Question

Imagine that two tunnels are bored completely through the Earth, passing through the center. Tunnel 1 is along the Earth's axis of rotation, and tunnel 2 is in the equatorial plane, with both ends at the Equator. Two identical balls, each with a mass of \(5.00 \mathrm{~kg}\), are simultaneously dropped into both tunnels. Neglect air resistance and friction from the tunnel walls. Do the balls reach the center of the Earth (point \(C\) ) at the same time? If not, which ball reaches the center of the Earth first?

Short answer

Answer: Yes, both balls reach the center of the Earth at the same time.

Step-by-step solution

Calculate gravitational force acting on the balls

To calculate the gravitational force acting on the balls, we will use Newton's law of gravitation, which states: \(F = G \frac{m_1 m_2}{r^2}\) Where \(F\) is the gravitational force, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the two objects, and \(r\) is the distance between them. In this problem, both balls have the same mass, \(m = 5.00 kg\), and we will use the variable \(x\) to represent the distance from the Earth's center to the position of the balls being calculated. Therefore, the gravitational force acting on the balls at any position \(x\) is: \(F = G \frac{mM}{x^2}\) Where \(M\) is the mass of the Earth.

Apply Newton's Second Law of Motion

Apply Newton's second law of motion, \(F = ma\), where \(F\) is the gravitational force acting on the ball, \(m\) is the mass of the ball, and \(a\) is the ball's acceleration. Substituting the gravitational force from Step 1, we have: \(G \frac{mM}{x^2} = ma\) Since \(m\) is a common term in both sides of the equation, we can cancel it out: \(a = G \frac{M}{x^2}\)

Determine the acceleration in both tunnels

Now, we will determine the acceleration of the balls in both tunnels: For Tunnel 1 (along the Earth's axis of rotation), the acceleration acting on the ball is directly toward Earth's center and is given by: \(a_1 = G \frac{M}{x^2}\) For Tunnel 2 (in the equatorial plane), the acceleration acting on the ball has two components, the gravitational acceleration toward the Earth's center (radial) and the Coriolis acceleration due to Earth's rotation. We will only consider the radial component of acceleration as the Coriolis acceleration is perpendicular to the radial acceleration, and it does not affect the time taken by the ball to reach the center: \(a_2 = G \frac{M}{x^2}\) Notice that the radial accelerations for both tunnels are the same.

Determine if the balls reach the center at the same time

Since the radial accelerations for both tunnels are the same, the motion of both balls is determined by the same equation of motion: \(a = G \frac{M}{x^2}\) This implies that both balls will be subjected to the same acceleration at any given distance from the Earth's center. As a result, they will reach the center at the same time. Therefore, both balls reach the center of the Earth at the same time.

Key concepts

Gravitational Force Calculation

Understanding the gravitational force calculation is crucial when delving into physics problems involving two masses attracting each other. According to Newton's law of gravitation, the force of gravity between two objects can be calculated using the equation:

\( F = G \frac{m_1 m_2}{r^2} \)

In this equation, \(F\) represents the gravitational force, \(G\) is the gravitational constant (approximately \(6.674 \times 10^{-11} N(m/kg)^2\)), \(m_1\) and \(m_2\) are the masses of the objects, and \(r\) is the distance between their centers. For our exercise, one mass is the Earth and the other is the ball. The simplification yielded from
\( F = G \frac{mM}{x^2} \)
shows a direct proportionality between the force and Earth's mass (\(M\)) and an inverse-square relation to the distance from Earth's center (\(x\)). This becomes foundationally important when considering the motion of objects under gravity alone, like the balls in the tunnels.

Acceleration Due to Gravity

When objects are influenced only by gravity, the acceleration due to gravity serves as a uniform field simplifying the analysis of their motions. Newton's second law (\(F = ma\)) provides a link between force (
\(F\)) and acceleration (
\(a\)), with \(m\) being the object's mass. Plugging in the gravitational force while cancelling out the mass of the object yields:

\(a = G \frac{M}{x^2}\)

This equation reveals that an object's acceleration towards Earth only depends on Earth's mass and the object's distance from Earth's center, not on the object's mass itself. Therefore, in our scenario with the two balls, despite the orientation of the tunnels, the balls' radial accelerations are identically affected by gravity as they travel towards the center of Earth, leading to their synchronized arrival.

Coriolis Effect

The Coriolis effect is a deflection pattern experienced by objects moving within a rotating frame of reference, such as Earth. It's essential when understanding weather patterns, ocean currents, and any motion over a large scale on Earth's surface. However, in the context of our exercise, the effect is still present but doesn’t affect the balls' travel time in the tunnels.

For the ball in Tunnel 2, its motion is subject to the Coriolis force due to Earth's rotation. This force acts perpendicular to the movement towards Earth's center, resulting in a curved path, but it does not change the speed of the object in the direction towards the center. Since we are only concerned with the time it takes for the balls to reach the center of Earth, the Coriolis force can be ignored in solving this problem—the radial acceleration (towards the Earth's center) remains unchanged, which assures the time for both balls to reach point C is equal.