Question

What's the angular speed of a point on the surface of the earth, if the radius of the earth is 6400km, what is it's linear velocity?

Answer 1

Answer:

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}ω=

T

2π

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

24×3600

2×

7

22

\implies \omega = \dfrac{44}{7\times 86400}⟹ω=

7×86400

44

\implies \omega = \dfrac{44}{604800}⟹ω=

604800

44

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

6.48

44×10

−5

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

−5

rad/sec

Now linear velocity :-

v = \omega rv=ωr

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

−5

×6.4×10

3

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

−2

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

As 1 km = 1000 m

\implies v = 465.6 \:m/s⟹v=465.6m/s

Explanation:

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