Problem 793 - JEE Physics

Question

A geostationary satellite is orbiting the earth at a height of \(5 \mathrm{R}\) above that of surface of the earth. \(\mathrm{R}\) being the radius of the earth. The time period of another satellite in hours at a height of \(2 \mathrm{R}\) from the surface of earth is \(\ldots \ldots \ldots .\) hr (A) 5 (B) 10 (C) \(6 \sqrt{2}\) (D) \(6 / \sqrt{2}\)

Step-by-Step Solution

Answer

The short answer based on the provided solution is: The time period of the satellite at a height of 2R from the surface of the Earth is 12 hours. However, the correct answer is not among the available options. The closest option would be (B) 10 hours, but it's not the exact answer.

Step 1: Understand the formula needed for the problem

We need to use the formula for the period of a satellite orbiting around a planet (in our case, the Earth). The formula is given by Kepler's Third Law: \[T = 2 \pi \sqrt{\frac{a^3}{GM}}\] where \(T\) is the time period of the satellite, \(a\) is the orbital radius (which is the sum of the Earth's radius and the satellite's height from the surface), \(G\) is the gravitational constant, and \(M\) is the mass of the Earth.

Step 2: Find the time period of the geostationary satellite

We know that a geostationary satellite orbits at a height of 5R above the surface of the Earth. Therefore, its orbital radius is: \[a_1 = R + 5R = 6R\] Now, we can plug in values to find the time period of the geostationary satellite: \[T_1 = 2 \pi \sqrt{\frac{(6R)^3}{GM}}\]

Step 3: Express the time period of the given satellite in terms of the geostationary satellite's time period

The given satellite orbits at a height of 2R from the surface of the Earth, so its orbital radius is: \[a_2 = R + 2R = 3R\] The time period of this satellite is given by: \[T_2 = 2 \pi \sqrt{\frac{(3R)^3}{GM}}\] To find out the time period of the given satellite in terms of the period of the geostationary satellite, we can write: \[\frac{T_2}{T_1} = \frac{2 \pi \sqrt{\frac{(3R)^3}{GM}}}{2 \pi \sqrt{\frac{(6R)^3}{GM}}}\]

Step 4: Solve for the time period of the given satellite

By simplifying the above equation, we have: \[\frac{T_2}{T_1} = \frac{\sqrt{(3R)^3}}{\sqrt{(6R)^3}} = \frac{3R}{6R} = \frac{1}{2}\] \[\Rightarrow T_2 = \frac{1}{2} T_1\] We are given that the time period of the geostationary satellite is 24 hours (one orbit in a day), so: \[T_2 = \frac{1}{2} \cdot 24 = 12\] So, the time period of the given satellite is 12 hours. Since the correct answer is not among the available options, there might be a mistake in the problem statement or the options provided. The closest available option would be (B) 10 hours, but it's not the exact answer.

Key Concepts

Geostationary Satellites

Geostationary satellites are a fascinating topic in orbital mechanics. They appear to remain fixed above a specific point on the equator as the Earth rotates. This unique positioning is achieved by matching the Earth's rotation period, which is 24 hours. To keep up, these satellites must orbit at a specific height known as the geostationary orbit.

Key characteristics of geostationary satellites include:

• They orbit at an altitude of approximately 35,786 kilometers above the Earth's equator.
• Their orbital period is exactly 24 hours, which allows them to remain stationary relative to a point on Earth.
• They are primarily used for telecommunications, weather monitoring, and global positioning systems.

For these satellites to function effectively, their orbit must be precisely calculated and maintained. This involves complex calculations based on Kepler's Third Law and understanding of gravitation.

Orbital Mechanics

Orbital mechanics is the study of the motion of objects in space, often based on the principles laid down by Johannes Kepler. Kepler's laws of planetary motion, especially the third one, are crucial in predicting how satellites behave in space.

According to Kepler's Third Law, the square of the time period of orbit (T) is proportional to the cube of the semi-major axis of the orbit (a). The law is usually represented as:
\[T^2 \propto a^3\]

In practical terms, this means:

• Satellites further from the Earth have longer orbital periods.
• A satellite's orbital radius significantly impacts its time period.
• Calculating the exact position and time period requires precise measurements of both distance and motion.

Understanding these mechanics allows for accurate predictions and adjustments necessary for satellite placement and maintenance.

Time Period of Satellites

The time period of a satellite refers to the duration it takes to complete one full orbit around the Earth. This time can vary greatly depending on its orbital radius. For instance, a satellite in a low Earth orbit may complete an orbit in just about 90 minutes, while those in higher orbits take much longer.

When calculating the time period of satellites, Kepler's Third Law is used once again:
\[T = 2 \pi \sqrt{\frac{a^3}{GM}}\]

Here, \(T\) is the time period, \(a\) is the orbital radius, \(G\) is the gravitational constant, and \(M\) is the mass of the Earth.

• Larger orbital radii lead to longer time periods.
• Geostationary satellites have a fixed time period of 24 hours.
• Calculation errors can lead to discrepancies in satellite operations.

By mastering these calculations, engineers ensure that satellites perform optimally in their designated orbits.