Problem 82 - A Complete Resource Book in Physics for JEE Main

Question

The height at which the acceleration due to gravity becomes \(\frac{g}{9}\) (where \(g=\) the acceleration due to gravity on the surface of the earth) in terms of \(R\), the radius of the earth, is (A) \(2 R\) (B) \(\frac{R}{\sqrt{3}}\) (C) \(\frac{R}{2}\) (D) \(\sqrt{2} R\)

Step-by-Step Solution

Answer

The height at which the acceleration due to gravity becomes \(\frac{g}{9}\) in terms of \(R\), the radius of the Earth, is (A) \(2 R\).

Step 1: Write down the equation for the acceleration due to gravity

The equation for gravitational acceleration, \(g'\) at a certain height \(h\) from the Earth's surface can be determined using the relationship: \[g'= \frac{GM}{(R+h)^2}\] Where: - \(G\) is the gravitational constant, - \(M\) is the mass of the Earth, - \(R\) is the Earth's radius, - \(h\) is the height above the Earth's surface.

Step 2: Set up the equation to represent \(\frac{g}{9}\)

Since we want to find the height at which the gravitational acceleration becomes \(\frac{g}{9}\), we can set up the following equation: \[\frac{g}{9} = \frac{GM}{(R+h)^2}\]

Step 3: Replace the acceleration due to gravity on the surface with its expression

On the surface of the Earth (\(h=0\)), the acceleration due to gravity, \(g\) is given by: \[g = \frac{GM}{R^2}\] Now, we can substitute this expression for \(g\) into our equation from Step 2: \[\frac{GM}{R^2} \cdot \frac{1}{9} = \frac{GM}{(R+h)^2}\]

Step 4: Solve the equation for \(h\)

We will now solve the equation for \(h\). Since \(GM\) is present in both numerators, we can cancel it out: \[\frac{1}{9} = \frac{1}{(R+h)^2}\] Next, we'll invert both sides of the equation: \[9 = (R+h)^2\] Now, take the square root: \[3 = R+h\] Finally, isolate \(h\): \[h = 3R - R\] \[h = 2R\] So, the height at which the acceleration due to gravity becomes \(\frac{g}{9}\) is \(2R\). The correct answer is (A) \(2 R\).