Question

A geo-stationary satellite is orbiting the earth of a height of \(6 \mathrm{R}\) above the surface of earth, \(\mathrm{R}\) being the radius of earth. The time period of another satellite at a height of \(2.5 \mathrm{R}\) from the surface for earth is \(=\ldots \ldots \ldots \ldots\) (A) 6 (B) \(6 \sqrt{2}\) (C) 10 (D) \(6 / \sqrt{2}\)

Short answer

The time period of another satellite at a height of \(2.5R\) from the surface for earth is calculated to be 12 hours. However, none of the given options are correct.

Step-by-step solution

Find the formula for the time period of orbiting satellites

Using Kepler's Third Law, we can derive an expression for the time period of a satellite in orbit. The law states that the square of the orbital period of a satellite (T) is directly proportional to the cube of its semi-major axis (r). Mathematically, this can be expressed as \[T^2 \propto r^3,\] where r is the total distance (height + Earth's radius) from the center of the Earth to the satellite.

Calculate the time period of the geo-stationary satellite

In the case of the geo-stationary satellite, we know that its height is \(6R\) above the Earth's surface. Therefore, its total distance from the Earth's center is \[r_1 = R + 6R = 7R.\] Since the satellite is geo-stationary, we know that its time period \(T_1\) is equal to 24 hours. Therefore, from Kepler's Law, we have \[T_1^2 \propto r_1^3 \Rightarrow (24)^2 \propto (7R)^3.\]

Calculate the distance of the second satellite from the Earth's center

We are given that the second satellite is at a height of \(2.5R\) from the Earth's surface. Therefore, its total distance from the Earth's center is \[r_2 = R + 2.5R = 3.5R.\]

Apply Kepler's Third Law to the second satellite

Let the time period of the second satellite be \(T_2\). From Kepler's Third Law, we have \[T_2^2 \propto r_2^3 \Rightarrow T_2^2 \propto (3.5R)^3.\]

Find the ratio of time periods of the two satellites

Dividing the equations obtained in Step 2 and Step 4, we get \[\frac{T_2^2}{T_1^2} = \frac{(3.5R)^3}{(7R)^3}.\]

Solve for the time period of the second satellite

Plugging in the values of \(T_1\) and simplifying, we get \[\frac{T_2^2}{(24)^2} = \frac{(3.5R)^3}{(7R)^3} \Rightarrow T_2 = 24 \times \frac{3.5}{7} = 12\] Since we only have options in multiples of 6, we can compare this value to the options. The closest answer is 10, which corresponds to option (C). However, since we calculated the time period to be 12, none of the given options are correct.

Key concepts

Geo-Stationary Satellite

A geo-stationary satellite is a unique type of satellite that appears to remain stationary in the sky from the perspective of an observer on Earth. It achieves this by orbiting the Earth at the same rotational speed that the planet revolves around its axis.
Geo-stationary satellites orbit at the equatorial plane at a very specific altitude, approximately 35,786 kilometers above Earth. This placement allows them to complete one orbit exactly every 24 hours.
Due to these characteristics, they are extremely useful for telecommunications, weather forecasting, and broadcasting, as they provide uninterrupted, consistent coverage of the same geographical area.

Orbital Period

The orbital period of a satellite is the time it takes to make one complete orbit around the Earth. Understanding the orbital period is crucial for planning satellite operations and ensuring consistent data transmission.
Based on Kepler's Third Law, the orbital period is proportional to the semi-major axis of the satellite's orbit. This means that satellites further from Earth generally have longer orbital periods. For a geo-stationary satellite, the orbital period is exactly 24 hours.
When calculating the orbital periods of other satellites, such as those at different altitudes (e.g., 2.5 times Earth's radius), the orbital period changes and can be derived using Kepler's mathematical principles.

Semi-Major Axis

The semi-major axis is one of the most critical parameters in understanding a satellite's orbit. It refers to the longest radius of an elliptical orbit and is a key factor in determining a satellite's orbital period.
For circular orbits, such as those used by many geo-stationary satellites, the semi-major axis is simply the average distance from the Earth's center to the satellite, identical to the radius of a circle. When calculating the orbital period, the semi-major axis plays a pivotal role depending on its power of three, according to Kepler's Third Law.
In the exercise, the semi-major axis helps determine the distance the satellite travels from the Earth's center.

Earth's Radius

Earth's radius is an important constant when considering calculations related to satellite orbits. It represents the mean distance from the center of the Earth to its surface, approximately 6,371 kilometers.
When calculating the distance from the Earth's center to a satellite, especially in problems involving Kepler’s Third Law, Earth's radius serves as a base measurement. In the exercise, multiplying Earth's radius by the given height ratios allows for determining the total distance involved in the satellite's orbit.
Understanding Earth's radius is essential for accurately solving problems involving satellite orbits, as it helps in precisely defining the total distance between the satellite and Earth's center, which is needed for applying formulas correctly.