Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively.
If $T_{A}$ and $T_{B}$ are the time periods of A and B respectively then the value of $T_{B} - T_{A}$ :

[Given : radius of earth = 6400 km, mass of earth $= 6 \times 10^{24} k g$ ]

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$4.24 \times 10^{2} s$

$1.33 \times 10^{3} s$

$3.33 \times 10^{2} s$

$4.24 \times 10^{3} s$

answer is B.

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Detailed Solution

$\begin{array}{l} Orbital speed, V = \sqrt{\frac{G M}{r}} \\ Time Period, T = \frac{2 π r}{\sqrt{\frac{G M}{r}}} = 2 π r \sqrt{\frac{r}{G M}} \\ \Rightarrow T = \sqrt{\frac{4 π^{2} r^{3}}{G M}} = 2 π \sqrt{\frac{r^{3}}{G M}} \\ ∴ T_{B} - T_{A} = 2 π \times \sqrt{\frac{{[8000 \times 10^{3}]}^{3}}{6.67 \times 10^{- 11} \times 6 \times 10^{24}}} - 2 π \times \sqrt{\frac{{(7000 \times 10^{3})}^{3}}{6.67 \times 10^{- 11} \times 6 \times 10^{24}}} = 1.289 \times 10^{3} s \\ Closest option is (2) \end{array}$