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# A geostationary satellite is orbiting the earth at height of $5R$ above the surface of the earth. $R$ being the radius of the earth. The time period of another satellite in hours at a height $2R$ from the surface of the earth is(A) $6\sqrt 2$(B) $6/\sqrt 2$(C) $5$(D) $10$

Last updated date: 09th Aug 2024
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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.