Question

A seconds pendulum is placed in a space laboratory orbiting round th earth at a height of 3R, where R is the radius of earth .the time period of the pendulum is:​

Answer 1

Answer: T=2π√16R²/GM

Explanation: Formula:   F=Gm₁m₂/r²

= F=GMm₂/(R+3R)²        =M is mass of earth and R is radius of earth

= m₂g=GMm₂/16R²        

= g=GM/16R²

for simple pendulum, time period is given by 2π√l/g

= T=2π√l/(GM/16R²)

= T=2π√1*16R²/GM

= T=2π√16R²/GM