Question

The radius of the earth is $6400km$ and $g=10m{s}^{-2}$. In order that a body of $5kg$ weights zero at the equator, the angular speed of the earth is (in rad/sec)

• A. $\frac{1}{80}$
• B. $\frac{1}{400}$
• C. $\frac{1}{800}$
• D. $\frac{1}{600}$

Answer 1

The correct option is C

$\frac{1}{800}$

Step 1: Given data

Radius of earth $\left(R\right)=6400km=6400\times {10}^{3}m$

Acceleration due to gravity $\left(g\right)$=$10m{s}^{-2}$

Weight of body $\left(m\right)=5kg$

At equator, body weighs zero

Step 2: Obtain the formula to compute angular speed

The acceleration at the equator because of the rotation of the earth,

$g^{\text{'}}=g-{\omega }^{2}R{\cos }^2\lambda$ , Where $g^{\text{'}}=$acceleration at equator , $g=$acceleration due to gravity, $\omega =$Angular velocity, $R=$radius $\lambda$ denotes latitude

At equator $\lambda =0$ . So,

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

For zero weight at equator as given $g^{\text{'}}=0$ . So,

$$\begin{gathered} 0=g-{\omega }^{2}R \\ \omega =\sqrt{\frac{g}{R}} \end{gathered}$$

Where $\omega =$angular velocity, $g=$gravity and $R=$radius.

Step 3: Compute the angular speed

The formula for a body to be weightless at the equator is $\omega =\sqrt{\frac{g}{R}}$

Substitute the known values in the relationship we get,

$$\begin{gathered} \omega =\sqrt{\frac{10}{6400\times {10}^3}} \\ \omega =\frac{1}{800}rad/sec \end{gathered}$$

Thus, the angular speed of the earth is $\frac{1}{800}rad/sec$

Hence, option C is the correct answer.