If the Earth has no rotational motion, the weight of a person on the equator is $$W$$. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh $$\frac{3}{4}W$$. The radius of the Earth is 6400 km and $$g = 10$$ m s$$^{-2}$$.

We are told that when the Earth is not rotating, the weight recorded by a person of mass $$m$$ at the equator is $$W$$. By definition, weight is the force with which the surface pushes up on the person, and in the absence of any other acceleration this equals the gravitational force, so we can write

$$W = mg,$$

where $$g = 10 \text{ m s}^{-2}$$ is the acceleration due to gravity.

Now imagine that the Earth begins to rotate with a uniform angular speed $$\omega$$. A person standing at the equator then describes a horizontal circle of radius $$R$$ (the Earth’s radius) and must continually change direction. The required centripetal force is

$$F_c = m\omega^{2}R.$$

This centripetal force is supplied by part of the gravitational pull, so the supporting surface has to supply a smaller upward force. The reading of the weighing machine (the apparent weight $$W'$$) is therefore

$$W' = mg - m\omega^{2}R = m\bigl(g - \omega^{2}R\bigr).$$

The problem states that we want this apparent weight to be only $$\dfrac34$$ of the original weight $$W$$, i.e.

$$W' = \dfrac34\,W.$$

Substituting $$W = mg$$ and $$W' = m\bigl(g - \omega^{2}R\bigr)$$, we have

$$m\bigl(g - \omega^{2}R\bigr) = \dfrac34\,mg.$$

We can divide both sides by $$m$$ to remove the mass:

$$g - \omega^{2}R = \dfrac34\,g.$$

Rearranging to isolate the term containing $$\omega$$ gives

$$\omega^{2}R = g - \dfrac34\,g = \dfrac14\,g.$$

Hence

$$\omega^{2} = \dfrac{g}{4R} \quad\Rightarrow\quad \omega = \sqrt{\dfrac{g}{4R}}.$$

We now substitute the numerical values. The radius of the Earth is given as $$R = 6400 \text{ km} = 6400 \times 10^{3} \text{ m}$$ and $$g = 10 \text{ m s}^{-2}$$, so

$$\omega = \sqrt{\dfrac{10}{4 \times 6400 \times 10^{3}}}\; \text{rad s}^{-1}.$$

First compute the denominator:

$$4 \times 6400 \times 10^{3} = 25600 \times 10^{3} = 2.56 \times 10^{7}.$$

Therefore

$$\omega^{2} = \dfrac{10}{2.56 \times 10^{7}} = 3.90625 \times 10^{-7},$$

and

$$\omega = \sqrt{3.90625 \times 10^{-7}}.$$

We take the square root in two parts. First, $$\sqrt{3.90625} \approx 1.976$$. Second, $$\sqrt{10^{-7}} = 10^{-3.5} = 0.31623 \times 10^{-3} = 3.1623 \times 10^{-4}.$$ Multiplying these factors, we obtain

$$\omega \approx 1.976 \times 3.1623 \times 10^{-4} \text{ rad s}^{-1} \approx 6.25 \times 10^{-4} \text{ rad s}^{-1}.$$

It is convenient to express this as

$$\omega \approx 0.625 \times 10^{-3} \text{ rad s}^{-1}.$$

Rounding to two significant figures gives

$$\omega \approx 0.63 \times 10^{-3} \text{ rad s}^{-1}.$$

Hence, the correct answer is Option A.