Question

What is the distance from the Earth’s center to a point in space where the gravitational acceleration due to the Earth is $\dfrac{1}{24}$ of its value at the Earth’s surface?

Answer 1

Hint: The gravitational acceleration acts on a body due to the force of gravity. The acceleration due to gravity depends on the mass of Earth and the inverse of the square of the distance from the center of the Earth. Substituting given values in the above equation, we can calculate the distance of the point.

Formulas used:
$a=\dfrac{GM}{{{R}^{2}}}$

Complete step-by-step solution:
Gravitational force is the attractive force that exists between two bodies with mass. The gravitational force when one of the bodies is Earth is called the force of gravity. The acceleration of a body that occurs due to the action of gravity on it is called acceleration due to gravity.
Let a point P in space be such that the gravitational acceleration at that point is $\dfrac{1}{24}$ of its value at the Earth. The acceleration due to gravity is calculated as,
$a=\dfrac{GM}{{{R}^{2}}}$
Here, $a$ is acceleration due to gravity
$G$ is the gravitational constant
$M$ is mass of the Earth
$R$ is the distance from the Earth’s center
Given that, $a=\dfrac{g}{24}$, $G=6.626\times {{10}^{-11}}N{{m}^{2}}\,k{{g}^{-2}}$, $M=6\times {{10}^{24}}kg$
In the above equation, substituting given values, we get,
$\begin{align}
  & \dfrac{g}{24}=\dfrac{6.6\times {{10}^{-11}}\times 6\times {{10}^{24}}}{{{R}^{2}}} \\
 & \Rightarrow \dfrac{10}{24}=\dfrac{6.6\times {{10}^{-11}}\times 6\times {{10}^{24}}}{{{R}^{2}}} \\
 & \Rightarrow R=\sqrt{\dfrac{6.6\times {{10}^{-11}}\times 6\times {{10}^{24}}\times 24}{10}} \\
 & \therefore R=30.82\times {{10}^{6}}m \\
\end{align}$
 The distance of point P from the Earth’s center is $30.82\times {{10}^{6}}m$.
 Therefore, the distance of a point from Earth’s center where the acceleration due to Earth’s gravity is $\dfrac{1}{24}$ of its value on Earth is $30.82\times {{10}^{6}}m$.

Note: The gravitational force is equal and opposite on the two bodies involved but the force on Earth due to objects on Earth is negligible due to difference in mass. Any change in acceleration due to gravity near the Earth’s surface is negligible. The Coulomb’s law of gravitation follows the inverse square law.