CUET Preparation Today
CUET
Physics
Gravitation
The time period of a simple pendulum in a satellite orbiting very close to earth surface is equal to |
2π√Rg 82 min ∞ none of these |
∞ |
Since the satellite is orbiting in a circular path, if the point of suspension P towards the centre of earth with an acceleration gp. The bob θ of the pendulum also is accelerating towards the centre of earth with an acceleration gQ ⇒ The relative acceleration grel=gP−gQ=0 because gP≈gQ=g=9.8m/s2 ∴ geffective =grel =0 ⇒T=2π√ℓgrel =∞ |