The acceleration due to gravity on the surface of the earth is given by:
\(g = \frac{GM}{R^2}\)
where \( G \) is the gravitational constant, \( M \) is the mass of the earth, and \( R \) is the radius of the earth.
If the diameter of the earth is reduced to half, the radius \( R \) will also be reduced to half, becoming \( \frac{R}{2} \). Substituting \( R' = \frac{R}{2} \) into the formula for \( g \), we get:
\(g' = \frac{GM}{(R/2)^2} = \frac{GM}{R^2/4} = 4 \cdot \frac{GM}{R^2} = 4g\)
Thus, the new acceleration due to gravity on the surface of the earth would be \( 4g \).
The acceleration due to gravity at a height of 6400 km from the surface of the earth is \(2.5 \, \text{ms}^{-2}\). The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth = 6400 km)
Statement-1: \( \text{ClF}_3 \) has 3 possible structures.
Statement-2: \( \text{III} \) is the most stable structure due to least lone pair-bond pair (lp-bp) repulsion.
Which of the following options is correct?
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
In the real world, everything is always in motion. Objects move at a variable or a constant speed. When someone steps on the accelerator or applies brakes on a car, the speed of the car increases or decreases and the direction of the car changes. In physics, these changes in velocity or directional magnitude of a moving object are represented by acceleration.