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JEE Main 2026 (22 Jan, Shift 2) · Satellites - Orbital Velocity & Period

Satellites - Orbital Velocity & Period question from JEE Main 2026 (22 Jan, Shift 2)

Q4JEE Main 2026 · 22 JAN · SHIFT 2Easy

Satellites - Orbital Velocity & Period

Given below are two statements:

Statement I: A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.

Statement II: The time period of revolution of the satellite is TT == 2πReg2\pi\sqrt{\frac{R_e}{g}} (for satellite very close to the earth surface), where ReR_e radius of earth and gg acceleration due to gravity. In the light of the above statements, choose the correct answer from the options given below:

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Answer: (c)

  1. Near the surface, gravity gives the centripetal force: mg=mω2Re⇒T=2πRegmg = m\omega^2 R_e \Rightarrow T = 2\pi\sqrt{\frac{R_e}{g}}. Statement II is true.
  2. With g=43πGρReg = \frac{4}{3}\pi G\rho R_e: T=3πGρT = \sqrt{\frac{3\pi}{G\rho}}, which depends only on the density. Statement I is true.
  3. So the answer is (c) both statements are true.
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