Question

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

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

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.

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: \(g_0 = \frac{GM}{R^2}\)

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: \(g_1 = g_0 - (0.01 \times g_0)\)

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: \(\frac{GM}{(R+h)^2} = g_1\)

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: \(\frac{GM}{(R+h)^2} = \frac{GM}{R^2} - \frac{GM}{100 R^2}\)

Solve for the height (\(h\))

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

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: \(h = \sqrt{\frac{100}{99} \times (6400 \times 10^3)^2} - 6400 \times 10^3\)

Calculate the height (\(h\))

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

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.

Key concepts

Acceleration Due to Gravity

Gravity is a force that pulls objects toward the center of the Earth. This force is due to the mass of the Earth. The acceleration due to gravity (\(g\)) is how quickly an object accelerates when it's in free fall. At the surface of the Earth, this value is approximately \(9.8 \text{ m/s}^2\).

• It can be affected by objects' altitude above the Earth.
• Measured by the formula \(g = \frac{GM}{(R+h)^2}\).
• The symbol \(G\) represents the gravitational constant, and \(M\) is Earth's mass.

As you go higher above Earth, reduce the pulling effect of Earth's mass on the object, slightly decreasing \(g\).Understanding this will help you calculate objects' falling speeds at different heights.

Gravitational Constant

The gravitational constant (\(G\)) is an important physical constant in physics. It appears in Newton's law of universal gravitation, providing the measure of the strength of gravity. This constant allows us to calculate the gravitational force between two masses.

• It is denoted by \(G\).
• Its value is approximately \(6.674 \times 10^{-11} \text{ N}(\text{m/kg})^2\).

Even though \(G\) is a very tiny number, it is essential for determining how strong gravity is at various locations. Without it, we couldn’t solve for gravitational effects in calculations.

Radius of Earth

The radius of Earth (\(R\)) is a crucial factor when solving problems related to gravity and physics calculations that involve the Earth. In most calculations, we use the average radius, which is approximately \(6400 \text{ km}\).

• The radius is used to find the distance of objects from Earth's center when they are above the surface.
• Included in calculations for the acceleration due to gravity: \(g = \frac{GM}{(R+h)^2}\).

The radius of Earth plays a part in defining the zone of influence where the Earth's gravity dominates. Understanding Earth's radius helps you evaluate potential decreases in gravitational pull, like our exercise.

Percent Decrease in Gravity

The percent decrease in gravity is how much less gravity affects an object when you change its location — typically by increasing its distance from Earth. In our exercise, we found what height above Earth makes gravity 1% less powerful.

• A 1% decrease means the gravity is \(g_1 = g_0 - (0.01 \times g_0)\).
• At this decrease, the object accelerates slightly slower due to gravity.

Solving for height when gravity decreases involves rearranging and simplifying the formula for acceleration due to gravity. This helps us understand how changing height impacts gravitational pull, so students grasp real-world applications of these concepts.