Problem 34 - College Physics

Question

What is the period of revolution of a satellite with mass \(m\) that orbits the earth in a circular path of radius \(7880 \mathrm{~km}\) (about \(1500 \mathrm{~km}\) above the surface of the earth)?

Step-by-Step Solution

Answer

The period of the satellite's revolution is approximately 1.77 hours.

Step 1: Understand the problem

We are given the radius of the satellite's orbit around Earth. The task is to find the period of revolution of the satellite, which indicates how long it takes to complete one orbit.

Step 2: Utilize the formula for the gravitational force

The gravitational force acting as the centripetal force on the satellite is given by:\[F = \frac{G \cdot M \cdot m}{r^2} = m \cdot \frac{v^2}{r}\]where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, \(r\) is the radius of orbit, and \(v\) is the orbital speed. Equate these to solve for \(v\).

Step 3: Derive the orbital velocity formula

Simplify the equation from the previous step to solve for the orbital velocity \(v\):\[v = \sqrt{\frac{G \cdot M}{r}}\]This formula calculates the velocity required to maintain the circular orbit.

Step 4: Calculate the orbital period

The orbital period \(T\) (the time taken for one complete orbit) is given by:\[T = \frac{2\pi r}{v}\]Substitute \(v\) from Step 3 into this equation to find \(T\).

Step 5: Plug in known values

Use known values: \(G = 6.67430 \times 10^{-11} \mathrm{~m^3~kg^{-1}~s^{-2}}\), \(M = 5.972 \times 10^{24} \mathrm{~kg}\), and \(r = 7880 \times 10^3 \mathrm{~m}\). Substitute these into the formulas to calculate \(v\) and \(T\).

Step 6: Calculate and conclude

First, calculate \(v\):\[v = \sqrt{\frac{6.67430 \times 10^{-11} \cdot 5.972 \times 10^{24}}{7880 \times 10^3}} \approx 7353.3 \mathrm{~m/s}\]Then, calculate \(T\):\[T = \frac{2\pi \cdot 7880 \times 10^3}{7353.3} \approx 6359.7 \mathrm{~s} \approx 1.77 \mathrm{~hours}\]

Key Concepts

Gravitational Force

Gravitational force is a fundamental natural force that attracts two objects with mass. This force keeps the satellite in orbit around the Earth and acts as the centripetal force ensuring a stable circular path. Here's a closer look at gravitational force:

• The formula for gravitational force is \[F = \frac{G \cdot M \cdot m}{r^2}\]where:
  • \(G\) is the gravitational constant \(6.67430 \times 10^{-11} \mathrm{~m^3~kg^{-1}~s^{-2}}\).
  • \(M\) is the Earth's mass \(5.972 \times 10^{24} \mathrm{~kg}\).
  • \(r\) is the distance from Earth's center to the satellite.

This equation shows that gravitational force decreases with the square of the distance. Understanding this principle helps calculate how the force changes as the satellite orbits at different altitudes.

Orbital Velocity

Orbital velocity is the speed that allows a satellite to maintain its orbit around a celestial body like Earth without escaping into space or crashing down. This speed ensures that the gravitational pull and the inertia of the satellite are balanced.

• The orbital velocity formula is given by:\[v = \sqrt{\frac{G \cdot M}{r}}\]

This formula shows that the orbital velocity depends on the mass of the central body (Earth in this scenario) and the radius of the orbit.

• Increasing the radius decreases the velocity required.
• A faster orbital velocity means a quicker completion of orbit.

It's crucial for satellites to have just the right velocity to prevent losing orbit stability.

Centripetal Force

Centripetal force is essential for any object moving in a circular path. For satellites, gravitational force serves as the centripetal force.

• The role of centripetal force is to continuously pull the satellite toward the Earth, preventing it from flying off tangentially.
• Without this force, the satellite would not follow a circular path.

The concepts of centripetal force and gravitational force are intertwined. When an object, like a satellite, orbits Earth, the centripetal force required is provided by Earth's gravity:

• Both forces satisfy this equation: \[m \cdot \frac{v^2}{r} = \frac{G \cdot M \cdot m}{r^2}\]

Understanding centripetal force helps explain why satellites remain in orbit and how gravitational forces are utilized to stabilize their paths.

Orbital Period Formula

The orbital period is the time it takes for a satellite to complete one full orbit around Earth. Calculating the orbital period involves combining knowledge of orbital velocity and the radius of the orbit.

• The formula for orbital period \(T\) is:\[T = \frac{2\pi r}{v}\]

This tells you the path length \(2\pi r\) and divides it by the speed of orbit \(v\).

• To calculate \(T\), you first need the orbital velocity \(v\), which you can find using the formula for orbital velocity.
• A smaller radius results in a shorter period, meaning the satellite orbits faster.
• Conversely, a larger radius leads to a longer orbital period.

It's practical for calculating how often a satellite will pass over the same point on Earth's surface and is crucial for designing satellite paths.