Answer

Verified

388.2k+ views

**Hint**Hint: A geostationary satellite is the one which has the same angular velocity, hence the same time period as that of the earth. The time period of the earth is equal to $24$ hours, so the time period of the given geostationary satellite is also equal to $24$ hours. For determining the time period of the second satellite, we need to use Kepler's third law which states that the square of the time period is proportional to the cube of the radius of orbit.

**Complete step-by-step solution:**

We know that a geostationary satellite orbits about the earth with the same angular velocity as that of the earth. So the geostationary satellite must have the same time period as that of the earth. We know that the time period of the earth is equal to $24$ hours, so the geostationary satellite has the time period of $24$ hours, that is,

${T_1} = 24hr$

Now, the height of the geostationary satellite above the earth’s surface is equal to $5R$. So the radius of its orbit is

${R_1} = 5R + R$

$ \Rightarrow {R_1} = 6R$

Also, the height of the second satellite is given as $2R$. So its orbital radius is given by

${R_2} = 2R + R$

$ \Rightarrow {R_2} = 3R$

Now, from the Kepler’s third law we know that the square of the time period is proportional to the cube of the orbital radius, that is,

\[{T^2} \propto {R^3}\]

$ \Rightarrow {\left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)^2} = {\left( {\dfrac{{{R_2}}}{{{R_1}}}} \right)^3}$

Putting \[{R_1} = 6R\] and \[{R_2} = 3R\], we get

${\left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)^2} = {\left( {\dfrac{{3R}}{{6R}}} \right)^3}$

$ \Rightarrow {\left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^3}$

Taking square root both the sides, we get

$\dfrac{{{T_2}}}{{{T_1}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{3}{2}}}$

$ \Rightarrow {T_2} = \dfrac{{{T_1}}}{{2\sqrt 2 }}$

Finally substituting ${T_1} = 24hr$ we get the time period of the second satellite as

${T_2} = \dfrac{{24}}{{2\sqrt 2 }}hr$

$ \Rightarrow {T_2} = 6\sqrt 2 hr$

Thus, the time period of the second satellite is equal to $6\sqrt 2 $ hours.

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

**Note:**Do not take the heights of the satellite as their orbital radius. This is because the radius is measured from the centre of the orbit. But the height is measured from the surface of the earth. Since the orbit of a satellite is concentric with the earth, so the orbital radius will be the sum of the height and the radius of earth.

Recently Updated Pages

Which of the following is correct regarding the Indian class 10 social science CBSE

Who was the first sultan of delhi to issue regular class 10 social science CBSE

The Nagarjuna Sagar project was constructed on the class 10 social science CBSE

Which one of the following countries is the largest class 10 social science CBSE

What is Biosphere class 10 social science CBSE

Read the following statement and choose the best possible class 10 social science CBSE

Trending doubts

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Why is the Earth called a unique planet class 6 social science CBSE