Question

Two satellites A and B revolve round the same planet in coplanar circular orbits lying in the same plane. Their periods of revolutions are 1 h and 8 h, respectively. The radius of the orbit of A is 10⁴ km. The speed of B relative to A, when they are close in km/h is [AIIMS 2018]

Moderate

Solution

Given,    $T_{A} = 1 h, T_{B} = 8 h$
and $r_{A} = 10^{4} km$
From Kepler's law,    $\frac{T_{A}^{2}}{T_{B}^{2}} = \frac{r_{A}^{3}}{r_{B}^{3}} \Rightarrow \frac{1^{2}}{8^{2}} = \frac{{(10^{4})}^{3}}{r_{B}^{3}}$
$\Rightarrow r_{B}^{3} = 64 \times {(10^{4})}^{3}$
$∴ r_{B} = 4 \times 10^{4} km$
Speed of satellIte $A, v_{A} = \frac{2 π r_{A}}{T_{A}} = \frac{2 π \times 10^{4}}{1}$
   $= 2 π \times 10^{4} km / h$
Speed of satellIte    $B, v_{B} = \frac{2 π r_{B}}{T_{B}} = \frac{2 π \times 4 \times 10^{4}}{8}$
$= π \times 10^{4} km / h$
The speed of B relative to A when they are close,
$v_{B A} = v_{A} - v_{R} = 2 π \times 10^{4} - π \times 10^{4}$
   $= π \times 10^{4} km / h$

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