Assume that the earth’s rate of rotation were increased to such an extent that the
apparent weight of an object on the equator was zero. What would then be the period
of rotation of the earth?
Radius of the earth = 6380 km
Answer
g′=g−Rω2g^{'}=g-R\omega^2g′=g−Rω2
Given they g′=0kg/m2g^{'}=0kg/m^2g′=0kg/m2
So
ω=gR\omega=\sqrt{\frac{g}{R}}ω=Rg
ω=9.86380000\omega=\sqrt{\frac{9.8}{6380000}}ω=63800009.8
And time period
T=2πω=2π9.86380000=4.1×106sT=\frac{2\pi}{\omega}=\frac{2\pi}{\sqrt{\frac{9.8}{6380000}}}=4.1\times10^6sT=ω2π=63800009.82π=4.1×106s
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