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JEE Main 202125 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Morning Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2021 · 25 February · Morning Shift · Physics · Gravitation

Two satellites A and B of masses 200kg and 400kg are revolving...

Two satellites AA and BB of masses 200kg200kg and 400kg400kg are revolving round the earth at height of 600km600km and 1600km1600km respectively. If TA{T}_{A} and TB{T}_{B} are the time periods of AA and BB respectively then the value of TB−TA{T}_{B} - {T}_{A} :

Two satellites A and B of masses 200kg and 400kg are revolving round the earth at height of 600km and 1600km

[ Given : radius of earth =6400km,= 6400km, mass of earth =6×1024kg= 6 \times {10}^{24}kg ]

Options

  1. A4.24×103s4 . 24 \times {10}^{3}s
  2. B3.33×102s3 . 33 \times {10}^{2}s
  3. C1.33×103s1 . 33 \times {10}^{3}s
  4. D4.24×102s4 . 24 \times {10}^{2}s

Correct answer

C. 1.33×103s1 . 33 \times {10}^{3}s

Step-by-step solution

⇒T=2πr3GM⇒ T = 2\pi\sqrt{\dfrac{{r}^{3}}{GM}}

⇒TA=2π(6400+600)×103GM⇒ {T}_{A} = 2\pi\sqrt{\dfrac{\left(6400 + 600\right) \times {10}^{3}}{GM}}

⇒TB=2π×10983GM⇒ {T}_{B} = 2\pi \times {10}^{9}\sqrt{\dfrac{{8}^{3}}{GM}}

⇒TB−TA=2π109GM[88−77]⇒ {T}_{B} - {T}_{A} = \dfrac{2\pi{10}^{9}}{\sqrt{GM}}\left[8\sqrt{8} - 7\sqrt{7}\right]

=314×4.107= 314 \times 4 . 107

=1289.64= 1289 . 64

=1.289×103s= 1 . 289 \times {10}^{3}s

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