If angular speed of Earth is $7.26 \times 10^{- 5} r a d / s$ and the radius of Earth is $6400 k m$ , calculate the change in weight of $1 k g$ mass taken from equator to pole.

Open in App

Solution

Given,
$R = 6.4 \times 10^{6} m$
$ω = 7.26 \times 10^{- 5} \frac{r a d}{s}$
$△ g = g_{p} - g_{q} = R ω^{2}$
$△ w = m △ g$
$∴ △ W = m (R w^{2}) = 1 \times 6.4 \times 10^{6} \times (7.26 \times 10^{- 5})^{2} = 3373 \times 10^{- 5} N$

Suggest Corrections

Similar questions

Q. A man has a weight

105 N

on a weighing machine kept near the north pole of the earth. What will be the reduction in the weight of the man if he moves to the equator? Take

g = 9.8 {m/s}^{2}

, radius of earth (

R) = 6400 km

, angular speed of the earth

ω = 7.27 \times 10^{- 5} rad/s

Q. If earth stops spinning about its own axis, what will be the change in acceleration due to gravity on its equator? the radius of earth is

6.4 \times 10^{6} m

and its angular speed is

7.27 \times 10^{- 5} r a d / s

Given radius of earth along the equator is 6378 km and that along the poles is 6357 km. Calculate the percentage change in the weight of a man when he moves from the equator to the poles.

Q. During motion of a man from equator to pole of earth, its weight will (neglect the effect of change in the radius of earth)

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

60

latitude becomes zero. (Radius of the earth

= 6400 k m

, gravitational acceleration at poles

= 10 m / s^{2}

cos 60 = 0.5

)

Join BYJU'S Learning Program

If angular speed of Earth is 7.26×10−5rad/s and the radius of Earth is 6400km, calculate the change in weight of 1 kg mass taken from equator to pole.

Given, R=6.4×106mω=7.26×10−5rads△g=gp−gq=Rω2△w=m△g∴△W=m(Rw2)=1×6.4×106×(7.26×10−5)2=3373×10−5N

If angular speed of Earth is $7.26 \times 10^{- 5} r a d / s$ and the radius of Earth is $6400 k m$ , calculate the change in weight of $1 k g$ mass taken from equator to pole.

Given,
$R = 6.4 \times 10^{6} m$
$ω = 7.26 \times 10^{- 5} \frac{r a d}{s}$
$△ g = g_{p} - g_{q} = R ω^{2}$
$△ w = m △ g$
$∴ △ W = m (R w^{2}) = 1 \times 6.4 \times 10^{6} \times (7.26 \times 10^{- 5})^{2} = 3373 \times 10^{- 5} N$