Two satellites A and B revolve round the same planet in coplanar … - Vidyadip

MCQ

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 $10^{4}\; km$. The speed of $B$ is relative to $A$. When they are closed in $k m / h$ is

A
$3 \pi \times 10^{4}$
B
zero
C
$2 \pi \times 10^{4}$
✓
$\pi \times 10^{4}$

✓

Answer

Correct option: D.

$\pi \times 10^{4}$

d
From Kepler's law,

$\frac{T_{A}^{2}}{T_{B}^{2}}=\frac{r_{A}^{3}}{r_{B}^{3}} \Rightarrow \frac{1^{2}}{8^{2}}=\frac{\left(10^{4}\right)^{3}}{r_{B}^{3}}$

$\Rightarrow r_{B}^{3}=64 \times\left(10^{4}\right)^{3}$

$\therefore r_{B}=4 \times 10^{4} km$

Speed of satellite $A, v_{A}=\frac{2 \pi T_{A}}{T_{A}}=\frac{2 \pi \times 10^{4}}{1}$

$=2 \pi \times 10^{4} km / h$

Speed of satellite $B, v_{B}=\frac{2 \pi r_{B}}{T_{B}}$

$\frac{2 \pi \times 4 \times 10^{4}}{8}$

$=\pi \times 10^{4} km / h$

The speed of $B$ relative to $A$ when they are closed.

$v_{B A}=v_{A}-v_{B}$

$=2 \pi \times 10^{4}-\pi \times 10^{4}$

$=\pi \times 10^{4} km / h$

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