Question

What must the angular speed of the rotation of earth so that the centrifugal force makes objects fly off its surface? Take g = 10 m s–2

Answer 1

Dear Student,

◆ Answer -

For objects to fly off earths surface, ω >= 1.237×10^-3 rad/s

● Explaination -

Magnitude of centifugal force acting on the body is -

a = v²/r = rω²

For object to just fly off earths surface, a = g and r = R,

g <= Rω²

Substitute values,

9.8 = 6.4×10^6 × ω²

ω² = 9.8 / 6.4×10^6

ω² = 1.53×10^-6

ω = √(1.53×10^-6)

ω = 1.237×10^-3 rad/s

Therefore, angular speed of rotation of earth should be more than or equal to 1.237×10^-3 rad/s.

Thanks dear...

Answer 2

Answer:

why 6.4× 10^6 is used as value R ?