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JEE Main 2024 (4 Apr, Shift 2) · Satellites - Orbital Velocity & Period

Satellites - Orbital Velocity & Period question from JEE Main 2024 (4 Apr, Shift 2)

Q21JEE Main 2024 · 4 APR · SHIFT 2Easy

Satellites - Orbital Velocity & Period

Correct formula for height of a satellite from earth's surface is:

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Answer: (a)

  1. Kepler's law: T2=4π2r3GMT^2 = \frac{4\pi^2 r^3}{GM} with r=R+hr = R + h.
  2. Use GM=gR2GM = gR^2: r3=T2gR24π2r^3 = \frac{T^2 g R^2}{4\pi^2}.
  3. h=r−R=(T2R2g4π2)1/3−Rh = r - R = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} - R
  4. So the answer is (a).
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