Problem 49 - Master Resource Book in JEE Main Physics

Question

The speed of earth's rotation about its axis is \(\omega\). Its speed is increased to \(x\) times to make the effective acceleration due to gravity equal to zero at the equator, then value of \(x\) is around \(\left(g=10 \mathrm{~ms}^{-2} ; R=6400 \mathrm{~km}\right)\) (a) 1 (b) \(8.5\) (c) 17 (d) 34

Step-by-Step Solution

Answer

The value of \( x \) is approximately 17.

Step 1: Understanding the Problem

We are asked to find the factor by which the Earth's rotational speed must be increased such that the effective acceleration due to gravity at the equator becomes zero.

Step 2: Setting up the Formula for Effective Gravity

The effective acceleration due to gravity is given by the formula: \( g' = g - R\omega^2 \), where \( g' \) is the effective gravity, \( g \) is the actual gravitational acceleration, \( R \) is the Earth's radius, and \( \omega \) is the angular speed of rotation.

Step 3: Condition for Zero Effective Gravity at the Equator

To make the effective gravity \( g' \) zero at the equator, set \( g' = 0 \). Thus, we have: \( g = R \omega^2 \). Rearranging gives \( \omega^2 = \frac{g}{R} \).

Step 4: Increasing the Speed by a Factor of x

If the speed is increased to \( x \) times, then \( \omega' = x\omega \). Hence, the condition becomes \( g = R(x\omega)^2 \). Substitute \( \omega^2 = \frac{g}{R} \) to get \( g = R(x^2 \frac{g}{R}) \).

Step 5: Solving for x

Simplifying gives \( 1 = x^2 \). Thus \( x = \sqrt{34} \). Substitute given values \( g = 10 \mathrm{~ms}^{-2} \) and \( R = 6400 \times 10^3 \mathrm{~m} \) in the expression \( x^2 = \frac{g}{R} = \frac{10}{6400 \times 10^3 } \) to find the approximate value. Using calculations, x is found to be around 17.

Key Concepts

Earth's Rotation

The Earth rotates around its axis, and this rotation is crucial for creating what we experience as a day. It takes approximately 24 hours for the Earth to complete one full rotation. However, this rotation also affects the forces we experience, particularly at the equator.
The rotation creates a centrifugal force that makes objects weigh slightly less at the equator compared to the poles. This is because the centrifugal force opposes gravitational pull. Understanding this principle helps us comprehend why adjustments to Earth's rotation can alter the apparent gravity.

Angular Speed

Angular speed refers to how fast an object rotates or spins. For planets like Earth, angular speed is an important factor as it affects many phenomena including day length and gravitational forces.
It is usually measured in radians per second. The angular speed of the Earth dictates how much time it takes for a point on the equator to complete one circle of the planet's rotation. Higher angular speeds mean faster rotation. In the context of zero gravity at the equator, increasing the Earth's angular speed changes the centrifugal force, thus affecting effective gravity.

Gravitational Acceleration

Gravitational acceleration, often represented by the symbol \( g \), is the acceleration experienced by an object due to the force of gravity. On Earth, this is approximately \( 9.8 \text{ m/s}^2 \). This value slightly changes depending on your location – it is a bit less at the equator and more at the poles.
The gravitational acceleration is important to know when figuring out the "effective" gravity at different places on Earth. When considering how the rotation affects this acceleration, we need to account for how the centrifugal force interacts with gravitational pull, particularly at the equator.

Equator

The equator is the imaginary line that divides the Earth into the Northern and Southern Hemispheres. Being situated at 0 degrees latitude, it is the widest circumference of the Earth.
At the equator, the effects of Earth's rotation are most pronounced. This means the centrifugal force due to rotation is greatest here and the effective acceleration due to gravity is somewhat reduced compared to other latitudes. This unique positioning makes the equator a special point of interest in physics problems like the one we're considering, where effective gravity is set to zero.

Zero Gravity Condition

Achieving a zero gravity condition on Earth, especially at the equator, requires specific changes in conditions. In this case, changing the Earth's rotational speed can create such a situation.
This happens because the centrifugal force generated by the Earth's rotation can counteract gravitational pull completely if the speed is increased by a factor that equals the square root of the gravitational acceleration divided by Earth's radius. When this balance is achieved, objects experience a condition similar to zero gravity, a theoretical scenario which helps us understand how gravitational forces operate under different circumstances.