Question

The height at which the acceleration due to gravity becomes g / 9 (where g = the
acceleration due to gravity on the surface of the earth) in terms of R, the radius of
(A) $R / 2$
(B) $\sqrt{2} R$
(C) 2 R
(D) $\dfrac{R}{\sqrt{2}}$

Answer 1

Hint: We know that in mechanics, acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities (in that they have magnitude and direction). The orientation of an object's acceleration is given by the orientation of the net force acting on that object. Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity. Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity. As velocity is an example of a vector, it has direction and magnitude, a change in both speed and direction.

Complete step by step answer
We know that the sensation of weightlessness, or zero gravity, happens when the effects of gravity are not felt. Technically speaking, gravity does exist everywhere in the universe because it is defined as the force that attracts two bodies to each other. But astronauts in space usually do not feel its effects. In the case of the earth, the force of gravity is greatest on its surface and gradually decreases as you move away from its centre (as a square of the distance between the object and the center of the Earth). Of course, the earth is not a uniform sphere so the gravitational field around it is not uniform.
Acceleration due to gravity at a height 'h' is given by: -
${\prime}^{\prime} \mathrm{g}^{\prime}=\mathrm{g}\left(\dfrac{\mathrm{R}}{\mathrm{R}+\mathrm{H}}\right)^{2}$
where $\mathrm{g} \rightarrow$ on the surface of the earth.
$\mathrm{R} \rightarrow$ Radius of earth.
As $\mathrm{g}^{\prime}=\mathrm{g} / 9,$ we get:
$\Rightarrow \dfrac{\mathrm{g}}{9}=\mathrm{g}\left(\dfrac{\mathrm{R}}{\mathrm{R}+\mathrm{H}}\right)^{2}$
$\Rightarrow \dfrac{\mathrm{R}}{\mathrm{R}+\mathrm{H}}=\dfrac{1}{3} \Rightarrow \mathrm{h}=2 \mathrm{R}$

Hence the correct answer is option C.

Note: We know that if the speed of the vehicle decreases, this is an acceleration in the opposite direction and mathematically a negative, sometimes called deceleration, and passengers experience the reaction to deceleration as an inertial force pushing them forward. Gravity is a force of attraction that exists between any two masses, any two bodies, any two particles. Gravity is not just the attraction between objects and the Earth. Artificial gravity can be created using a centripetal force. A centripetal force directed towards the center of the turn is required for any object to move in a circular path. In the context of a rotating space station it is the normal force provided by the spacecraft's hull that acts as centripetal force.