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Gravitation question

2021 · 18 Mar · Shift 1 · Q61
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  5. /2021 · 18 Mar · Shift 1 · Q61

Gravitation question

2021 · 18 Mar · Shift 1 · Q61

JEE MainPhysicsGravitationMCQ+4 / −1
The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :
  1. A
    9 T
  2. B
    27 T
  3. C
    12 T
  4. D
    3 T
View written solutionFree

Correct answer: B

  1. For a satellite in a circular orbit, Kepler’s third law gives

T2∝r3T^2 \propto r^3T2∝r3

So,

T∝r3/2T \propto r^{3/2}T∝r3/2

  1. Let the first satellite have:
  • orbital radius RRR
  • time period TTT

For the second satellite:

  • orbital radius 9R9R9R
  • time period T′T'T′

Using

T′T=(9RR)3/2\frac{T'}{T} = \left(\frac{9R}{R}\right)^{3/2}TT′​=(R9R​)3/2

we get

T′T=93/2\frac{T'}{T} = 9^{3/2}TT′​=93/2

  1. Now,

93/2=(9)3=33=279^{3/2} = (\sqrt{9})^3 = 3^3 = 2793/2=(9​)3=33=27

Thus,

T′=27TT' = 27TT′=27T

  1. Therefore, the correct option is:

B: 27T27T27T

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