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MHT CET 2026 (20 May, Shift 1) · Satellites - Orbital Velocity & Period

Satellites - Orbital Velocity & Period question from MHT CET 2026 (20 May, Shift 1)

Q5MHT CET 2026 · 20 MAY · SHIFT 1Easy

Satellites - Orbital Velocity & Period

The geostationary satellite is revolving round the earth of mean radius RR. If gg is the acceleration due to gravity at earth's surface, ω\omega is the angular speed of rotation of earth about its own axis and rr is the radius of orbit of a satellite, then the correct relation between rr, ω\omega, gg and RR is

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Answer: (b)

  1. For the orbit: GMmr2=mω2r⇒GM=ω2r3\frac{GMm}{r^2} = m\omega^2 r \Rightarrow GM = \omega^2 r^3.
  2. At Earth's surface GM=gR2GM = gR^2.
  3. gR2=ω2r3⇒R2=r3ω2ggR^2 = \omega^2 r^3 \Rightarrow R^2 = \frac{r^3\omega^2}{g}
  4. So the answer is (b) R2=r3ω2gR^2 = \frac{r^3\omega^2}{g}.
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