Determine, as a function of the latitude λ, the normal acceleration of the point P on the surface of the Earth due to the spin ωE of Earth about its axis. In addition, determine the normal acceleration of the Earth due to its rotation about the Sun. Using these results, determine the latitude above

Question

Determine, as a function of the latitude [latex]\lambda[/latex] , the normal acceleration of the point P on the surface of the Earth due to the spin [latex]{ \omega }_{ E [/latex] of Earth about its axis. In addition, determine the normal acceleration of the Earth due to its rotation about the Sun. Using these results, determine the latitude above which the acceleration due to the orbital motion of the Earth is more significant than the acceleration due to the spin of the Earth about its axis. Use [latex]{ R }_{ E }[/latex] = 6371 km for the mean radius of the Earth, and assume the Earth’s orbit about the Sun is circular with radius [latex]{ R }_{ O }=1.497 imes { 10 }^{ 8 }[/latex] km.

Accepted Answer

The tangential velocity of P due to the rotation of Earth is [latex]v_P=R_E \omega_E \cos \lambda[/latex]. The radius of curvature as a function of [latex]\lambda ext { is } ho_P=R_E \cos \lambda[/latex]. The normal acceleration of P due to rotation is

[latex]a_P=\frac{v_P^2}{ ho_P}=\frac{\omega_E^2 R_E^2 \cos ^2 \lambda}{R_E \cos \lambda} \Rightarrow \omega_E^2 R_E \cos \lambda,[/latex] (1)

where [latex]R_E=6371 \ km =6371 imes 10^3 \ m[/latex] and

[latex]\omega_E=(1 \ rev / ext { day })(2 \pi \ rad / rev )(24 \ h / ext { day })^{-1}(3600 \ s / h )^{-1}=72.72 imes 10^{-6} \ rad / s[/latex] (2)

The tangential velocity of P due to the motion of the Earth arounf the Sun is [latex]v_E=R_O \omega_O[/latex], where

[latex]a_O=\frac{v_E^2}{ ho_O}=\frac{\omega_O^2 R_O^2}{R_O} \Rightarrow a_O=\omega_O^2 R_O=5.94 imes 10^{-3} \ m / s ^2,[/latex] (3)

where we have used the fact that [latex]R_O=1.497 imes 10^8 \ km =1497 imes 10^8 \ m[/latex]. Consequently, the latitude [latex]\lambda[/latex] above which the acceleration due to the motion of the Earth around the Sun is greater than the acceleration of due to the Earth’s spin about its own axis is

[latex]\lambda=\cos ^{-1}\left(\frac{\omega_O^2 R_O}{\omega_E^2 R_E} ight)=79.8^{\circ},[/latex]

where we have used the fact that [latex]R_O=1.497 imes 10^8 \ km =1497 imes 10^8 \ m , R_E=6371 \ km =6,371,000 \ m[/latex], and the values of [latex]\omega_E ext { and } \omega_O[/latex] in Eqs. (2) and (3), respectivley.

Question 2.245: Determine, as a function of the latitude λ, the normal accel...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions