How many time faster than the present speed would the earth have to rotate about its axis so that the apparent weight of the bodies on the equator…

Question

How many time faster than the present speed would the earth have to rotate about its axis so that the apparent weight of the bodies on the equator becomes zero? (The radius of the earth R=6.4×10^6 m)

MARKS Content Team · Accepted Answer

Correct answer: (D) 17 times. Let ω_0 be the new angular velocity. At the equator g_E=g-Rω^2 The weight mg_E should be zero. g_E=0 ∴Rω02=g or ω_0=√gR ∴ω_0=√(9.86.4×10^6)=1.24×10^(-3)rad/s The present angular velocity of the earth is ω=(2π)/T=(2×3.14)/(24×60×60) ∴ ω=(3.14×10^(-2))/(12×36)=7.27×10^(-5) ∴(ω_0)/(ω)=(1.24×10^(-3))/(7.27×10^(-5))=17 ∴ω_0=17ω It should rotate faster by 17 times the present value.

How many time faster than the present speed would the earth have to rotate about its axis so that the apparent weight of the bodies on the equator becomes zero? (The radius of the earth $R = 6 .4 \times 10^{6} m$ )

Solution

More Gravitation Questions

How many time faster than the present speed would the earth have to rotate about its axis so that the apparent weight of the bodies on the equator becomes zero? (The radius of the earth R=6.4×106 m)

Solution

More Gravitation Questions

How many time faster than the present speed would the earth have to rotate about its axis so that the apparent weight of the bodies on the equator becomes zero? (The radius of the earth $R = 6 .4 \times 10^{6} m$ )