Question
Question: If \({r_2} = 3{r_1}\) and time period of revolution for \(B\) be \(T\) than time taken by \(A\) in m...
If r2=3r1 and time period of revolution for B be T than time taken by A in moving from position 1 to position 2 is:
(A). T23
(B). T23
(C). 33T2
(D). 3T2
Solution
You can start by calculating the average radius of satellite A when it moves from position 1 to position 2 by using the equation a=2r1+r2. Then use the T2T12=r23a3 to find the value of T1. Then divide the time period of satellite A by half to reach the solution.
Complete step-by-step answer:
In the given problem we are given two satellites A and B. B is revolving around the earth with a radius r2 and A is revolving around the earth with a radius r1.
Given r2=3r1
Let the average radius of satellite A (when it goes from position 1 to position 2) bea.
We know that a will be the average radius of satellite A, so
a=2r1+r2
a=2r1+3r1
a=2r1
Kepler’s law are very important laws in physics, these laws are used to describe the motion of planets and some other celestial bodies.
Let’s take the time taken of the period of revolution of satellite A in the elliptical path as T1
We know by Kepler’s law
T2T12=r23a3
T2T12=(3r1)3(2r1)3=27r138r13
T2T12=278
T1=278T
If we see the movement of satellite A from position 1 to position 2 is half of the whole revolution so the time taken by the satellite in moving from position 1 to position 2 is 2T1=332T.
Hence, option C is the correct choice.
Note: In the problem there is no specific mention that the objects A and B are satellites. But in these types of questions, it is very important to figure out what the objects are. We figured it out in this case as in this case, the image provided in the problem has earth in the center and satellites revolve around the earth.