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Gravitation question

2012 · Q164
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Gravitation question

2012 · Q164

NEETPhysicsGravitationMCQ+4 / −1
A geostationary satellite is orbiting the earth at a height of 5R above the 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
  1. A
    5
  2. B
    10
  3. C
    62\sqrt 22​
  4. D
    62{6 \over {\sqrt 2 }}2​6​
View written solutionFree

Correct answer: C

According to Kelpner’s law of period T2 ∝\propto∝ R3

T12T22=R13R23=(6R)3(3R)3=8{{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}} = {{{{\left( {6R} \right)}^3}} \over {{{\left( {3R} \right)}^3}}} = 8T22​T12​​=R23​R13​​=(3R)3(6R)3​=8

24×24T22=8{{24 \times 24} \over {T_2^2}} = 8T22​24×24​=8

T22=24×248=72=36×2T_2^2 = {{24 \times 24} \over 8} = 72 = 36 \times 2T22​=824×24​=72=36×2

∴\therefore∴ T2=62{T_2} = 6\sqrt 2 T2​=62​

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