Earth Satellites - Time Period of Satellite

The time period of a satellite is the time taken by a satellite to complete one full revolution around the Earth. When a satellite is projected to a certain height and given the right horizontal velocity (critical velocity), it moves in a circular orbit. The radius of this orbit is the sum of Earth's radius (R) and the height (h) of the satellite above Earth's surface. Understanding the time period helps in planning satellite launches, communication systems, and GPS technologies.

T = \[2\pi\sqrt{\frac{(R+h)^3}{GM}}\]

Where:

• T = Time period of the satellite (in seconds)
• R = Radius of the Earth
• h = Height of the satellite above Earth's surface
• G = Universal gravitational constant
• M = Mass of the Earth
• (R + h) = r = Radius of the satellite's orbit

• The time period does not depend on the mass of the satellite.
• The time period depends on:
  Mass of the Earth (M)
  Radius of the Earth (R)
  Height of the satellite (h)
• If the height of projection increases, the time period increases.
• Kepler's Third Law: T² ∝ r³ — the square of the time period is directly proportional to the cube of the orbital radius.

• Helps in calculating orbital parameters for satellite launches.
• Essential for designing communication satellites and GPS systems.
• Proves Kepler's Third Law mathematically for artificial satellites.
• Useful for determining geostationary orbit requirements.
• Helps scientists predict satellite positions at any given time.

Problem: An artificial satellite revolves around a planet in a circular orbit close to its surface. Obtain the formula for the period of the satellite in terms of density ρ and radius R of the planet.

Solution:

1. Start with the period formula:

• T = \[2\pi\sqrt{\frac{(R+h)^3}{GM}}\]

2. Since the satellite is close to the surface: R + h ≈ R

3. Express mass in terms of density:

• \[\mathrm{Density}(\rho)=\frac{\mathrm{Mass}(M)}{\mathrm{Volume}(V)}\]
• $M=\rho \times V$

4. Volume of a spherical planet:

• V = \[\frac{4}{3}\pi R^{3}\]

5. Substitute mass:

• M = \[\frac{4}{3}\pi R^{3}\rho\]

6. Put in period formula:

• T = \[2\pi\sqrt{\frac{R^3}{G\times\frac{4}{3}\pi R^3\rho}}\]

7. Simplify:

• T = \[T=\sqrt{\frac{3\pi}{G\rho}}\]

• GPS Satellites: GPS satellites orbit Earth at specific heights with calculated time periods to provide accurate location data to users worldwide.
• Communication Satellites: Geostationary satellites have a time period of exactly 24 hours, matching Earth's rotation, so they appear stationary over one location — used for TV broadcasting and weather monitoring.
• International Space Station (ISS): The ISS orbits close to Earth (about 400 km) and completes one revolution in approximately 90 minutes.
• Weather Satellites: Polar satellites with ~85-minute periods orbit from pole to pole, scanning the entire Earth as it rotates beneath them.
• Moon as a Natural Satellite: The Moon takes about 27.3 days to complete one revolution around Earth, following the same principle of the time period formula.