Problem 784 - JEE Physics

Question

The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to four times the previous value, the new time period will become \(\ldots \ldots \ldots\) hours (A) 10 (B) 120 (C) 40 (D) 80

Step-by-Step Solution

Answer

The new time period of the satellite will be 40 hours when the separation between the Earth and the satellite is increased to four times the previous value.

Step 1: Write the proportionality equation

Since the square of the period is proportional to the cube of the semi-major axis, we can write the equation as: \(T_1^2 / T_2^2 = a_1^3 / a_2^3\) where - \(T_1\) is the initial time period, - \(T_2\) is the new time period, - \(a_1\) is the initial separation between the Earth and the satellite, and - \(a_2\) is the new separation (four times the initial value).

Step 2: Plug in values and solve for the new time period

We are given the initial time period \(T_1 = 5 \) hours and the new separation \(a_2 = 4a_1\). We need to find \(T_2\). Now we can set up the proportionality equation: \(\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}\) Substitute the given values: \(\frac{(5)^2}{T_2^2} = \frac{a_1^3}{(4a_1)^3}\) Simplify the equation: \(\frac{25}{T_2^2} = \frac{a_1^3}{64a_1^3}\) Now cross multiply: \(25 \cdot 64a_1^3 = T_2^2 \cdot a_1^3\) Divide by \(a_1^3\) and simplify: \(25 \cdot 64 = T_2^2\) Calculate the square root of both sides: \(T_2 = \sqrt{25 \cdot 64}\) \(T_2 = 5 \cdot 8 = 40\) So the new time period will be 40 hours. The correct answer is (C) 40.

Key Concepts

Time Period of Satellite

The time period of a satellite refers to the time it takes for the satellite to complete one full orbit around the Earth. This is an important concept in satellite motion, as it helps us understand the satellite's operating cycle and its position relative to Earth over time. The time period is influenced by several factors, including the altitude of the satellite and the gravitational pull exerted by the Earth.

The time period (\( T \)) is closely related to the distance from the Earth, measured as the semi-major axis of the orbit. For a satellite orbiting Earth, the period is directly affected by this separation, which, in simpler terms, means the higher the altitude, the longer the time period. This relationship can be mathematically expressed using Kepler's laws, which allow us to predict how changes in the orbit affect the time it takes for a satellite to circle the Earth once.

In practical terms, understanding the time period is crucial for synchronizing communication satellites, GPS operations, and even in planning space missions where timing is critical.

Kepler's Third Law

Kepler's Third Law is a fundamental principle in orbital mechanics that describes the relationship between the time period of an orbiting object and the semi-major axis of its orbit. Specifically, it states that the square of the orbital period (\( T^2 \)) is proportional to the cube of the semi-major axis (\( a^3 \)) of its orbit. Mathematically, this is expressed as:
\[ T^2 \propto a^3 \]
This law reveals that there is a predictable pattern between how far a satellite is from the planet it orbits and how long it takes to complete one orbit. Essentially, if you know the distance from the planet, you can determine how long the orbit will take, and vice versa.

Kepler's Third Law is used extensively in calculating satellite trajectories and in understanding the dynamics of celestial bodies, making it an indispensable tool in astronomy and astrophysics. By applying this law, scientists and engineers ensure that satellites achieve their desired paths and function correctly in orbit.

The significance of Kepler's law lies in its simplicity and the accuracy with which it can describe real-world orbital systems, ranging from the satellites around our planet to the planets orbiting distant stars.

Orbital Mechanics

Orbital mechanics is the study of the motions of artificial satellites and natural celestial bodies under the influence of gravitational forces. It encompasses the principles and equations that govern the paths of these bodies as they orbit planets or other stars.

The fundamental forces at play include gravity, which acts as the central force pulling a satellite towards the planet, and inertia, which keeps the satellite moving along its orbital path. The balance between these forces determines the nature of the orbit, whether it is circular, elliptical, or another shape entirely.

In the context of satellite motion, orbital mechanics is crucial for various applications including:

• Designing satellite trajectories and ensuring they reach their intended orbits.
• Optimizing fuel consumption during launch and maneuvers.
• Predicting satellite positions for communication and observation purposes.
• Understanding potential orbital decay and collision avoidance.

Grasping the basics of orbital mechanics allows scientists and engineers to develop efficient, safe, and economical methods for space exploration and for maintaining the complex networks of satellites that support modern life.