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Motion In Straight Line

Question
CBSEENPH11020591

A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The  time period of another satellite in hours at a height of 2R from the surface of the earth is

  • 5

  • 10

  • 6 square root of 2
  • 6

Solution

C.

6 square root of 2 From space Keplar space third apostrophe straight s space law
straight T squared proportional to space straight r cubed
Hence comma
straight T subscript 1 superscript 2 space proportional to space straight r subscript 1 superscript 3
and
straight T subscript 2 superscript 2 space proportional to space straight r subscript 2 superscript 3
So comma
fraction numerator straight T subscript 2 superscript 2 over denominator straight T subscript 1 superscript 2 end fraction space equals fraction numerator straight r subscript 2 superscript 3 over denominator straight r subscript 1 superscript 3 end fraction
space equals space fraction numerator open parentheses 3 straight R close parentheses cubed over denominator left parenthesis 6 straight R right parenthesis cubed end fraction
Or
fraction numerator straight T subscript 2 superscript 2 over denominator straight T subscript 1 superscript 2 end fraction space equals space 1 over 8
straight T subscript 2 superscript 2 space equals space 1 over 8 straight T subscript 1 superscript 2
straight T subscript 2 space equals space fraction numerator 24 over denominator 2 square root of 2 end fraction space equals space 6 square root of 2 space straight h