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Assuming the radius of the earth to be 6.4 X 10⁶ m. What is the time period T and speed of satellite for equatorial orbit at 1.4 X 10³ km above the surface of the earth?

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6831 s and 2144 ms^-1

34155 s and 3204 ms^-1

2431 s and 3514 ms^-1

6842 s and 7163 ms^-1

answer is A.

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Detailed Solution

$r = R_{earth} + h = (6.4 \times 10^{6} + 1.4 \times 10^{6}) m$

$\Rightarrow r = 7.8 \times 10^{6} m$

$∴ T = {[\frac{4 π^{2} r^{3}}{G M_{earth}}]}^{1 / 2} = 6842 s$

Speed of satellite, $v = \sqrt{\frac{G M_{earth}}{r}} = 7162.93 = 7163 {ms}^{- 1}$

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