Question #172495

Take the radius of the Earth to be 6309 km


1)The angular speed of a point on Earth’s surface at latitude 30° N is ____ x 10-4 rad/s


2)The linear speed of a point on Earth’s surface at latitude 30° N is


3)The latitude where your linear speed is 10 m/s is ±


1
Expert's answer
2021-03-23T08:08:16-0400

1) The angular speed at this point is the same as the angular speed of the Earth:


ωEarth=2π rad0.997 d24 h1 d3600 s1 h=7.29105 rads.\omega_{Earth}=\dfrac{2\pi\ rad}{0.997\ d\cdot\dfrac{24\ h}{1\ d}\cdot\dfrac{3600\ s}{1\ h}}=7.29\cdot10^{-5}\ \dfrac{rad}{s}.

2) We can find the linear speed of a point on Earth’s surface at latitude 30° N as follows:


v=ωEarthREarthcosθ,v=\omega_{Earth}R_{Earth}cos\theta,v=7.29105 rads6.37106 mcos30=402 ms.v=7.29\cdot10^{-5}\ \dfrac{rad}{s}\cdot6.37\cdot10^6\ m\cdot cos30^{\circ}=402\ \dfrac{m}{s}.

3) We can find the latitude as follows:


v=ωEarthREarthcosθ,v=\omega_{Earth}R_{Earth}cos\theta,θ=cos1 (vωEarthREarth),\theta=cos^{-1}\ (\dfrac{v}{\omega_{Earth}R_{Earth}}),θ=cos1 (10 ms7.29105 rads6.37106 m)=±88.76.\theta=cos^{-1}\ (\dfrac{10\ \dfrac{m}{s}}{7.29\cdot10^{-5}\ \dfrac{rad}{s}\cdot6.37\cdot10^6\ m})=\pm88.76^{\circ}.

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