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Physical World

Question

Near earth's surface, time period of a satelliteis 4 hrs. Find its time period at height '4R' from the centre of earth

32 hrs

$(\frac{1}{8 \sqrt[3]{2}}) hrs$

8 $\sqrt[3]{2}$ hrs

16hrs

Solution

32 hrs

We know $T^{2} = (\frac{4 π^{2}}{{GM}_{s}}) R^{3}$

T²∝ R³

$∴ {(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{4 R}{R})}^{3 / 2} \Rightarrow T_{2} = 4 \times 8 = 32 hrs$