Q. An artificial satellite is describing an equatorial orbit at $3600 \,km$ above the earth's surface. Calculate its period of revolution ? Take earth radius $6400 \,km$.

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A

$8.71\, hrs$

B

$9.71\, hrs$

C

$10.71\, hrs$

D

$11.71 \,hrs$

Solution:

The time period of satellite is given by
$T^{2} =\frac{4 \pi^{2}}{G M}(R+h)^{3}=\frac{4 \pi^{2}}{g} \frac{(R+h)^{3}}{R^{2}}$
$\Rightarrow T =\frac{2 \pi}{R} \sqrt{\frac{(R+h)^{3}}{g}} $
$\Rightarrow T =\frac{2 \pi}{6400 \times 10^{3}} \sqrt{\frac{10^{21}}{9.8}} $
$\Rightarrow T =31360.78 \sec =8.71\, hrs$

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