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Chapter 6: Problem 764

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) \(24 \sqrt{2}\)

Short Answer

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

Step by step solution

01

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: \[ \frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3} \]
02

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 \(T_2\).
03

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) \(24 \sqrt{2}\).

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Most popular questions from this chapter

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