{E_{total} = K + U_g = \dfrac{G M m}{2r} – \dfrac{G M m}{r} = -\dfrac{G M m}{2r}} \)
Hence,
\( \mathrm{E_{total} = -\tfrac{1}{2} \dfrac{G M m}{r}} \)
This shows that the total energy is negative, signifying that the satellite is gravitationally bound to the central object. To remove the satellite from orbit (escape to infinity), external work equal to \( |\mathrm{E_{total}}| \) must be done on it.
Summary of Derived Equations
| Quantity | Equation | Relationship |
|---|---|---|
| Gravitational Potential Energy | \( \mathrm{U_g = -\dfrac{G M m}{r}} \) | Defined as zero at infinite separation. |
| Kinetic Energy | \( \mathrm{K = \dfrac{G M m}{2r}} \) | Half the magnitude of potential energy, but positive. |
| Total Energy | \( \mathrm{E_{total} = -\dfrac{G M m}{2r}} \) | Total energy is negative — system is bound. |
Example
A \( \mathrm{1500\,kg} \) satellite orbits Earth in a circular orbit at an altitude of \( \mathrm{3.0\times10^7\,m} \) above the surface. Calculate (a) the satellite’s kinetic energy, (b) gravitational potential energy, and (c) total mechanical energy.
▶️ Answer / Explanation
Given: \( \mathrm{M_E = 5.97\times10^{24}\,kg} \), \( \mathrm{R_E = 6.37\times10^6\,m} \), \( \mathrm{h = 3.0\times10^7\,m} \)
Step 1: Distance from Earth’s center:
\( \mathrm{r = R_E + h = 6.37\times10^6 + 3.0\times10^7 = 3.637\times10^7\,m} \)
Step 2: Kinetic energy:
\( \mathrm{K = \dfrac{G M_E m}{2r} = \dfrac{(6.67\times10^{-11})(5.97\times10^{24})(1500)}{2(3.637\times10^7)} = 8.21\times10^9\,J} \)
Step 3: Gravitational potential energy:
\( \mathrm{U_g = -\dfrac{G M_E m}{r} = -1.64\times10^{10}\,J} \)
Step 4: Total mechanical energy:
\( \mathrm{E_{total} = K + U_g = 8.21\times10^9 – 1.64\times10^{10} = -8.21\times10^9\,J} \)
Result: – \( \mathrm{K = 8.21\times10^9\,J} \) – \( \mathrm{U_g = -1.64\times10^{10}\,J} \) – \( \mathrm{E_{total} = -8.21\times10^9\,J} \)
The satellite’s total energy is negative, indicating that it is gravitationally bound to Earth.
Escape Velocity
The escape velocity of a satellite is the minimum velocity needed for it to move infinitely far away from a central massive body without ever falling back — that is, to escape the gravitational field completely.
At escape velocity, the total mechanical energy (kinetic + gravitational potential) of the satellite–central-body system is zero:
\( \mathrm{E_{total} = K + U_g = 0} \)
This condition defines the threshold between a bound system (negative total energy) and an unbound system (zero or positive total energy).
Behavior of a Satellite at Escape Velocity
When the only force acting on a satellite is the gravitational force from the central object:
- A satellite launched with velocity less than escape velocity will remain bound in an orbit (circular or elliptical).
- A satellite launched at exactly escape velocity will move away, slowing continuously, and come to rest at infinite distance.
- A satellite launched faster than escape velocity will escape completely with positive total energy and nonzero speed at infinity.
At infinite distance \( \mathrm{(r \to \infty)} \), the gravitational potential energy \( \mathrm{U_g \to 0} \) and the satellite’s kinetic energy \( \mathrm{K \to 0} \). Thus, the escape condition is defined by \( \mathrm{E_{total} = 0} \).
Derivation of Escape Velocity
Using conservation of energy between launch point (distance \( \mathrm{r} \)) and infinity:
\( \mathrm{K_i + U_i = K_f + U_f} \)
At infinity, \( \mathrm{K_f = 0} \) and \( \mathrm{U_f = 0} \), so:
\( \mathrm{\tfrac{1}{2} m v_e^2 – \dfrac{G M m}{r} = 0} \)
Solving for \( \mathrm{v_e} \):
\( \mathrm{v_e = \sqrt{\dfrac{2 G M}{r}}} \)
Derived Equation:
\( \mathrm{\boxed{v_e = \sqrt{\dfrac{2 G M}{r}}}} \)
- \( \mathrm{v_e} \): escape velocity
- \( \mathrm{G} \): universal gravitational constant (\( \mathrm{6.67\times10^{-11}\,N·m^2/kg^2} \))
- \( \mathrm{M} \): mass of the central body
- \( \mathrm{r} \): distance from the center of the central body
The escape velocity depends only on the mass of the central body and the distance from its center, not on the mass of the satellite.
Energy Perspective
At escape velocity:
- \( \mathrm{K_i = -U_i} \)
- \( \mathrm{E_{total} = 0} \)
- All the initial kinetic energy is used to overcome the gravitational potential energy barrier.
Example
Calculate the escape velocity for a satellite launched from Earth’s surface. Given: \( \mathrm{M_E = 5.97\times10^{24}\,kg} \), \( \mathrm{R_E = 6.37\times10^6\,m} \).
▶️ Answer / Explanation
Step 1: Use the escape velocity equation:
\( \mathrm{v_e = \sqrt{\dfrac{2 G M_E}{R_E}}} \)
Step 2: Substitute values:
\( \mathrm{v_e = \sqrt{\dfrac{2(6.67\times10^{-11})(5.97\times10^{24})}{6.37\times10^6}}} \)
Step 3: Simplify:
\( \mathrm{v_e = \sqrt{1.25\times10^8} = 1.12\times10^4\,m/s} \)
Result: The escape velocity from Earth’s surface is approximately \( \mathrm{11.2\,km/s} \).