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Chapter 7: Problem 82

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) \(\sqrt{2} R\)

Short Answer

Expert verified
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 by step solution

01

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.
02

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: \[\frac{g}{9} = \frac{GM}{(R+h)^2}\]
03

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: \[\frac{GM}{R^2} \cdot \frac{1}{9} = \frac{GM}{(R+h)^2}\]
04

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) \(2 R\).

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