Satellites orbiting massive central objects follow specific motion patterns governed by gravitational forces. These orbits can be circular or elliptical, with each type having unique energy characteristics. Conservation laws play a crucial role in determining satellite behavior and constraints.

Understanding satellite motion involves analyzing energy relationships, orbital shapes, and escape velocity. These concepts are essential for predicting satellite trajectories, designing space missions, and comprehending the dynamics of natural celestial bodies in our universe.

Central Object Motion

When a satellite orbits a massive central object like a planet or star, the central object appears to remain stationary due to the enormous difference in masses. 🪐

• The center of mass of the system is extremely close to the center of the massive object
• For example, Earth's mass is approximately 6 × 10^24 kg while a typical satellite might be only 1000 kg
• This mass ratio (10^21:1) means the central object's motion is negligible for most calculations
• In physics problems involving satellite motion, we typically treat the central object as fixed at the origin of our coordinate system

This simplification allows us to focus on the satellite's motion without needing to account for the minimal movement of the central body.

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Satellite Orbit Constraints

Conservation laws establish fundamental constraints on satellite motion, creating predictable orbital patterns regardless of the satellite's mass.

• Conservation of Angular Momentum: Ensures that a satellite sweeps out equal areas in equal times (Kepler's Second Law)
• Conservation of Energy: Dictates the total mechanical energy remains constant throughout the orbit
• Conservation of Linear Momentum: Applies to the system as a whole

In circular orbits, values remain constant for:

• Total mechanical energy
• Gravitational potential energy
• Satellite's kinetic energy
• Angular momentum

In elliptical orbits, only these remain constant:

• Total mechanical energy
• Angular momentum

The gravitational potential energy between two masses is defined as zero at infinite separation and becomes increasingly negative as the objects approach each other. This convention allows us to quantify the energy needed to completely separate the objects.

Energy in Circular Orbits

In a circular orbit, the satellite maintains a constant distance from the central object, resulting in balanced energy components. 🛰️

The gravitational potential energy of the system is: $U = -\frac{GMm}{r}$

The kinetic energy of the satellite is: $K = \frac{1}{2}mv^2$

For circular orbits, a special relationship exists: $K = -\frac{1}{2}U$

This means the total energy is: $E_{total} = K + U = -\frac{1}{2}U + U = \frac{1}{2}U = -\frac{GMm}{2r}$

This relationship derives from the fact that in circular orbits, the centripetal force is provided entirely by gravity: $\frac{mv^2}{r} = \frac{GMm}{r^2}$

Energy in Elliptical Orbits

Elliptical orbits introduce varying distances between the satellite and central object, creating a dynamic energy exchange throughout the orbit.

• At periapsis (closest approach):
  • Highest satellite speed
  • Maximum kinetic energy
  • Minimum (most negative) gravitational potential energy
  • Minimum distance from central object
• At apoapsis (farthest point):
  • Lowest satellite speed
  • Minimum kinetic energy
  • Maximum (least negative) gravitational potential energy
  • Maximum distance from central object

Throughout this orbital dance, the total mechanical energy remains constant: $E_{total} = K + U = constant$

The total energy of an elliptical orbit depends on the semi-major axis (a) of the ellipse: $E_{total} = -\frac{GMm}{2a}$

This equation shows that larger elliptical orbits (greater semi-major axis) have less negative total energy, approaching zero as the orbit size increases toward infinity.

Escape Velocity

Escape velocity represents the minimum initial speed a satellite needs to break free from a gravitational field completely. 🚀

When a satellite achieves escape velocity, its total energy becomes exactly zero: $E_{total} = K + U = 0$

This means: $\frac{1}{2}mv_{escape}^2 - \frac{GMm}{r} = 0$

Solving for escape velocity: $v_{escape} = \sqrt{\frac{2GM}{r}}$

Key insights about escape velocity:

• Independent of the satellite's mass
• Depends only on the central object's mass and the starting distance
• Decreases with increasing distance from the central object
• For Earth at the surface: approximately 11.2 km/s

A satellite with exactly escape velocity will:

• Continue moving away from the central object indefinitely
• Gradually slow down as it travels outward
• Approach zero velocity as distance approaches infinity
• Never return to the central object

Practice Problem 1: Circular Orbit Energy

Solution

First, we need to convert the radius to meters: 8000 km = 8 × 10^6 m

a) The gravitational potential energy is: $U = -\frac{GMm}{r}$ $U = -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(1500)}{8 \times 10^6}$ $U = -7.48 \times 10^{10}$ J

b) For a circular orbit, the kinetic energy is half the magnitude of the potential energy: $K = -\frac{1}{2}U = -\frac{1}{2}(-7.48 \times 10^{10})$ $K = 3.74 \times 10^{10}$ J

c) The total energy is: $E_{total} = K + U = 3.74 \times 10^{10} + (-7.48 \times 10^{10})$ $E_{total} = -3.74 \times 10^{10}$ J

Alternatively, we could have used $E_{total} = \frac{1}{2}U = -\frac{GMm}{2r}$ directly.

Practice Problem 2: Escape Velocity

Solution

The escape velocity is given by: $v_{escape} = \sqrt{\frac{2GM}{r}}$

Substituting the values: $v_{escape} = \sqrt{\frac{2(6.67 \times 10^{-11})(6.42 \times 10^{23})}{3.39 \times 10^6}}$ $v_{escape} = \sqrt{\frac{8.56 \times 10^{13}}{3.39 \times 10^6}}$ $v_{escape} = \sqrt{2.53 \times 10^7}$ $v_{escape} = 5.03 \times 10^3$ m/s

Therefore, the escape velocity from Mars' surface is approximately 5.03 km/s, which is less than Earth's escape velocity (11.2 km/s) due to Mars' smaller mass and size.