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Question: At what height above the earth’s surface does the acceleration due to gravity fall to \(1\% \)of its...

At what height above the earth’s surface does the acceleration due to gravity fall to 1%1\% of its value at the earth’s surface?
A. 9R9R
B. 8R8R
C. 11R11R
D. None of these

Explanation

Solution

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.g.

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

Complete step by step answer:
Acceleration owing to gravity is the acceleration gained by an object due to gravitational force. Its SI unit isms−2m{s^{ - 2}}. Because it contains both magnitude and direction, it is a vector quantity. The letter denotes the acceleration due to gravity.gg. GGhas a standard value of 9.8 ms−29.8{\text{ }}m{s^{ - 2}}on the earth's surface at sea level.

Given: gh=1%{g_h} = 1\% and gh=0.01gs{g_h} = 0.01{g_s}.
Using :
gh=gs(RR+h)2{g_h} = {g_s}{\left( {\dfrac{R}{{R + h}}} \right)^2}
⇒0.01gs=gs(RR+h)2\Rightarrow 0.01{g_s} = {g_s}{\left( {\dfrac{R}{{R + h}}} \right)^2}
⇒0.1=(RR+h)\Rightarrow 0.1 = \left( {\dfrac{R}{{R + h}}} \right)
⇒0.1R+0.1h=R\Rightarrow 0.1R + 0.1h = R
∴h=9R\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 gg 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.