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Physics

Mechanics

Easy

Question

$text Two satellites end text S subscript 1 text and end text S subscript 2 text are revolving round a planet in coplanar and concentric circular orbit end text text of radii end text R subscript 1 text and end text R subscript 2 text in the same direction respectively. Their respective periods of revolution are end text 1 hr text and end text 8 hr text . The radius of the orbit of satellite end text S subscript 1 text is equal to end text 10 to the power of 4 km text . Find the relative speed end text text in kmph when they are closest. end text$

$2 π \times 10^{4} kmph$
$π \times 10^{4} kmph$
$\frac{π}{2} \times 10^{4} kmph$
$\frac{π}{3} \times 10^{4} kmph$

The correct answer is: $text end text pi cross times 10 to the power of 4 kmph$

$text By Kepler's end text 3 to the power of text rd end text end exponent text law end text comma T squared over R cubed equals text constant end text therefore fraction numerator T subscript 1 superscript 2 over denominator R subscript 1 superscript 3 end fraction equals fraction numerator T subscript 2 superscript 2 over denominator R subscript 2 superscript 3 end fraction text or end text 1 over open parentheses 10 to the power of 4 close parentheses cubed equals fraction numerator 64 over denominator R subscript 2 superscript 3 end fraction text or end text R subscript 2 equals 4 cross times 10 to the power of 4 k m text Distance travelled in one revolution, end text S subscript 1 equals 2 pi R subscript 1 equals 2 pi cross times 10 to the power of 4 text and end text S subscript 2 equals 2 pi R subscript 2 equals 2 pi cross times 4 cross times 10 to the power of 4 v subscript 1 equals S subscript 1 over t subscript 1 equals fraction numerator 2 pi cross times 10 to the power of 4 over denominator 1 end fraction equals 2 pi cross times 10 to the power of 4 kmph text and end text v subscript 2 equals S subscript 2 over t subscript 2 equals fraction numerator 2 pi cross times 4 cross times 10 to the power of 4 over denominator 8 end fraction equals pi cross times 10 to the power of 4 kmph straight v subscript 1 minus straight v subscript 2 equals 2 straight pi cross times 10 to the power of 4 minus straight pi cross times 10 to the power of 4 equals straight pi cross times 10 to the power of 4 kmph$

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