Problem 55 - University Physics with Modern Physics

Question

(a) Suppose you are at the earth's equator and observe a satellite passing directly overhead and moving from west to east in the sky. Exactly 12.0 hours later, you again observe this satellite to be directly overhead. How far above the earth's surface is the satellite's orbit? (b) You observe another satellite directly overhead and traveling east to west. This satellite is again overhead in 12.0 hours. How far is this satellite's orbit above the surface of the earth?

Step-by-Step Solution

Answer

(a) The satellite's orbit is about 35,900 km above Earth's surface. (b) The other satellite's orbit is approximately 13,800 km above Earth's surface.

Step 1: Understanding the Satellite's Motion (Part a)

First, realize that a satellite moving from west to east and observed directly overhead every 12 hours is a geostationary satellite with twice the Earth's rotational period. The Earth rotates 360 degrees in 24 hours, so in 12 hours, it rotates 180 degrees. Therefore, the satellite completes one full orbit relative to the surface in 12 hours.

Step 2: Determine the Angular Velocity (Part a)

For a satellite to appear overhead every 12 hours, the angular velocity \( \omega \) of the satellite with respect to the Earth must be such that: \( \omega = \frac{2\pi}{43200} \) rad/s (since 12 hours is 43,200 seconds).

Step 3: Apply the Circular Orbit Formula (Part a)

Using the formula for centripetal force, which balances gravitational force \( \frac{G M m}{r^2} = m \omega^2 r \), we solve for \( r \), the distance from the Earth's center. Here, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \), \( M = 5.98 \times 10^{24} \, \text{kg} \). Simplifying gives: \[ r = \left( \frac{G M}{\omega^2} \right)^{1/3} \].

Step 4: Calculate the Satellite's Orbital Altitude (Part a)

Substitute \( G \), \( M \), and \( \omega \) into the formula: \[ r = \left( \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{\left(\frac{2\pi}{43200}\right)^2} \right)^{1/3} \]. After computation, \( r \approx 4.23 \times 10^7 \) meters. The altitude of the satellite from the Earth's surface is \( r - R_E \), where \( R_E \approx 6.37 \times 10^6 \) meters (Earth's radius). So, the altitude is \[ 4.23 \times 10^7 - 6.37 \times 10^6 = 3.59 \times 10^7 \] meters.

Step 5: Understanding the Satellite's Motion (Part b)

A satellite moving from east to west and observed directly overhead every 12 hours means it needs to "outpace" the Earth's rotation, effectively "catching up" to being directly overhead again in 12 hours through an apparent retrograde motion.

Step 6: Satellite Orbital Period Calculation (Part b)

The satellite's actual period is less than 12 hours, as it must move against the direction of Earth's rotation. Using the same approach, but knowing the satellite's observed period is 12 hours, you'll apply relative motion to determine that actual rotational period.

Step 7: Orbital Calculation (Part b)

Using the reduced period \( T' < 24 \text{ hours (since relative to Earth's rotation it is 12 hours)} \), and substituting into \( \omega = \frac{2\pi}{T'} \), calculate using the forwarded steps from part (a) while considering the relative period.

Step 8: Calculate the Satellite's Orbital Altitude (Part b)

For simplicity, assume the approximate solution for familiar periodic orbit calculations such as setting \( T' \approx 8 \text{ hours} \) after interplay, and follow the same steps as in (a) to solve for \( r \). For an overhead observation, \( r \approx 2.02 \times 10^7 \) meters, leading to an altitude of \( 1.38 \times 10^7 \) meters.

Key Concepts

Geostationary Satellite

Imagine watching a satellite that seems to hover in the sky above a fixed point on Earth. This is no ordinary feat; it is the essence of a geostationary satellite. These satellites orbit in sync with Earth's rotation, making a complete orbit every 24 hours. This means that to an observer on Earth, it appears to be stationary. A unique feature of geostationary satellites is that they must be placed at an altitude of approximately 35,786 kilometers above the equator, where their orbital period matches the Earth's rotational period.
They are incredibly valuable for applications such as weather monitoring and telecommunications, providing constant coverage over the same geographical area. A satellite following this specific orbital path must maintain a precise speed and trajectory to counteract gravitational pull while keeping up with Earth's rotation.

Orbital Mechanics

Orbital mechanics is the science of predicting the trajectories of objects, like satellites, as they move around celestial bodies. These principles help us understand how to place satellites into desired orbits, like the ones that allow for geostationary positions.
Two major forces come into play in orbital mechanics: gravity, which pulls the satellite towards Earth, and the centrifugal force generated by the satellite's motion that pushes it outward. For a stable orbit, these forces must balance perfectly.

• Understanding elliptical, circular, and parabolic orbits is crucial.
• Calculations often involve Kepler's laws and Newton's laws of motion.

By mastering these mechanics, engineers ensure satellites can stay aligned over a particular spot or travel as needed.

Angular Velocity

In simple terms, angular velocity is the rate of change of an object's angle as it moves along a circular path. For satellites, it indicates how quickly they complete an orbit around Earth. Specifically, angular velocity is calculated by the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the orbital period in seconds.
A satellite that completes an orbit in 12 hours has a different angular velocity compared to one completing its path in 24 hours. This distinction is vital:

• A faster angular velocity means the satellite takes less time to orbit.
• For geostationary satellites, matching Earth's period of 24 hours is essential.

This precise rate ensures consistent positioning and functionality necessary for communication and data relay.

Centripetal Force

Centripetal force is the "center-seeking" force that ensures an object follows a curved path, like a satellite orbiting Earth. For satellites, this force must be perfectly balanced with gravitational pull, ensuring a stable orbit. The centripetal force can be calculated using the formula \( F = m \omega^2 r \), where \( m \) is the satellite's mass, \( \omega \) is its angular velocity, and \( r \) is the radius of its orbit.
Understanding how centripetal force works helps in designing orbits:

• Greater force is required for a satellite in a lower orbit due to increased gravitational effects.
• In balance, it guarantees satellites don't drift away or crash back to the planet.

Mastering this concept allows engineers to create satellites that can maintain their intended paths, providing reliable services from space.