An earth's satellite moves in a circular orbit with an orbital speed $6280$ ms $^{- 1}$ . Find the time of revolution.

130 m i n

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145 m i n

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155 m i n

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175 m i n

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Solution

The correct option is B $175 m i n$

$v_{0} = \sqrt{\frac{G M}{r}}$ or $r = \frac{G M}{v_{0}^{2}}$ and

$T = \frac{2 π r^{3 / 2}}{\sqrt{G M}} = \frac{2 π G M}{v_{0}^{3}}$

$T = \frac{2 π \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{(6280)^{2} \times 6280 \times 60} = 175 m i n .$

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An earth's satellite moves in a circular orbit with an orbital speed 6280 ms−1. Find the time of revolution.

The correct option is B 175min v0=√GMr or r=GMv20 and T=2πr3/2√GM=2πGMv30 T=2π×6.67×10−11×6×1024(6280)2×6280×60=175min.

An earth's satellite moves in a circular orbit with an orbital speed $6280$ ms $^{- 1}$ . Find the time of revolution.

The correct option is B $175 m i n$

$v_{0} = \sqrt{\frac{G M}{r}}$ or $r = \frac{G M}{v_{0}^{2}}$ and

$T = \frac{2 π r^{3 / 2}}{\sqrt{G M}} = \frac{2 π G M}{v_{0}^{3}}$

$T = \frac{2 π \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{(6280)^{2} \times 6280 \times 60} = 175 m i n .$