Answer

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**Hint:**To answer this question we have to use the formula of the orbital velocity of a revolving satellite. From there we can easily find out the expression for the time period of the satellite. On substituting the values in that expression, we will get the time period of the satellite.

**Formula used:**

The formula used to solve this question are given by

$v = \sqrt {\dfrac{{GM}}{r}} $

$g = \dfrac{{GM}}{{{R^2}}}$

Here $v$ is the orbital velocity of a satellite, $G$ is the gravitational constant, $r$ is the radius of revolution of the satellite, $g$ is the acceleration due to gravity, $M$ is the mass of the earth, and $R$ is the radius of the earth.

**Complete step-by-step answer:**

We know that a satellite revolves around the earth with a velocity known as the orbital velocity, which is given by

$v = \sqrt {\dfrac{{GM}}{r}} $ …………………….(1)

As we know that the angular velocity is given by

$\omega = \dfrac{v}{r}$

Substituting (1)

$\omega = \dfrac{1}{r}\sqrt {\dfrac{{GM}}{r}} $

$ \Rightarrow \omega = \sqrt {\dfrac{{GM}}{{{r^3}}}} $ ………………….(2)

Now, we know that the time period of revolution is given by

$T = \dfrac{{2\pi }}{\omega }$

From (2)

$T = 2\pi \sqrt {\dfrac{{{r^3}}}{{GM}}} $ ………………..(3)

Let the satellite be revolving at a height $h$ above the earth’s surface. So the radius of revolution of the satellite is

$r = R + h$ ……………...(4)

Substituting (4) in (3) we get

$T = 2\pi \sqrt {\dfrac{{{{\left( {R + h} \right)}^3}}}{{GM}}} $

According to the question, the satellite is revolving very near to the earth’s surface. So the height of the satellite above the earth’s surface is negligible compared to the radius of the earth. Therefore substituting $h = 0$ above we get

$T = 2\pi \sqrt {\dfrac{{{R^3}}}{{GM}}} $ …………………..(5)

Now, we know that the acceleration due to gravity is given by

$g = \dfrac{{GM}}{{{R^2}}}$

$ \Rightarrow \dfrac{{{R^2}}}{{GM}} = \dfrac{1}{g}$ ……………………..(6)

Substituting (6) in (5) we get

$T = 2\pi \sqrt {\dfrac{R}{g}} $ ……………………….(7)

Now, we know that the radius of the earth is approximately $6400km$. So we have

$R = 6400km$

$ \Rightarrow R = 64 \times {10^5}m$ ………………...(8)

Also we know that the value of the acceleration due to gravity is approximately equal to $10m{s^{ - 2}}$. So we have

$g = 10m{s^{ - 2}}$ …………………….(9)

Substituting (8) and (9) in (7) we get

$T = 2\pi \sqrt {\dfrac{{64 \times {{10}^5}}}{{10}}} $

\[T = 2\pi \sqrt {64 \times {{10}^4}} \]

On solving we get

$T = 5026.55s$

We know that $1s = \dfrac{1}{{60}}\min $. So we get

$T = \dfrac{{5026.55}}{{60}}\min $

$T = 83.77\min = 60\min + 23.77\min $

We know that $1h = 60\min $. So the time period of the satellite is

$T = 1h{\text{ }}23.77\min $

\[ \Rightarrow T \approx 1h{\text{ }}24\min \]

So the time period of the satellite is nearly equal to $1h$ $24\min $.

So the Assertion is correct.

From (8) we can see that the time period only depends upon the radius of the earth, and the height of the satellite above the earth’s surface is negligible.

So the Reason is incorrect.

**Hence, the correct answer is option C.**

**Note:**We should not assume the radius of the earth to be the height of the satellite. We should remember, the height of a satellite is measured from the surface of the earth, while the radius of the earth is measured from the centre of the earth.

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