Question

Calculate that imaginary angular velocity of the Earth for which effective acceleration due to gravity at the equator becomes zero. In this condition, find the length (in hours) of a day? Radius of Earth = 6400 km. g = 10 ms^(-2).

Answer 1

Answer:

w = 1.25×10-³ .....

Explanation:

....

Answer 2

Answer:

ω=1.25×10^-3 rad/s

T=5024s=1.4h

Explanation:

The apparent weight of a person on the equator (Lattitude α=0) is given by

W' =W−mRëω²

g'=g−Rëω²

g'=0

g−Rω² =0

ω= √g/√Rë

ω=√10/√6400*10³

ω=1.25×10^-3 rad/s

hence the new angular velocity of imaginary earth is 1.25×10^-3 rad/s

T= 2π/ω

T= (2×3.14)/(1.25×10 ^−3)

T=5024s=1.4h

Hence, new time period of this imaginary earth is 1.4hours