Question

At what height above the earth’s surface does the acceleration due to gravity fall to $1\mathrm{\%}$of its value at the earth’s surface?
A. $9R$
B. $8R$
C. $11R$
D. None of these

Answer 1

Hint:Let us know about the acceleration due to gravity. The net acceleration imparted to objects as a result of the combined effects of gravitation (from the Earth's mass distribution) and centrifugal force (from the Earth's rotation) is symbolised by $$g.$$

Formula used:
$g_h=g_s{\left({\displaystyle \frac{R}{R+h}}\right)}^2$
Where, $g_h=$ Acceleration due to gravity at a height, $g_s=$ Acceleration at earth's surface, $R=$radius and $h=height$.

Complete step by step answer:
Acceleration owing to gravity is the acceleration gained by an object due to gravitational force. Its SI unit is$$m{s}^{-2}$$. Because it contains both magnitude and direction, it is a vector quantity. The letter denotes the acceleration due to gravity.$$g$$. $$G$$has a standard value of $$9.8m{s}^{-2}$$on the earth's surface at sea level.

Given: $g_h=1\mathrm{\%}$ and $g_h=0.01g_s$.
Using :
$g_h=g_s{\left({\displaystyle \frac{R}{R+h}}\right)}^2$
$\Rightarrow 0.01g_s=g_s{\left({\displaystyle \frac{R}{R+h}}\right)}^2$
$\Rightarrow 0.1=\left({\displaystyle \frac{R}{R+h}}\right)$
$\Rightarrow 0.1R+0.1h=R$
$\therefore h=9R$

So, option A is correct.

Additional Information: The weight of an object on Earth's surface is defined as the downward force on that object by Newton's second law of motion, or (force = mass x acceleration). Other factors, such as the Earth's spin, contribute to the total gravitational acceleration and, as a result, influence the object's weight. Gravity does not normally account for the gravitational attraction of the Moon and Sun, which is usually explained in terms of tidal processes. It's a vector (physics) quantity with the same direction as a plumb bob.

Note: In order to solve this problem there are some important points which we should keep on our fingertips. The different conditions and equations of $g$ when the object is at a certain height from the earth surface and the object is down to the earth surface. By remembering this concept this problem will become a very easy one.