Answer in Brief

**Solve the problem.**

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 \[2\sqrt{2} T\] 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 h

_{1}, 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 h

When the satellite takes \[2\sqrt{2} T\] time to revolve around the Earth, let it be at height h

_{2}. 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

\[2\sqrt{2} = \frac{\left( R + h_2 \right)^\frac{3}{2}}{\left( R + h_1 \right)^\frac{3}{2}}\]

\[ \Rightarrow \frac{R + h_2}{R + h_1} = 2\]

\[ \Rightarrow h_2 = R + 2 h_1\]

\[ \Rightarrow h_2 = R + 2 h_1\]

Concept: High Earth Orbits : (Height from the earth’s surface > 35780 km)

Is there an error in this question or solution?

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