Question

At what height above the earth’s surface does the acceleration due to gravity fall to $1\% $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:
${g_h} = {g_s}{\left( {\dfrac{R}{{R + h}}} \right)^2}$
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 \[9.8{\text{ }}m{s^{ - 2}}\]on the earth's surface at sea level.

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