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Question

A geostationary satellite orbits around the earth in a circular orbit of radius $$36000 km$$. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth's surface will approximately be $$($$ Given: $$R_{Earth}=6400$$ km $$)$$


A
12h
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B
1h
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C
2h
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D
4h
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Solution

The correct option is A $$2 h$$
For a satellite of mass $$m$$ moving with a velocity $$v$$ in a circular orbit of radius $$r$$ around the earth of mass $$M$$, we have,
$$\dfrac {mv^2}{r}=\dfrac {GMm}{r^2}$$ or $$v=\sqrt {\dfrac {GM}{r}}$$
But $$v=\dfrac {2\pi r}{T}$$.
Thus $$\dfrac {2\pi r}{T}=\sqrt {\dfrac {GM}{r}}$$ or $$T\propto r^{\dfrac {3}{2}}$$
$$\therefore \dfrac {T_2}{T_1}=\left(\dfrac {r_2}{r_1}\right)^{\dfrac {3}{2}}$$ ...................$$(1)$$
Given $$r_2=6400$$ km and $$r_1=36000$$ km. For a geostationary satellite $$T_1=24$$ h. Using these values in $$(1)$$, we have get,
$$T_2=24\times \left(\dfrac {64}{360}\right)^{\dfrac {3}{2}}=1.8$$ hours.
Hence the closest choice is (c).

Physics

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