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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:**

${T^2} \propto {R^3}$

**Complete answer:**

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

**Additional information:**

Here, as we know that according to Kepler's first law each planet's orbit about the Sun is an ellipse shape. This shows that the Sun's center is always located at one focus of the orbital ellipse. The planet follows the ellipse in its orbit, this means that the planet to Sun distance is constantly changing when the planet goes around its orbit.

Next when we come on Kepler's Second Law, it tells that the imaginary line joining a planet and the Sun sweeps equal areas of space during equal time intervals as the planet orbits.

Here, we know that planets do not move with constant speed along their orbits. But, their speed varies with the line joining the centers of the Sun and the planet sweeps out equal parts of an area in equal times.

**Note:**

Note that the planets orbit the Sun in a counterclockwise direction as viewed from above the Sun's North Pole. We should also remember that planet Mercury takes only 88 days to orbit the Sun, our earth takes 365 days, while Saturn requires 10,759 days to orbit the Sun.

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