Question

Physics :-

What is the angular speed of a point on the surface of the earth? If the radius of the earth is 6400 km, what is its linear velocity?

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Answer 1

Solution :-

To find out we will be using

v = $\omega r$

Omega = 2π/T

Now as we know

r = radius of earth = 6.4 × 10³ km

Time = 24 hrs = 24 × 3600 sec

Now for angular velocity :-

$\omega = \dfrac{2\pi}{T}$

$\implies \omega = \dfrac{2 \times \dfrac{22}{7}}{24 \times 3600}$

$\implies \omega = \dfrac{44}{7\times 86400}$

$\implies \omega = \dfrac{44}{604800}$

$\implies \omega = \dfrac{44 \times 10^{-5}}{6.48}$

$\implies \omega = 7.275 \times 10^{-5}\: rad/sec$

Now linear velocity :-

$v = \omega r$

$\implies v = 7.275 \times 10^{-5} \times 6.4 \times 10^{3}$

$\implies v = 46.56 \times 10^{-2}$

$\implies v = 0.4656 \:km/s$

As 1 km = 1000 m

$\implies v = 465.6 \:m/s$