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MHT CET 2023 (9 May, Shift 2) · Satellites - Orbital Velocity & Period

Satellites - Orbital Velocity & Period question from MHT CET 2023 (9 May, Shift 2)

Q34MHT CET 2023 · 9 MAY · SHIFT 2Medium

Satellites - Orbital Velocity & Period

Periodic time of a satellite revolving above the earth's surface at a height equal to radius of the earth 'RR' is [ g=g = acceleration due to gravity]

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Answer: (b)

  1. T=2πr3GM=2πr3gR2T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{r^3}{gR^2}}
  2. At height RR: r=2Rr = 2R, so r3=8R3r^3 = 8R^3.
  3. T=2π8Rg=4π2RgT = 2\pi\sqrt{\frac{8R}{g}} = 4\pi\sqrt{\frac{2R}{g}}
  4. So the answer is (b).
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