Problem 56 - Fundamentals of Physics

Question

An Earth satellite moves in a circular orbit \(640 \mathrm{~km}\) (uniform circular motion) above Earth's surface with a period of \(98.0 \mathrm{~min}\). What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

Step-by-Step Solution

Answer

The speed is approximately 7482 m/s, and the centripetal acceleration is about 7.98 m/s².

Step 1: Understand the Problem

We are given a satellite orbiting at a height of \(640\, \text{km}\) above Earth's surface with an orbital period of \(98.0\, \text{min}\). We need to find the satellite's speed and its centripetal acceleration.

Step 2: Calculate Orbital Radius

The radius of the Earth is approximately \(6371\, \text{km}\). The orbital radius \(r\) is the sum of Earth's radius and the altitude of the satellite: \[ r = 6371\, \text{km} + 640\, \text{km} = 7011\, \text{km} = 7011000\, \text{m} \]

Step 3: Convert Orbital Period to Seconds

The given orbital period is \(98.0\, \text{min}\). Convert this into seconds: \[ T = 98.0\, \text{min} \times 60\, \text{s/min} = 5880\, \text{s} \]

Step 4: Calculate the Speed of the Satellite

The orbital speed \(v\) can be found using the formula for circular motion: \[ v = \frac{2\pi r}{T} \] Substitute the known values: \[ v = \frac{2\pi \times 7011000\, \text{m}}{5880\, \text{s}} \approx 7482\, \text{m/s} \]

Step 5: Calculate the Centripetal Acceleration

The centripetal acceleration \(a_c\) is given by the formula: \[ a_c = \frac{v^2}{r} \] Substitute the known values: \[ a_c = \frac{(7482\, \text{m/s})^2}{7011000\, \text{m}} \approx 7.98\, \text{m/s}^2 \]

Step 6: Review the Results

The satellite's speed is approximately \(7482\, \text{m/s}\), and its centripetal acceleration is about \(7.98\, \text{m/s}^2\).

Key Concepts

Circular Motion

When we talk about circular motion, we are describing the movement of an object along the circumference of a circle. This type of motion is characterized by a constant speed and a direction that continuously changes to keep the object traveling in a curved path.
For satellites orbiting Earth, they move in a nearly perfect circle, resulting in what we call uniform circular motion. In uniform circular motion:

• The speed of the object remains constant, but the velocity changes direction.
• Acceleration is directed towards the center of the circle, known as centripetal acceleration.

A good real-world example of circular motion is the hands of a clock. They move around in a circle at a constant rate, maintaining circular motion.

Satellite

Satellites are objects that orbit larger celestial bodies due to gravitational forces. The satellite in our exercise is moving around Earth. It stays in orbit because of the balance between its forward velocity and the gravitational pull from the planet.
There are different types of satellites, including natural satellites like moons and artificial satellites like the one described in the exercise.

• Artificial satellites are used for communication, weather monitoring, and scientific research.
• They require a specific speed and altitude to maintain orbit without falling back to Earth.

The satellite's orbit is calculated meticulously to accomplish its mission without interruption.

Centripetal Acceleration

Centripetal acceleration is essential for any object in circular motion. It keeps the object moving along its curved path by always pointing towards the center of the circle.
For our satellite, centripetal acceleration ensures it doesn't drift off into space or crash into Earth.

• The formula to calculate centripetal acceleration is \( a_c = \frac{v^2}{r} \).
• Here, \(v\) is the speed of the object, and \(r\) is the radius of the orbit.

Understanding centripetal acceleration helps explain how satellites stay in their paths around planets and why they don't need engines to keep moving.

Orbital Speed

Orbital speed is the speed required for a satellite to stay in its circular path around Earth. This speed ensures that the satellite's inertia and the gravitational pull from Earth are perfectly balanced.
To find it, we use the formula: \( v = \frac{2\pi r}{T} \), where:

• \( v \) is the orbital speed.
• \( r \) is the radius of the orbit, calculated by adding Earth's radius to the altitude of the satellite.
• \( T \) is the period of orbit.

In the exercise, the satellite's orbital speed is calculated to be approximately \(7482\, \text{m/s}\). This speed is critical for ensuring that the satellite maintains its orbit without the risk of descending to Earth's surface or flying off into space.