To find the height \( h \) above the Earth where the gravitational acceleration \( g \) reduces to \( \frac{g}{2} \), we can use the formula for gravitational acceleration at a distance from the center of the Earth:
The gravitational acceleration \( g' \) at a distance \( r \) from the center of the Earth is given by:
\(g' = \frac{GM}{r^2}\)
Where:
At the surface of the Earth:
\(g = \frac{GM}{R^2}\)
Now, at height \( h \), the distance from the center of the Earth is \( R + h \) and gravitational acceleration is \( \frac{g}{2} \). Hence, we can write:
\(\frac{g}{2} = \frac{GM}{(R + h)^2}\)
Equating the two expressions:
\(\frac{GM}{(R + h)^2} = \frac{1}{2} \cdot \frac{GM}{R^2}\)
Cancel the common terms and simplify:
\(\frac{1}{(R + h)^2} = \frac{1}{2R^2}\)
Take the reciprocal of both sides:
\((R + h)^2 = 2R^2\)
Taking the square root of both sides:
\(R + h = \sqrt{2}R\)
Solving for \( h \):
\(h = \sqrt{2}R - R\)
\(h = (\sqrt{2} - 1)R\)
Therefore, the height \( h \) at which the gravitational acceleration becomes \( \frac{g}{2} \) is \( (\sqrt{2} - 1)R \).
The correct answer is: \((\sqrt{2} - 1)R\)
The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is \(\_\_\)? (Where \(g\) is the acceleration due to gravity on the surface of the Earth and \(R\) is the radius of the Earth.)