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# If the Height of a Satellite Completing One Revolution Around the Earth in T Seconds is H1 Meter, Then What Would Be the Height of a Satellite Taking 2 √ 2 T Seconds for One Revolution? - Science and Technology 1

Answer in Brief

Solve the problem.

If the height of a satellite completing one revolution around the earth in T seconds is h1 meter, then what would be the height of a satellite taking  $2\sqrt{2} T$ seconds for one revolution?

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#### Solution

Time period of the satellite is given as

$T = \frac{2\pi(R + h)\sqrt{R + h}}{\sqrt{GM}}$
When the height of the satellite is h1, it takes T time to revolve around the Earth.
Thus,
$T = \frac{2\pi(R + h_1 )\sqrt{R + h_1}}{\sqrt{GM}} = \frac{2\pi(R + h_1 )^\frac{3}{2}}{\sqrt{GM}}$  ....(i)
When the satellite takes $2\sqrt{2} T$  time to revolve around the Earth, let it be at height h2. Thus,
$2\sqrt{2}T = \frac{2\pi(R + h_2 )\sqrt{R + h_2}}{\sqrt{GM}} = \frac{2\pi \left( R + h_2 \right)^\frac{3}{2}}{\sqrt{GM}}$
Dividing (ii) by (i), we get
$2\sqrt{2} = \frac{\left( R + h_2 \right)^\frac{3}{2}}{\left( R + h_1 \right)^\frac{3}{2}}$
$\Rightarrow \frac{R + h_2}{R + h_1} = 2$
$\Rightarrow h_2 = R + 2 h_1$
Concept: High Earth Orbits : (Height from the earth’s surface > 35780 km)
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