(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the orbital radius $(r)$: $

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(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the orbital radius $(r)$: $T^2 \propto r^3$,which implies $T \propto r^{3/2}$.Given that the initial time period $T_1 = 24$ hours (for a geostationary orbit) and the new radius $r_2 = 2r_1$.The new time period $T_2$ is given by: $T_2 = T_1 \times (r_2/r_1)^{3/2}$.Substituting the values: $T_2 = 24 \times (2)^{3/2} = 24 \times 2\sqrt{2} = 48\sqrt{2}$ hours.

Class 11 Physics Gravitation Kepler’s laws of Planetary Motion

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$A$ satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of the Earth in the new orbit is $2$ times that of the earlier orbit. The time period in the second orbit is:

A
$4.8$ hours
B
$48\sqrt{2}$ hours
C
$24$ hours
D
$24\sqrt{2}$ hours

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Gravitation

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Kepler’s laws of Planetary Motion

The time period of a geostationary satellite is $24 \; h$,at a height $6 R_{E}$ ($R_{E}$ is the radius of the Earth) from the surface of the Earth. The time period of another satellite whose height is $2.5 R_{E}$ from the surface will be:

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$A$ satellite moves in a circle around the earth. The radius of this circle is equal to one half of the radius of the moon's orbit. The satellite completes one revolution in

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When a planet revolves around the Sun,in general,for the planet:

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If the distance of the earth from the Sun is $1.5 \times 10^8 \, km$. Then the distance of an imaginary planet from the Sun,if its period of revolution is $2.83$ years,is $............. \times 10^8 \, km$.

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$A$ satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of the Earth in the new orbit is $2$ times that of the earlier orbit. The time period in the second orbit is:

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The time period of a geostationary satellite is $24 \; h$,at a height $6 R_{E}$ ($R_{E}$ is the radius of the Earth) from the surface of the Earth. The time period of another satellite whose height is $2.5 R_{E}$ from the surface will be:

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When a planet revolves around the Sun,in general,for the planet:

If the distance of the earth from the Sun is $1.5 \times 10^8 \, km$. Then the distance of an imaginary planet from the Sun,if its period of revolution is $2.83$ years,is $............. \times 10^8 \, km$.

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