Question

What should be the period of rotation of earth so as to make any object on the equator weigh half of its present value? [Radius of earth = 6.4 x 106 m; Acceleration due to gravity at earth surface go = 9.8 m/s?]

2 hrs

24 hrs

8 hrs

12 hrs​

Answer 1

Answer:

Earth rotates once in about 24 hours with respect to the Sun, but once every 23 hours, 56 minutes, and 4 seconds with respect to other, distant, stars (see below). Earth's rotation is slowing slightly with time; thus, a day was shorter in the past. This is due to the tidal effects the Moon has on Earth's rotation.

Answer 2

option(1) 2 hrs

Explanation:

Given:

• g' =$\frac{g}{2}$

• g = 9.8 m/s

• R = 6.4×$10^6$m

To find:

Period of rotation

Solution:

At the equator, gravity is given by :

$\bf \boxed{g'=g-\omega^2R}$

Where,

g' = apparent value of the acceleration due to

gravity.

g = acceleration due to gravity.

$\omega$ = angular velocity of the earth.

R = radius of the earth.

Now, in the question it's given that weight of an object is to be halved which means we will be halving g'.

$\implies \frac{g'}{2}=g-R\omega^2$

$\implies R\omega^2=g-\frac{g'}{2}$

$\implies \omega^2 = \frac{g}{2R}$

$\implies \omega = \sqrt{\frac{g}{2R}}$

Time period is given by;

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

$\implies T=2\pi \sqrt{\frac{2R}{g}}$

Now, putting values in the above formula;

=$2\times 3.14\times \sqrt{\frac{2\times 6.4\times 10^6}{9.8}}$

= 7.177×$10^3$

= 7177 sec

Now, as we know 1 hour = 60×60 =3600 seconds

$\implies T= \frac{7177}{3600}$

T = 1.99 hours

→ T = 2 hours