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# If the height of a satellite completing one Revolution around the earth in T seconds is ${h_1}$ meter, then what would be the height of a satellite taking $2\sqrt 2$ seconds for one revolution?

Last updated date: 21st Jul 2024
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Hint: In this question we will use the third law of Kepler. We will also know about the basics of Kepler's third law. Also, by substituting the given values in the formula we get the required result. Further, we will discuss all the three laws of Kepler.
Formula used:
${T^2} \propto {R^3}$

As we know, there are three laws given by Kepler. These laws describe the orbit of planetary bodies about the sun.
In this question we will use the Kepler’s third law, according to this law the square of the orbital periods T of the planet is directly proportional to the cube of the semi major axes of their orbits R. Also, Kepler's third Law shows that the period for a planet to orbit around the Sun increases rapidly with the radius of the planet’s orbit.
In this question we have,
\eqalign{& {T_1} = T \cr & {R_1} = {h_1} \cr & {T_2} = 2\sqrt 2 T \cr}
We know that, according to Kepler’s Law we get:
${T^2} \propto {R^3}$
Now, by applying Kepler’s law we get:
$\dfrac{{T_1^2}}{{T_2^2}} = \dfrac{{R_1^3}}{{R_2^3}}$
Substituting the given values in above equation, we get:
$\dfrac{{{T^2}}}{{T_2^2}} = \dfrac{{{{(R + {h_1})}^3}}}{{{{(R + {h_2})}^3}}}$
Further putting the value of T we get:
\eqalign{& \dfrac{{{T^2}}}{{8T_{}^2}} = \dfrac{{{{(R + {h_1})}^3}}}{{{{(R + {h_2})}^3}}} \cr & \Rightarrow {(R + {h_2})^3} = 8{(R + {h_1})^3} \cr}
\eqalign{& \Rightarrow R + {h_2} = 2(R + {h_1}) \cr & \therefore {h_2} = R + 2{h_1} \cr}
Therefore, we get the required result of the height of a satellite taking $2\sqrt 2$ seconds for one revolution