Problem 764 - JEE Physics

Question

4 A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the center of earth in new orbit is two times of the earlier orbit. The time period in second orbit is \(\ldots \ldots \ldots \ldots\) hours. (A) \(4.8\) (B) \(48 \sqrt{2}\) (C) 24 (D) \(24 \sqrt{2}\)

Step-by-Step Solution

Answer

The time period of the satellite in its second orbit is \(24 \sqrt{2}\) hours.

Step 1: Recall Kepler's Third Law of Planetary Motion

According to Kepler's Third Law, the square of the orbital period (T) of a satellite is directly proportional to the cube of the semi-major axis (a) of its orbit. Mathematically, it is represented as: \[ T^2 \propto a^3 \] For any two orbits, we can write this relation as: \[ \frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3} \]

Step 2: Apply the Information Given

In the exercise, it's given that the distance from the center of Earth in the new orbit (second orbit) is two times the distance in the earlier orbit (initial orbit). So, we can write: \[ a_2 = 2a_1 \] Now, the time period of a geostationary satellite in its initial orbit is 24 hours. So, \[ T_1 = 24 \] Our goal is to find the time period in the second orbit, which we'll represent as \(T_2\).

Step 3: Substitute Values into the Relation and Solve for T_2

Plug the values into the relation we got from Step 1: \[ \frac{24^2}{T_2^2} = \frac{a_1^3}{(2a_1)^3} \] Simplify the equation: \[ \frac{576}{T_2^2} = \frac{a_1^3}{8a_1^3} \] Now, we can see that the \(a_1^3\) term cancels out: \[ \frac{576}{T_2^2} = \frac{1}{8} \] Now, isolate the term \(T_2^2\): \[ T_2^2 = 576 \cdot 8 \] Solve for \(T_2\): \[ T_2 = \sqrt{4608} = 24 \sqrt{2} \] The time period of the satellite in its second orbit is \(24 \sqrt{2}\) hours, so the correct option is (D) \(24 \sqrt{2}\).

Key Concepts

Geostationary Satellite

A geostationary satellite is a fascinating concept in satellite technology. These satellites are positioned at a specific point over the Earth's equator and rotate in sync with the Earth's rotation. This means they appear to be "stationary" over one particular spot on the Earth’s surface. This unique characteristic is achieved by placing the satellite in a circular orbit at around 35,786 kilometers above the Earth’s equator.
Geostationary satellites are primarily used for communications, weather monitoring, and broadcasting. Since they stay over the same geographic location, they are excellent for providing consistent coverage over specific regions. For instance, they can provide uninterrupted transmission of satellite TV or continuous real-time weather updates.
To maintain a geostationary orbit, the satellite must match the Earth's rotational speed, which is exactly 24 hours a day. Such satellites are crucial for the modern world's connectivity, making them vital to our daily lives.

Orbital Period

The orbital period, often called "period", is the time a satellite takes to complete one full orbit around the Earth or any other celestial body. For satellites, this typically means how long it takes to go around the Earth once.
When considering a geostationary satellite, the orbital period is precisely 24 hours. This synchronization is what allows the satellite to remain fixed above a particular point on Earth. For satellites in non-geostationary orbits, the periods can be much shorter or longer, depending on their distance from Earth.
It's important to understand that closer orbits have shorter periods, due to stronger gravitational pulls, resulting in faster speeds. Conversely, satellites in higher orbits take longer to travel around the Earth due to weaker gravitational forces. Kepler's Third Law helps students understand the mathematical relationship between a satellite's orbital period and its distance from the center of the Earth.

Planetary Motion

Planetary motion refers to the movement of planets and other celestial objects, including satellites, as they orbit due to the gravitational forces. Johannes Kepler formulated three famous laws of planetary motion in the 17th century, which describe how planets move around the Sun. These laws apply not only to planets but also to any orbiting bodies, such as satellites.
Kepler's Third Law is particularly useful for calculating the relationship between the orbital period of a satellite and the size of its orbit. It states that the square of the orbital period of a satellite is proportional to the cube of the semi-major axis of its orbit. This mathematical relationship offers a deeper understanding of how different orbits work and allows scientists to predict satellite motions with great accuracy.
By observing planetary motion principles, engineers can design and position satellites effectively to ensure they fulfill their intended purposes, like communications or weather observation.

Semi-Major Axis

The semi-major axis is a crucial element in understanding orbits, including those of satellites. It is essentially the longest radius of an elliptical orbit, extending from the center of the orbit to its perimeter. This measurement helps determine the size of the orbit itself.
In the context of a geostationary satellite, the semi-major axis is the same as the satellite's distance from the Earth’s center in its circular orbit, as these orbits are circular rather than elliptical for geostationary purposes. This distance is pivotal because it directly influences the satellite's orbital period, according to Kepler's Third Law.
To put it simply, increasing the semi-major axis (the orbit's size around the Earth) leads to a longer orbital period. Therefore, for the satellite in the problem, when it moves to an orbit twice as far from Earth's center, its time to complete an orbit also increases, as demonstrated through the mathematical calculations in the original solution.