Problem 70 - College Physics

Question

Using a development similar to Kepler's law of periods for planets orbiting the Sun, find the required altitude of geosynchronous satellites above the Earth. [Hint: The period of such satellites is the same as that of the Earth.]

Step-by-Step Solution

Answer

The altitude of a geosynchronous satellite is approximately 35,793 km above Earth's surface.

Step 1: Understanding Kepler's Third Law

Kepler's Third Law states that the square of the orbital period \( T \) of a planet is directly proportional to the cube of the semi-major axis \( a \) of its orbit. Mathematically, \( T^2 \propto a^3 \). For objects orbiting a common central body, \( \frac{T^2}{a^3} = k \), where \( k \) is the same for all satellites orbiting Earth.

Step 2: Define Period of Geosynchronous Satellite

Geosynchronous satellites have an orbital period \( T \) equal to the Earth's rotational period, which is 24 hours or 86400 seconds. This means that the time it takes for the satellite to orbit the Earth once is the same as the time it takes for the Earth to complete one full rotation on its axis.

Step 3: Use Kepler’s Law in Terms of Earth’s Gravitational Constant

The gravitational force provides the necessary centripetal force for circular orbits, so \( \frac{GMm}{r^2} = \frac{mv^2}{r} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, \( r \) is the orbit radius, and \( v \) is the satellite’s velocity. For circular orbits, \( v = \frac{2\pi r}{T} \).

Step 4: Rearrange to Find Orbital Radius

Substitute \( v = \frac{2\pi r}{T} \) into the centripetal force formula: \( \frac{GM}{r^2} = \left(\frac{2\pi}{T}\right)^2 r \). Solving for \( r \) gives \( r^3 = \frac{GMT^2}{4\pi^2} \).

Step 5: Calculate Radius for Geosynchronous Satellite

Substitute \( G = 6.674 \times 10^{-11} \text{ m}^3/\text{kg} \cdot \text{s}^2 \), \( M = 5.972 \times 10^{24} \text{ kg} \), and \( T = 86400 \text{ s} \) into the equation. Calculate \( r \): \( r = \left(\frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(86400)^2}{4\pi^2}\right)^{1/3} \approx 42,164 \text{ km} \).

Step 6: Find Altitude Above Earth’s Surface

The orbital radius \( r \) calculated includes the Earth’s radius. Subtract this from the orbital radius to find the altitude: Earth’s radius is approximately 6,371 km. Thus, altitude = 42,164 km - 6,371 km = 35,793 km.

Key Concepts

Kepler's Third Law

Kepler's Third Law is a fundamental principle that helps us understand the motion of celestial bodies. It states that the square of the orbital period (\( T \) ) of an orbiting body is directly proportional to the cube of the average distance (\( a \) ) from the body to its central orbited body. In mathematical terms, this is expressed as:

• \( T^2 \propto a^3 \)

For bodies orbiting around the same central body, the proportionality becomes an equation involving a constant (\( k \) ), such that \( \frac{T^2}{a^3} = k \). This constant (\( k \) ) remains the same for all satellites orbiting Earth. This law is crucial because it implies that if you know the period of orbit, you can predict the size of the orbit, and vice versa. In solving problems such as determining the altitude of geosynchronous satellites, applying Kepler's Third Law allows us to relate their period to their distance from Earth.

Orbital Period

The orbital period is a term used to describe the time it takes for a satellite or any celestial body to make one complete orbit around another body. For geosynchronous satellites, their orbital period is particularly important because it is equal to Earth's rotational period.

• For Earth, this period is 24 hours or 86,400 seconds.

This means that a geosynchronous satellite appears to remain fixed over one spot of the Earth's surface, as it completes its orbit in the same time the Earth rotates once on its axis. This characteristic is highly valued for telecommunications and weather satellites, offering consistent coverage of a particular area on Earth. Understanding this concept allows us to design satellites that precisely fulfill specific functional requirements, such as constant, reliable communication signals in Internet and television broadcasting.

Gravitational Force

Gravitational force plays a pivotal role in the motion of satellites. It is the attractive force that acts between any two masses. For an object orbiting Earth, gravitational force is the key factor that keeps it in orbit. The gravitational force between two objects of masses (\( M \) ) and (\( m \) ), separated by a distance (\( r \) ), is given by:

• \( F = \frac{G M m}{r^2} \)

where \( G \) is the gravitational constant. This force must equal the required centripetal force for circular motion, ensuring the satellite stays in orbit. In our scenario, it was used along with centripetal force formulas to derive the necessary conditions for a satellite's orbit. Without gravitational force, satellites would not be able to maintain a stable path around Earth.

Centripetal Force

Centripetal force is responsible for keeping an object moving in a circular path. For satellites in orbit, this force must be provided by the gravitational attraction between the satellite and the Earth. The formula for centripetal force is:

• \( F_c = \frac{mv^2}{r} \)

where \( m \) is the mass of the satellite, \( v \) is its velocity, and \( r \) is the radius of its orbit. For a geosynchronous satellite, in a perfectly circular orbit, the gravitational force equates to the centripetal force:

• \( \frac{GMm}{r^2} = \frac{mv^2}{r} \)

Substituting velocity in terms of period \( (v = \frac{2\pi r}{T}) \), allows calculation of the orbit radius, crucial for identifying the altitude above Earth's surface. It is through understanding and applying the concept of centripetal force that one can determine orbits that are stable and sustainable over long durations for practical use.