Question

How far away from the center of the earth does the acceleration due to gravity be reduced by $1\%$ of its value at the earth’s surface? The radius of earth is $R = 6400\;{\text{km}}$.

Answer 1

Hint In this question, first use the acceleration due to gravity formula and find the relation between the distance of the point where the gravity reduces and the radius of the Earth. After that substitute the values to obtain the desired result.

Complete step by step answer
Before the solving we talk about the acceleration due to gravity we know that acceleration gained by an object by the gravitational force .its SI unit is ${\text{m/}}{{\text{s}}^{\text{2}}}$. The acceleration due to gravity has both magnitude and direction, so it is a vector quantity. We also know that acceleration due to gravity is represented by $g$. It has a specific value on the surface of the earth at sea level is $9.8\;{\text{m/}}{{\text{s}}^{\text{2}}}$.

At distance $r$ from the center of the earth value of ${g^1}$ which is $1\%$ of its value at the earth’s surface that means $0.99g$.
By using the formula, obtain the relation between the distance of the point where the gravity reduces and the radius of the Earth as,
${g^1} = \dfrac{{g{R^2}}}{{{r^2}}}$
$0.99g = \dfrac{{g{R^2}}}{{{r^2}}} \\\ \sqrt {0.99} = \dfrac{R}{r} \\\$
$r = \dfrac{R}{{\sqrt {0.99} }} \\\ = 1.005R \\\$

Now, substitute the value of the radius of the Earth as $R = 6400\;{\text{km}}$

$r = 1.005\left( {6400} \right) \\\ = 6432\;{\text{km}} \\\$

Thus, the distance of the point where the gravity reduces $1\%$ is $6432\;{\text{km}}$.

Note:
The acceleration of gravity decreases as the distance increases from the surface of the Earth while the acceleration due to gravity increases as we move towards the core of the earth that is towards the center of Earth.