The numerical value of the angular velocity of rotation of the earth should be ________ rad/s in ord

Q: Why does the rotation of the Earth reduce the value of g?

In the rotating frame of the Earth, any object experiences an outward centrifugal force ($m\omega^2 r$) directed away from the axis of rotation. At the equator, this force acts directly opposite to gravity, reducing the net downward force and thus lowering the effective acceleration due to gravity.

Q: What is the latitude angle at the poles, and does rotation affect gravity there?

At the poles, the latitude angle is $\phi = 90^\circ$. Since $\cos(90^\circ) = 0$, the term $\omega^2 R \cos^2 \phi$ becomes zero. Therefore, the rotation of the Earth has absolutely no effect on the value of gravity at the poles.

Q: Why did we convert the radius of the Earth from kilometers to meters?

To maintain consistency in SI units. The acceleration due to gravity $g$ is given in $\text{m/s}^2$, so the radius $R$ must be in meters ($6.4 \times 10^6 \text{ m}$) rather than kilometers ($6400 \text{ km}$) to get the correct value of $\omega$ in $\text{rad/s}$.

Q: If the Earth spins 17 times faster, will objects fly off into space?

If the Earth spins exactly at this critical speed, the normal force from the ground becomes zero, meaning objects will float just above the surface. If it spins any faster, the centrifugal force will exceed gravity, and objects will indeed be launched outwards into space.

Q: Does this weightlessness affect the mass of the object?

No, mass is an intrinsic property of matter and remains completely unchanged. Only the effective weight (the force exerted on the surface) becomes zero.

Question

The numerical value of the angular velocity of rotation of the earth should be \_\_\_\_\_\_\_\_ $	ext{rad/s}$ in order to make the effective acceleration due to gravity at equator equal to zero. (Take $g = 9.8 	ext{ m/s}^2$ and radius of earth $R = 6400 	ext{ km}$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on the equator of the Earth. Under normal circumstances, you feel firmly anchored to the ground. This anchoring force is what we call your **weight**, which is a direct result of the Earth's gravitational pull.

However, the Earth is not static; it is a spinning sphere. Because of this rotation, any object on its surface is in a rotating frame of reference and experiences an outward **centrifugal force**.

At the equator, this centrifugal force acts directly opposite to the force of gravity. Therefore, the effective acceleration due to gravity ($g'$) is slightly less than the actual gravitational acceleration ($g$).

### The Master Equation

The variation of the effective acceleration due to gravity with latitude $\phi$ is mathematically described by the formula:

$$g' = g - \omega^2 R \cos^2 \phi$$

Where:
- $g$ is the acceleration due to gravity at the surface if the Earth were stationary ($9.8 	ext{ m/s}^2$).
- $\omega$ is the angular velocity of the Earth's rotation.
- $R$ is the radius of the Earth ($6400 	ext{ km} = 6.4 	imes 10^6 	ext{ m}$).
- $\phi$ is the latitude angle.

At the equator, we are at the widest circle of the Earth, which corresponds to a latitude of $\phi = 0^\circ$. Since $\cos(0^\circ) = 1$, the equation simplifies directly to:

$$g' = g - \omega^2 R$$

### The Condition for Weightlessness

For an object at the equator to experience complete weightlessness, the effective acceleration due to gravity must drop to zero ($g' = 0$).

Setting $g' = 0$ in our simplified equation gives:

$$0 = g - \omega^2 R$$

$$\omega^2 R = g$$

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

This elegant relationship shows that the required angular velocity depends solely on the ratio of the local gravitational acceleration to the Earth's radius.

### Final Calculation

Now, let's substitute the standard physical constants into our derived formula:

$$\omega = \sqrt{\frac{9.8}{6.4 	imes 10^6}}$$

$$\omega = \sqrt{1.53125 	imes 10^{-6}}$$

$$\omega \approx 1.24 	imes 10^{-3} 	ext{ rad/s}$$

Thus, the numerical value of the required angular velocity is **$1.24 	imes 10^{-3} 	ext{ rad/s}$**.

### A Deeper Physical Insight

To put this number into perspective, let's compare it to the Earth's actual current angular velocity ($\omega_0$). The Earth completes one rotation in $24$ hours ($86400$ seconds):

$$\omega_0 = \frac{2\pi}{86400} \approx 7.27 	imes 10^{-5} 	ext{ rad/s}$$

Dividing our calculated value by the current value:

$$\frac{\omega}{\omega_0} = \frac{1.24 	imes 10^{-3}}{7.27 	imes 10^{-5}} \approx 17$$

This means that if the Earth were to spin **$17$ times faster** than it currently does, the centrifugal force at the equator would perfectly balance gravity, and objects there would float freely in a state of weightlessness!

Animated Solution for Physics - Gravitation: The numerical value of the angular velocity of rotation of the earth should be \_\_\_\_\_\_\_\_ $rad/s$ in order to make the effective acceleration due to gravity at equator equal to zero. (Take $g = 9.8 m/s^{2}$ and radius of earth $R = 6400 km$ )

Enter Numerical Value:

Visualized Solution

Visualizing the Rotating Earth

The Formula for Effective Gravity

Applying the Equator Condition

Setting Effective Gravity to Zero

Solving for Angular Velocity $ω$

Substituting the Numerical Values

Calculating the Final Value

The Way Forward

The Sigma Insight: Acceleration due to Gravity and its Variation

Analyzing the Setup

The Master Equation

The Condition for Weightlessness

Final Calculation

A Deeper Physical Insight

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