Question

What is the magnitude of the free-fall acceleration of a ball (mass \(m)\) due to the Earth's gravity at an altitude of \(2 R,\) where \(R\) is the radius of the Earth? Ignore the rotation of the Earth.

Short answer

Answer: The magnitude of the free-fall acceleration at an altitude of 2R is approximately 0.98 m/s².

Step-by-step solution

Write down the formula for gravitational force

The gravitational force between two masses \(m_1\) and \(m_2\) separated by a distance \(r\) is given by the formula: $$F = G \frac{m_1 m_2}{r^2}$$ where \(G\) is the gravitational constant \((6.674 \times 10^{-11}\,\text{m}^3\cdot\text{kg}^{-1}\cdot \text{s}^{-2})\).

Find the distance between the ball and the Earth's center

The altitude of the ball is \(2R\), so the distance between the ball and the Earth's center is \((R+2R) = 3R\).

Determine the mass of the Earth

The mass of the Earth (\(M_\oplus\)) is approximately \(5.972 \times 10^{24}\,\text{kg}\).

Calculate the gravitational force on the ball

Plug the values of the Earth's mass (\(M_\oplus\)), the ball's mass (\(m\)), and the distance between them (\(3R\)) into the formula for gravitational force: $$F = G \frac{m M_\oplus}{(3R)^2}$$

Use Newton's Second Law to find the acceleration

Newton's Second Law states that the force acting on an object is equal to its mass times its acceleration (\(F=ma\)). Therefore, the acceleration due to gravity can be found by dividing the gravitational force by the mass of the ball : $$a = \frac{F}{m}$$

Substitute the gravitational force equation into the acceleration formula

Replace the \(F\) in the acceleration formula with the expression for gravitational force: $$a = \frac{G \frac{m M_\oplus}{(3R)^2}}{m}$$

Simplify the equation

The mass of the ball (\(m\)) cancels out in the numerator and denominator: $$a = \frac{G M_\oplus}{(3R)^2}$$

Plug in the values of G, Earth's mass, and Earth's radius

Now we can plug in the values for \(G\), \(M_\oplus\), and \(R\): $$a = \frac{(6.674 \times 10^{-11}\,\text{m}^3\cdot\text{kg}^{-1}\cdot \text{s}^{-2})(5.972 \times 10^{24}\,\text{kg})}{(3 \times 6371 \times 10^3\,\text{m})^2}$$

Calculate the acceleration

Perform the calculations to find the magnitude of the free-fall acceleration at an altitude of \(2R\): $$a \approx 0.98\,\text{m/s}^2$$ The magnitude of the free-fall acceleration due to the Earth's gravity at an altitude of \(2R\) is approximately \(0.98\,\text{m/s}^2\).