Question

At what rate should the earth rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?

Answer 1

Explanation:

• It is given that the apparent g at the equator,  $g_{\text {apparent}}=g-\omega^{2} R=0$.  Here,  ω is the rate of rotation.
• The rate at which the earth should rotate to make the apparent g at the equator to be zero:

          $\omega=\sqrt{\frac{g}{R}}=\frac{2 \pi}{T}=\sqrt{\frac{10}{6400 * 10^{3}}}=1.25 * 10^{-3} \frac{\mathrm{rad}}{\mathrm{s}}$

• The length of the day in this situation:
• We know that the Length of the day is nothing but the time equal to the time taken for one complete rotation.

         $\mathrm{T}=2 \pi \sqrt{\frac{R}{g}}=16 \pi * 10^{2}=5026.5 \mathrm{s}=1.4 \mathrm{h}$

• Thus the length of the day is 1.4h.

Answer 2

For apparent g at equator to be 0,

mω  

2

R−mg=0

ω=  

R

g

​

 

​

=  

6400000

9.8

​

 

​

=1.24X10  

−3

rad/s

Also, length of day is equal to the time of one rotation.

Hence, T=  

ω

2π

​

=  

1.24X10  

−3

 

2π

​

=5067.74s=1.41hr

Answer is 1.41hr.