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Question: In the following four (i) The time period of revolution of a satellite just above the earth’s sur...

In the following four
(i) The time period of revolution of a satellite just above the earth’s surface (Tst)\left( {{T_{st}}} \right)
(ii) The time period of oscillation of ball inside the tunnel bored along the diameter of the earth (Tma)\left( {{T_{ma}}} \right)
(iii) The time period of a simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 newton/kg (Tsp)\left( {{T_{sp}}} \right)
(iv) The time period of an infinite length simple pendulum in the earth’s gravitational field (Tis)\left( {{T_{is}}} \right)

Which of the following is true
a. Tst>Tma{T_{st}} > {T_{ma}}
b. Tst<Tma{T_{st}} < {T_{ma}}
c. Tsp>Tis{T_{sp}} > {T_{is}}
d. All the time periods are equal.

Explanation

Solution

The time period is defined as the time taken by a complete cycle of revolving or oscillating objects to pass a point. The time period is affected by gravity therefore the time period in the above conditions may differ. In the first statement, the satellite is revolving just above the earth’s surface. In this case, we will consider the radius of the orbit equal to the radius of the earth. Use the time period formula in every case to compare them.

Complete step by step answer:
Step 1: The time period of the revolution of a satellite just above the earth’s surface can be calculated by taking the radius of orbit equal to the radius of the earth. Therefore the time period of the satellite is given by
Tst=2πRv{T_{st}} = \dfrac{{2\pi R}}{v}
But we know that the orbital velocity of a revolving object can be written as v=GMRv = \sqrt {\dfrac{{GM}}{R}} ,
Tst=2πRRGM\therefore {T_{st}} = 2\pi R\sqrt {\dfrac{R}{{GM}}}
Tst=2πR\raise0.5ex3/\lower0.25ex2GM\Rightarrow {T_{st}} = 2\pi \dfrac{{{R^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}{{\sqrt {GM} }} …….equation (1)
We know that the acceleration due to the gravity of a satellite is given by g=GMR2g = \dfrac{{GM}}{{{R^2}}} . Put this value in the first equation,
Tst=2πRg\therefore {T_{st}} = 2\pi \sqrt {\dfrac{R}{g}}

Step 2: Now calculate the time period of oscillation of the ball inside the tunnel bored along the diameter of the earth (Tma)\left( {{T_{ma}}} \right) . First express the relation among the angular speed, amplitude, and acceleration for a simple harmonic motion.
a=Aω2\therefore a = A{\omega ^2} , where aa is the acceleration, AA is the amplitude, and ω\omega is the angular speed.
Here, as the body reaches the end of the tunnel the acceleration becomes equal to the earth’s acceleration and the amplitude becomes equal to the radius of the earth.
ω=gR\therefore \omega = \sqrt {\dfrac{g}{R}} …..equation (2)
But we have the other expression for the angular speed ω\omega in terms of the time period
ω=2πTma\therefore \omega = \dfrac{{2\pi }}{{{T_{ma}}}} , where Tma{T_{ma}} is the time period.
Substituting the value of ω\omega into the second equation
2πTma=gR\therefore \dfrac{{2\pi }}{{{T_{ma}}}} = \sqrt {\dfrac{g}{R}}
Tma=2πRg{T_{ma}} = 2\pi \sqrt {\dfrac{R}{g}}
Step 3: Calculate the time period of a simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 newton/kg (Tis)\left( {{T_{is}}} \right) . Write the expression for the time period of the pendulum moving in simple harmonic motion
Tsp=2π1g(1l+1R)\therefore {T_{sp}} = 2\pi \sqrt {\dfrac{1}{{g\left( {\dfrac{1}{l} + \dfrac{1}{R}} \right)}}} , where ll is the length of the simple pendulum.
But in the question the length of the pendulum is equal to the radius of the earth, therefore
Tsp=2π1g(1R+1R)\therefore {T_{sp}} = 2\pi \sqrt {\dfrac{1}{{g\left( {\dfrac{1}{R} + \dfrac{1}{R}} \right)}}}
Tsp=2πR2g\Rightarrow {T_{sp}} = 2\pi \sqrt {\dfrac{R}{{2g}}}

Step 4: Now let us calculate the time period of an infinite length simple pendulum in the earth’s gravitational field (Tis)\left( {{T_{is}}} \right) . The formula for the time period will be the same as in step 3. The only difference is that the length of the pendulum is infinity. Therefore,
Tis=2π1g(1+1R)\therefore {T_{is}} = 2\pi \sqrt {\dfrac{1}{{g\left( {\dfrac{1}{\infty } + \dfrac{1}{R}} \right)}}}
We know that the value of 1\dfrac{1}{\infty } is zero
Tis=2π1g(1R)\therefore {T_{is}} = 2\pi \sqrt {\dfrac{1}{{g\left( {\dfrac{1}{R}} \right)}}}
Tis=2πRg\Rightarrow {T_{is}} = 2\pi \sqrt {\dfrac{R}{g}}

Step 5: Now consider option A. Since Tst{T_{st}} and Tma{T_{ma}} both are equal therefore option A is incorrect. Option B is also incorrect for the same reason.

Step 6: Consider option C. Clearly, Tsp>Tis{T_{sp}} > {T_{is}} therefore the option C is correct.

Step 7: Option D cannot be correct as Tsp{T_{sp}} is not equal to Tis{T_{is}}.

Hence, the correct answer is option (C).

Note: Oscillation is a harmonic motion in which the object vibrates. The motion is called harmonic because musical instruments make such vibrations that in turn cause corresponding sound waves in the air. Therefore the oscillation should not be confused with another type of motion.