Dive into the fascinating world of physics with our detailed explanation of Newton’s Law of Gravitation. This fundamental law governs the interactions between any two objects with mass, from apples falling from trees to planets orbiting stars. Understanding Newton’s Law of Gravitation is crucial for comprehending the forces that shape our universe. We’ll explore how this law works and see a practical example to solidify your understanding.

This explanation of Newton’s Law of Gravitation will walk you through the concept and provide a clear example. We’ll use a straightforward problem involving two masses and calculate the gravitational force between them. Moreover, we’ll break down the solution step-by-step, making the process easy to follow. Consequently, you’ll gain a solid grasp of how to apply Newton’s Law of Gravitation in various scenarios.

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Newton’s Law of Universal Gravitation – Numerical Example

This example demonstrates the calculation of gravitational force between two masses using Newton’s Law of Universal Gravitation. This fundamental law in physics is crucial for understanding the interactions between celestial bodies and objects on Earth.

Problem Statement

Two masses, ## m_1 = 10 \, \text{kg} ## and ## m_2 = 5 \, \text{kg} ##, are separated by a distance of ## r = 2 \, \text{m} ##. Calculate the gravitational force acting between them. The universal gravitational constant, ## G ##, is approximately ## 6.674 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2 ##.

Solution

Understanding the Problem

This problem involves applying Law of Universal Gravitation to calculate the force of attraction between two objects. The key is to correctly substitute the given values into the formula and perform the necessary calculations.

Solving the Problem

Step 1: Formula

The formula for gravitational force is:

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

Step 2: Substituting Values

Substituting the given values into the formula:

### F = (6.674 \times 10^{-11}) \frac{(10 \, \text{kg})(5 \, \text{kg})}{(2 \, \text{m})^2} ###

Step 3: Simplifying

Simplifying the expression:

### F = (6.674 \times 10^{-11}) \frac{50 \, \text{kg}^2}{4 \, \text{m}^2} ###

Step 4: Calculation

Performing the calculation:

### F = (6.674 \times 10^{-11}) \times 12.5 \, \text{kg}^2/\text{m}^2 ### ### F \approx 83.425 \times 10^{-11} \, \text{N} ###

Step 5: Expressing in Scientific Notation

Expressing the result in scientific notation:

### F \approx 8.3425 \times 10^{-10} \, \text{N} ###

Final Answer

The gravitational force between the two masses is approximately:

##8.34 \times 10^{-10} \, \text{N}##

This example demonstrates a straightforward application of Newton’s Law of Universal Gravitation. The result shows the extremely small magnitude of gravitational forces between objects with moderate masses at typical distances.

Concept	Formula/Equation	Calculation/Data
Newton’s Law of Gravitation	### F = G \frac{m_1 m_2}{r^2} ###
Universal Gravitational Constant (G)		## 6.674 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2 ##
Mass 1 (m1)		## 10 \, \text{kg} ##
Mass 2 (m2)		## 5 \, \text{kg} ##
Separation Distance (r)		## 2 \, \text{m} ##
Gravitational Force (F)		## 8.34 \times 10^{-10} \, \text{N} ##

This detailed explanation of Newton’s Law of Gravitation offers a comprehensive understanding of this fundamental principle in physics. We’ve explored how this law dictates the interactions between any two objects with mass, from everyday objects to celestial bodies. The example provided clearly illustrates the application of the formula and the calculations involved, making the concept more accessible and tangible.

The numerical example demonstrates a practical application of Newton’s Law of Gravitation. It showcases the calculation of the gravitational force between two masses, highlighting the importance of precise calculations and understanding the units involved. This example also underscores the extremely small magnitude of gravitational forces between objects with moderate masses at typical distances. This small magnitude is often overlooked but is crucial in understanding the dynamics of the universe.

• Newton’s Law of Gravitation: This law explains the attractive force between any two objects with mass. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
• Gravitational Constant (G): A fundamental constant in physics, G represents the strength of the gravitational force. Its value is crucial in calculations involving Newton’s Law of Gravitation.
• Units of Measurement: Careful attention to units (e.g., kilograms for mass, meters for distance, and Newtons for force) is essential for accurate calculations in Newton’s Law of Gravitation.
• Practical Applications: Newton’s Law of Gravitation is essential for understanding planetary motion, satellite orbits, and even the forces acting on objects on Earth.
• Practical Applications: Newton’s Law of Gravitation is essential for understanding planetary motion, satellite orbits, and even the forces acting on objects on Earth.

Understanding Newton’s Law of Gravitation is a cornerstone of physics, laying the groundwork for more advanced concepts like Einstein’s theory of general relativity. The provided example serves as a solid foundation for further exploration and problem-solving in this fascinating field. Further exploration of the law can involve considering the effects of varying masses and distances on the gravitational force.