Satellite Motion as Uniform Circular Motion

Introduction

This section explores the motion of satellites (artificial, the Moon, and planets) by modeling their orbits as uniform circular motion. We will examine the relationship between gravitational force, centripetal force, orbital speed, and orbital radius.

Uniform Circular Motion

• Definition: Uniform circular motion is the movement of an object at a constant speed along a circular path.
• Key Characteristics:
  • Constant speed ($v$)
  • Constant radius ($r$)
  • Centripetal acceleration ($a_c$) directed towards the center of the circle.
• Centripetal Acceleration: The acceleration required to keep an object moving in a circle.
  $$a_c = \frac{v^2}{r}$$
• Centripetal Force: The net force required to keep an object moving in a circle. It’s always directed towards the center of the circle.
  $$F_c = ma_c = \frac{mv^2}{r}$$

KEY TAKEAWAY: In uniform circular motion, the velocity is constant in magnitude but constantly changing direction, resulting in centripetal acceleration.

Gravitational Force

• Newton’s Law of Universal Gravitation: Describes the attractive force between two objects with mass.
  $$F_g = G\frac{Mm}{r^2}$$
  Where:
  • $F_g$ is the gravitational force
  • $G$ is the gravitational constant (\$6.674 \times 10^{-11} Nm^2/kg^2$)
  • $M$ and $m$ are the masses of the two objects
  • $r$ is the distance between the centers of the two objects

REMEMBER: Gravitational force is always attractive and acts along the line connecting the centers of the two masses.

Satellite Motion

• Modeling Satellite Motion: We model the orbits of satellites (artificial, Moon, planets) as uniform circular motion. This is an approximation, as real orbits are elliptical, but it simplifies the analysis.
• Centripetal Force Provided by Gravity: In satellite motion, the gravitational force between the satellite and the central body (e.g., Earth) provides the centripetal force required for the circular motion.
  $$F_c = F_g$$

Equations for Satellite Motion

• Equating Centripetal and Gravitational Forces:
  $$\frac{mv^2}{r} = G\frac{Mm}{r^2}$$
  Where:
  • $m$ is the mass of the satellite
  • $v$ is the orbital speed of the satellite
  • $r$ is the orbital radius (distance from the center of the central body to the satellite)
  • $G$ is the gravitational constant
  • $M$ is the mass of the central body (e.g., Earth)
• Solving for Orbital Speed:
  $$v = \sqrt{\frac{GM}{r}}$$
• Relationship between Speed and Radius: This equation shows that the orbital speed of a satellite depends only on the mass of the central body and the orbital radius. A smaller radius implies a higher orbital speed.
• Period of Orbit: The time it takes for a satellite to complete one orbit.
  $$T = \frac{2\pi r}{v}$$
• Substituting for v:
  $$T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}} = 2\pi \sqrt{\frac{r^3}{GM}}$$
• Relationship between Period and Radius: This equation shows that the period of orbit depends on the orbital radius. A larger radius implies a longer orbital period.

EXAM TIP: When solving problems involving satellite motion, start by equating the centripetal force and gravitational force. Then, solve for the unknown variable.

Key Variables and Relationships

Applications

• Artificial Satellites: Used for communication, navigation (GPS), weather monitoring, and scientific research.
• Moon: Earth’s natural satellite, influencing tides and stabilizing Earth’s axial tilt.
• Planets: Orbiting the Sun, governed by the same principles of gravitational force and centripetal motion.

APPLICATION: Understanding satellite motion is crucial for designing and maintaining satellite orbits for various purposes, such as communication and Earth observation.

Example Problem

A satellite orbits the Earth at an altitude of 500 km. The Earth’s radius is \$6.371 \times 10^6$ m, and its mass is \$5.972 \times 10^{24}$ kg. Calculate the satellite’s orbital speed and period.

1. Calculate the orbital radius:
   $r = 6.371 \times 10^6 m + 500 \times 10^3 m = 6.871 \times 10^6 m$
2. Calculate the orbital speed:
   $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.674 \times 10^{-11} Nm^2/kg^2)(5.972 \times 10^{24} kg)}{6.871 \times 10^6 m}} \approx 7615 m/s$
3. Calculate the orbital period:
   $T = \frac{2\pi r}{v} = \frac{2\pi (6.871 \times 10^6 m)}{7615 m/s} \approx 5670 s \approx 94.5 minutes$

COMMON MISTAKE: Forgetting to add the altitude of the satellite to the radius of the Earth when calculating the orbital radius.

Limitations of the Model

• Elliptical Orbits: Real satellite orbits are elliptical, not perfectly circular. This model is an approximation.
• Other Forces: Other forces, such as atmospheric drag and the gravitational pull of other celestial bodies, are ignored in this simplified model.

STUDY HINT: Practice solving various problems involving satellite motion to become comfortable with the equations and concepts.

VCAA Physics Study Design Points Addressed:

• model satellite motion (artificial, Moon, planet) as uniform circular orbital motion: $F_c = \frac{mv^2}{r} = G\frac{Mm}{r^2}$

VCAA FOCUS: VCAA exam questions often involve calculations of orbital speed, orbital period, and the relationships between these variables and the orbital radius. Make sure you understand how to manipulate the equations.