(D) According to Kepler's Third Law,the time period $T$ of a satellite in a circular orbit of radius $a$ is given by $T = 2\pi \sqrt{\frac{a^3}{GM}}$,

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(D) According to Kepler's Third Law,the time period $T$ of a satellite in a circular orbit of radius $a$ is given by $T = 2\pi \sqrt{\frac{a^3}{GM}}$,where $M$ is the mass of the planet.We know that the acceleration due to gravity at the surface of the planet is $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.Substituting $GM$ into the time period formula: $T = 2\pi \sqrt{\frac{a^3}{gR^2}} = 2\pi a^{3/2} g^{-1/2} R^{-1}$.Comparing this with the given expression $T \propto a^{3/2} g^x R^y$,we get $x = -1/2$ and $y = -1$.

Class 11 Physics Gravitation Motion of Satellites in Circular Orbits and Planets in Elliptical Orbits

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An artificial satellite is revolving around a planet of radius $R$ in a circular orbit of radius $a$. If the time period of revolution of the satellite $T \propto a^{3/2} g^x R^y$,then the values of $x$ and $y$ are respectively. [Note: $g$ is the acceleration due to gravity at the surface of the planet.]

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A
$1, 1/2$
B
$1/2, 1$
C
$-1/2, 1/2$
D
$-1/2, -1$

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Gravitation

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Motion of Satellites in Circular Orbits and Planets in Elliptical Orbits

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The orbital velocity of an artificial satellite in a circular orbit very close to the Earth is $v$. The velocity of a geostationary satellite orbiting in a circular orbit at an altitude of $3R$ from the Earth's surface will be

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Two satellites $A$ and $B$ revolve around the same planet in coplanar circular orbits. Their periods of revolution are $1\, h$ and $8\, h$,respectively. The radius of the orbit of $A$ is $10^{4}\, km$. The speed of $B$ relative to $A$ when they are closest is (in $km/h$):

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The time period of a $1500 \,kg$ satellite is equal to the time period of rotation of the earth. The altitude of the satellite is nearly

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$A$ geostationary satellite is orbiting around an arbitrary planet $P$ at a height of $11R$ above the surface of $P$,where $R$ is the radius of $P$. The time period of another satellite in hours at a height of $2R$ from the surface of $P$ is $........$. The planet $P$ has a rotation period of $24\, \text{hours}$.

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