Problem 25 - University Physics with Modern Physics

Question

A student with mass 45 \(\mathrm{kg}\) jumps off a high diving board. Using \(6.0 \times 10^{24} \mathrm{kg}\) for the mass of the earth, what is the acceleration of the earth toward her as she accelerates toward the earth with an acceleration of 9.8 \(\mathrm{m} / \mathrm{s}^{2}\) ? Assume that the net force on the earth is the force of gravity she exerts on it.

Step-by-Step Solution

Answer

Earth's acceleration is approximately \(7.35 \times 10^{-23} \, \text{m/s}^2\).

Step 1: Understand the Problem

We are given the mass of the student and the acceleration towards Earth. We must find the Earth's acceleration toward the student as they influence each other through gravity.

Step 2: Apply Newton's Third Law

According to Newton's third law, the gravitational force exerted by the Earth on the student is equal and opposite to the force exerted by the student on the Earth. Thus, the force can be calculated using the formula: \[ F = m \times a \]where \( m \) is the mass of the student (45 kg) and \( a \) is the acceleration (9.8 m/s²).

Step 3: Calculate the Force

Calculate the force using the student's mass and acceleration: \[ F = 45 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 441 \, \text{N} \]This is the force the student exerts on the Earth.

Step 4: Use Newton's Second Law to Find Earth's Acceleration

Using Newton's second law \( F = M \times A \), where \( M \) is the mass of the Earth (\(6.0 \times 10^{24} \text{ kg}\)) and \( F \) is the calculated force (441 N), solve for the Earth's acceleration \( A \):\[ A = \frac{F}{M} = \frac{441 \, \text{N}}{6.0 \times 10^{24} \, \text{kg}} \]

Step 5: Calculate the Earth's Acceleration

Perform the division to find the acceleration:\[ A = \frac{441}{6.0 \times 10^{24}} \approx 7.35 \times 10^{-23} \, \text{m/s}^2 \]

Step 6: Conclude the Calculation

The small numerical result indicates that the Earth's acceleration due to her force is extremely small, almost negligible in practical scenarios.

Key Concepts

Gravitational Force

In the universe, everything that has mass attracts everything else that has mass. This is called gravitational force. It's a force of attraction between two masses, such as the Earth and a student jumping off a diving board. The gravitational force is what pulls objects towards each other.

The strength of the gravitational force depends on two main factors:

• The mass of the objects involved: Greater mass means a stronger force.
• The distance between the objects: Greater distance means a weaker force.

In our scenario, the Earth and the student are relatively close, and both have mass. But because Earth's mass is so much larger, we can focus on the student’s acceleration towards Earth. Newton's Third Law tells us that the student and the Earth pull on each other with equal force, but in opposite directions.

Force Calculation

The concept of force calculation is vital in understanding how objects interact through Newton's Third Law. In our exercise, we need to calculate the gravitational force that the student exerts on Earth. This is done using the formula: \[ F = m \times a \]where \( F \) stands for force, \( m \) is the student's mass (45 kg), and \( a \) is the acceleration due to gravity (9.8 m/s²).

Performing the calculation gives us:\[ F = 45 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 441 \, \text{N} \]

This means the force exerted by the student on Earth is 441 Newtons. Even though the force is fairly large, due to Earth's massive size, its acceleration is practically imperceptible.

Newton's Second Law

Newton's Second Law of Motion is a fundamental principle that explains the relationship between force, mass, and acceleration. It can be articulated with the formula:\[ F = M \times A \]where \( F \) is force, \( M \) is mass, and \( A \) is acceleration. This law is crucial when calculating how objects move under various influences.

In our case, we need to find the Earth's acceleration caused by the gravitational force the student exerts on it. We've already calculated that force as 441 N. Using the formula, and knowing Earth’s mass is a whopping \(6.0 \times 10^{24} \text{ kg}\), we solve for \( A \):\[ A = \frac{441 \, \text{N}}{6.0 \times 10^{24} \, \text{kg}} \approx 7.35 \times 10^{-23} \, \text{m/s}^2 \]

This incredibly tiny acceleration showcases how immense Earth is compared to everyday objects. Newton's Second Law thus helps illustrate why Earth’s movement under such small forces is effectively invisible to us.