Question

The earth spins on its axis once a day and orbits the sun once a year (365 1/4 days). Determine the average angular velocity (in rad/s) of the earth as it (a) spins on its axis and (b) orbits the sun. In each case, take the positive direction for the angular displacement to be the direction of the earth's motion.

Answer 1

Answer:

Given that

The earth spins on its axis once a day and orbits the sun once a year (365 1/4 days)

a)

When earth spins on its axis

We know that earth take 1 day to complete one revolution around its own axis.

T= 1 day = 24 hr = 24 x 3600 s

T=86400 s

We know that

T=2π/ω

ω= 2π/T

ω= 2π/86400

ω=7.27 x 10⁻5 rad/s

b)

When earth revolve around earth

T =365 1/4 days = 365.25 days

T= 365.24 x 86400 s

T=31557600

We know that

T=2π/ω

ω= 2π/T

ω= 2π/31557600

ω=1.99 x 10⁻⁷ rad/s

Answer 2

Answer:

(a) $7.27\times 10^{- 5}\ rad/s$

(b) $1.99\times 10^{- 7}\ rad/s$

Solution:

As per the question:

The time period of the Earth's spin on its axis, T = 1 day

1 day = $24\times 60\times 60 = 86400\ s$

Now,

(a) The Earth's average angular velocity is given by:

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

$\omega = \frac{2\pi}{86400} = 7.27\times 10^{- 5}\ rad/s$

(b) Time period for orbiting the sun, T':

T' = $365 \frac{1}{4} = 365.25\ days$

T' = $365.25\times 24\times 60\times 60 = 3.1557\times 10^{7}\ s$

Angular velocity, $\omega'$:

$\omega' = \frac{2\pi}{T'} = \frac{2\pi}{3.1557\times 10^{7}} = 1.99\times 10^{- 7}\ rad/s$