Two satellites 'A' and 'B' go around the earth in circular orbits at heights 6R and 4R respectively from the surface of the earth. Assuming the earth to be a uniform sphere of radius 'R', the ratio of the magnitudes of the velocity of satellite A to satellite B is

\sqrt{\frac{3}{2}}

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\sqrt{\frac{5}{7}}

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\frac{5}{7}

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\frac{3}{2}

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Solution

The correct option is B $\sqrt{\frac{5}{7}}$
The velocity of satellite A is
$V_{A} = \frac{2 π r_{A}}{T_{A}}$
Similarly, the velocity of satellite B is
$V_{B} = \frac{2 π r_{B}}{T_{B}}$
Taking the ratio, we get
$\frac{V_{A}}{V_{B}} = \frac{r_{A}}{r_{B}} \times \frac{T_{B}}{T_{A}}$ .........(1)
From Kepler's third law, we have
$T^{2} α r^{3}$
Hence, here we can write
$(\frac{T_{A}}{T_{B}})^{2} = (\frac{r_{A}}{r_{B}})^{3}$
$⟹ \frac{T_{A}}{T_{B}} = (\frac{r_{A}}{r_{B}})^{\frac{3}{2}}$
Substituting in (1), we get
$\frac{V_{A}}{V_{B}} = \frac{r_{A}}{r_{B}} \times (\frac{r_{B}}{r_{A}})^{3 / 2}$
Next, substituting values
$r_{A} = 7 R$
$r_{B} = 5 R$
$\frac{V_{A}}{V_{B}} = \frac{7}{5} \times \frac{5}{7} \times \sqrt{\frac{5}{7}}$
$\frac{V_{A}}{V_{B}} = \sqrt{\frac{5}{7}}$

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