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MHT CET 2026 (19 Apr, Shift 1) · Satellites - Orbital Velocity & Period

Satellites - Orbital Velocity & Period question from MHT CET 2026 (19 Apr, Shift 1)

Q7MHT CET 2026 · 19 APR · SHIFT 1Medium

Satellites - Orbital Velocity & Period

A geostationary satellite is orbiting the earth at a height of 4R4R above the surface of the earth, where RR is the radius of the earth. Another satellite is orbiting the earth at a height 1.5R1.5R from the surface of the earth with periodic time 'T' in hour. The value of 'T' in hour is

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Answer: (b)

  1. Orbit radii from the centre: geostationary r1=5Rr_1 = 5R (T1=24T_1 = 24 h), other r2=2.5Rr_2 = 2.5R.
  2. T∝r3/2T \propto r^{3/2}: T2=24(2.5R5R)3/2=24×122T_2 = 24\left(\frac{2.5R}{5R}\right)^{3/2} = 24\times\frac{1}{2\sqrt{2}}
  3. T2=122=62T_2 = \frac{12}{\sqrt{2}} = 6\sqrt{2} h
  4. So the answer is (b).
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