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Chapter 6: Problem 702

At what height over the earth's pole, the free fall acceleration decreases by one percent \(=\ldots \ldots \ldots \mathrm{km}(\mathrm{Re}=6400 \mathrm{~km})\) (A) 32 (B) 80 (C) \(1.253\) (D) 64

Short Answer

Expert verified
The height above the Earth's pole where the free fall acceleration decreases by one percent is approximately \(64~km\). Therefore, the correct answer is (D) 64.

Step by step solution

01

Write down the formula for acceleration due to gravity

The formula for acceleration due to gravity is given by: \(g = \frac{GM}{(R+h)^2}\) where \(g\) is the acceleration due to gravity at height \(h\), \(G\) is the gravitational constant, \(M\) is the mass of Earth, and \(R\) is the radius of Earth.
02

Determine the acceleration due to gravity at the Earth's surface

At the Earth's surface, the height \(h\) is equal to 0. So the acceleration due to gravity at the Earth's surface, \(g_0\), is given by: \(g_0 = \frac{GM}{R^2}\)
03

Calculate the acceleration due to gravity at one percent less than the surface acceleration

We are given that we need to find the height where the acceleration due to gravity has decreased by one percent of the surface acceleration (\(g_0\)). So we need to calculate \(g_1\), which is given by: \(g_1 = g_0 - (0.01 \times g_0)\)
04

Set up an equation for the height at which the acceleration due to gravity is one percent less than the surface acceleration

We can now set up an equation to solve for the height (\(h\)) when \(g = g_1\). This equation can be written as: \(\frac{GM}{(R+h)^2} = g_1\)
05

Substitute \(g_0\) in terms of \(R\) and \(M\) into the equation for one percent less acceleration

Substituting \(g_0 = \frac{GM}{R^2}\) into our equation, we get: \(\frac{GM}{(R+h)^2} = \frac{GM}{R^2} - \frac{GM}{100 R^2}\)
06

Solve for the height (\(h\))

Now, we can rearrange the equation to solve for the height (\(h\)): \(h = \sqrt{\frac{GM}{\frac{99 GM}{100R^2}}} - R\)
07

Substitute the radius of Earth (\(R\)) and simplify the equation

We are given the radius of Earth (\(R=6400\) km) and can substitute this into our equation to get: \(h = \sqrt{\frac{GM}{\frac{99 GM}{(6400 \times 10^3)^2}}} - 6400 \times 10^3\) Notice that the \(GM\) terms cancel each other out, and we are left with: \(h = \sqrt{\frac{100}{99} \times (6400 \times 10^3)^2} - 6400 \times 10^3\)
08

Calculate the height (\(h\))

Now, we can calculate the height (\(h\)) using the above equation: \(h \approx 64000~m\)
09

Convert the height to kilometers and make a comparison with answer options

Converting the height to kilometers, we get: \(h \approx 64~km\) Looking at the answer options, we can see that the closest answer to our solution is (D) 64 km.

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