0. A student with mass 45 kg jumps off a high diving board. Using 6.0 x 1024 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 9.81 m/s?? (Hint: Use Newton's Third Law to determine the net gravitational force the student exerts on the earth.)

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Chapter1: Units, Trigonometry. And Vectors
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**Problem:**
A student with a mass of 45 kg jumps off a high diving board. Using \(6.0 \times 10^{24}\) 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 9.81 m/s\(^2\)? *(Hint: Use Newton’s Third Law to determine the net gravitational force the student exerts on the Earth.)*

**Solution Explanation:**

To solve this problem, we need to understand that according to Newton's Third Law, every action has an equal and opposite reaction. When the student exerts a force on the Earth by falling, the Earth exerts an equal and opposite force on the student.

1. **Force exerted by the student:**
   - The force due to gravity on the student is given by \( F = m \times g \), where \( m = 45 \) kg and \( g = 9.81 \) m/s\(^2\).
   - Calculate the force: \( F = 45 \, \text{kg} \times 9.81 \, \text{m/s}^2 \).

2. **Acceleration of the Earth:**
   - According to Newton’s Third Law, the same force \( F \) acts on the Earth.
   - Use the formula \( F = M \times a \), where \( M = 6.0 \times 10^{24} \) kg is the mass of the Earth, to find the Earth's acceleration \( a \).
   - Rearrange the formula to solve for \( a \): \( a = \frac{F}{M} \).

By substituting the values calculated in the first step into this equation, you can determine the acceleration of the Earth.
Transcribed Image Text:**Problem:** A student with a mass of 45 kg jumps off a high diving board. Using \(6.0 \times 10^{24}\) 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 9.81 m/s\(^2\)? *(Hint: Use Newton’s Third Law to determine the net gravitational force the student exerts on the Earth.)* **Solution Explanation:** To solve this problem, we need to understand that according to Newton's Third Law, every action has an equal and opposite reaction. When the student exerts a force on the Earth by falling, the Earth exerts an equal and opposite force on the student. 1. **Force exerted by the student:** - The force due to gravity on the student is given by \( F = m \times g \), where \( m = 45 \) kg and \( g = 9.81 \) m/s\(^2\). - Calculate the force: \( F = 45 \, \text{kg} \times 9.81 \, \text{m/s}^2 \). 2. **Acceleration of the Earth:** - According to Newton’s Third Law, the same force \( F \) acts on the Earth. - Use the formula \( F = M \times a \), where \( M = 6.0 \times 10^{24} \) kg is the mass of the Earth, to find the Earth's acceleration \( a \). - Rearrange the formula to solve for \( a \): \( a = \frac{F}{M} \). By substituting the values calculated in the first step into this equation, you can determine the acceleration of the Earth.
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