Find the imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero. In this condition, find the length (in hours) of a day also.
$(T a k e, g = 10 m s^{- 2}$ for the acceleration due to gravity if the earth were at rest and radius of earth equal to 6400 Km.)

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Solution

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

$W^{'} = W - m R_{e} ω^{2}$

$g^{'} = g - R_{e} ω^{2}$

$g^{'} = 0$

$g - R ω^{2} = 0$

$ω = \sqrt{\frac{g}{R}}$

$= \sqrt{\frac{10}{6400 \times 10^{3}}}$

$= 1.25 \times 10^{- 3} r a d / s$

$T = \frac{2 π}{ω} = \frac{2 \times 3.14}{1.25 \times 10^{- 3}} = 5024 s = 1.4 h$

Hence, new time period of this imaginary earth is $1.4 h o u r s$

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Similar questions

Q. The imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to

Q. The angular velocity of rotation of the earth in order to make the effective acceleration due to gravity equal to zero at equator should be

g = 10 m / s^{2}, R = 6400 k m

Q. In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be

(g = 10 m s^{- 2} a n d r a d i u s o f e a r t h i s 6400 k m)

Q. Calculate angular velocity of earth so that acceleration due to gravity at

60^{o}

latitude becomes zero. (Radius of earth

= 6400

km, gravitational acceleration at poles

= 10 m / s^{2}, cos 60^{o} = 0.5

)

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^{\circ}$ latitude becomes zero is (Radius of earth = 6400 At the poles g = 10 $m s^{- 2}$ )

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The apparent weight of a person on the equator (Lattitude α=0) is given by W′=W−mReω2 g′=g−Reω2 g′=0 g−Rω2=0 ω=√gR =√106400×103 =1.25×10−3rad/s T=2πω=2×3.141.25×10−3=5024s=1.4h Hence, new time period of this imaginary earth is 1.4hours