Problem 57 - University Physics with Modern Physics

Question

Antenna emf. A satellite, orbiting the earth at the equator at an altitude of 400 \(\mathrm{km}\) , has an antenna that can be modeled as a \(2.0-\mathrm{m}-\) long rod. The antenna is oriented perpendicular to the earth's surface. At the cquator, the earth's magnetic field is cssentially horizontal and has a value of \(8.0 \times 10^{-5} \mathrm{T}\) ; ignore any changes in \(B\) with altitude. Assuming the orbit is circular, determine the induced emf between the tips of the antenna.

Step-by-Step Solution

Answer

The induced emf is approximately 1.23 volts.

Step 1: Understand the Given Problem

The problem involves finding the induced electromotive force (emf) on a satellite antenna as it orbits Earth. We are given information about the antenna's length, the value of the Earth's magnetic field, and the satellite's altitude and orientation.

Step 2: Identify Relevant Formulas

To find the induced emf (\( \epsilon \)), we can use the formula: \( \epsilon = B \cdot v \cdot L \cdot \sin \theta \), where \(B\) is the magnetic field, \(v\) is the orbital velocity of the satellite, \(L\) is the length of the antenna, and \(\theta\) is the angle between the velocity and magnetic field. Since the magnetic field is horizontal and the antenna is vertical, \(\theta = 90^\circ\), making \(\sin \theta = 1\).

Step 3: Calculate Orbital Velocity

The orbital velocity \(v\) of the satellite can be calculated with the formula \(v = \sqrt{\frac{GM}{R}}\), where \(G = 6.674 \times 10^{-11} \,\mathrm{m}^3/\mathrm{kg\cdot s}^2\), \(M = 5.972 \times 10^{24} \, \mathrm{kg}\) is the mass of Earth, and \(R = 6.371 \times 10^{6} \, \mathrm{m} + 400 \times 10^3 \, \mathrm{m}\) is the radius of the orbit. Calculate \(R\) first, then substitute values to find \(v\).

Step 4: Calculate Radius of Orbit

The radius \(R\) of the orbit is the sum of Earth's radius and the altitude of the satellite: \(R = 6.371 \times 10^{6} + 400 \times 10^3 = 6.771 \times 10^6\, \mathrm{m}\). This is the distance from the center of the Earth to the satellite.

Step 5: Compute the Orbital Velocity

Substitute the radius \(R\) into the velocity formula: \( v = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.771 \times 10^6}} \), to find \(v\). Calculating gives \( v \approx 7.67 \times 10^3 \, \mathrm{m/s}\).

Step 6: Calculate Induced Emf

Now use the formula to find the emf: \( \epsilon = B \cdot v \cdot L \cdot \sin 90^\circ \). Substitute \(B = 8.0 \times 10^{-5} \, \mathrm{T}\), \(v = 7.67 \times 10^3 \, \mathrm{m/s}\), and \(L = 2.0 \, \mathrm{m}\): \( \epsilon \approx 8.0 \times 10^{-5} \times 7.67 \times 10^3 \times 2.0 = 1.23 \, \mathrm{V} \).

Step 7: Conclusion

The induced emf between the tips of the satellite's antenna as it orbits the Earth is approximately 1.23 volts.

Key Concepts

Orbital Velocity

Orbital velocity refers to the speed at which a satellite must travel to maintain orbit around a celestial body, like Earth. This velocity keeps the satellite moving in a stable, circular path due to the balance between gravitational pull and inertia.

• To calculate orbital velocity, we use the formula: \[ v = \sqrt{ \frac{GM}{R} } \] where:
  • \( G \) is the gravitational constant \(6.674 \times 10^{-11} \, \mathrm{m^3/kg \cdot s^2}\)
  • \( M \) is the mass of Earth \(5.972 \times 10^{24} \, \mathrm{kg}\)
  • \( R \) is the orbital radius, the distance from Earth's center to the satellite.
• The orbital radius accounts for Earth's radius and the altitude of the satellite, giving \[ R = 6.371 \times 10^6 + 400 \times 10^3 = 6.771 \times 10^6 \, \mathrm{m}. \]

Substituting values into the formula grants us the satellite's orbital velocity, \( v \), which in this case is approximately 7,670 meters per second. This speed carefully balances the gravitational force pulling the satellite inward and the inertia wanting it to fly straight.

Earth's Magnetic Field

Earth's magnetic field plays a significant role in the induced electromotive force experienced by antennas on orbiting satellites.

• This geomagnetic field resembles that of a giant bar magnet centered on the Earth, influencing compasses and affecting satellites in orbit.
• Near the equator, the magnetic field is essentially horizontal. In this context, the strength of Earth's magnetic field is given as \( 8.0 \times 10^{-5} \, \mathrm{T} \) or Tesla.

The magnetic field interacts with the satellite's motion to generate an electromotive force. Understanding this influence helps in determining key variables for satellite operations and communication systems.

Satellite Antenna

The satellite antenna in this scenario is modeled as a 2-meter long rod, oriented vertically perpendicular to the Earth's horizontal magnetic field.

• As an important aspect of satellite design, antennas are pivotal in facilitating communication between the satellite and Earth.
• The orientation is crucial because the induced electromotive force depends on the angle between the antenna and the magnetic field.
• The formula used here for induced emf is: \[ \epsilon = B \cdot v \cdot L \cdot \sin \theta \] where \( \sin 90^\circ = 1 \), simplifying calculation as this angle results from the vertical antenna and horizontal field configuration.

This configuration ensures optimal interaction with Earth's magnetic field, maximizing the induced voltage.

Circular Orbit

A circular orbit is characterized by a path where a satellite maintains a constant distance from the Earth's center.

• Satellites in circular orbits travel around Earth in a stable, predictable manner, making them ideal for communication, weather monitoring, and GPS systems.
• The consistency of altitude around the Earth ensures uniform gravitational force and uniform speed, simplifying satellite operation.
• In this exercise, the assumption of a circular orbit allows calculations to focus on constant speed and direction, maximizing the predictability of the satellite’s path.

Understanding circular orbits is essential for designing effective satellite systems and ensuring seamless data transmission and precision in applications. The stable nature of circular orbits aids in maintaining balance and efficiency in communication networks.