/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 The drawing (not to scale) shows... [FREE SOLUTION] | 91Ó°ÊÓ

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The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force \(\overrightarrow{\mathbf{F}}_{\mathrm{sM}}\) that the sun exerts on the moon is perpendicular to the force \(\overrightarrow{\mathbf{F}}_{\mathrm{EM}}\) that the earth exerts on the moon. The masses are: mass of sun \(=1.99 \times 10^{30} \mathrm{kg},\) mass of earth \(=5.98 \times 10^{24} \mathrm{kg},\) mass of moon \(=\) \(7.35 \times 10^{22} \mathrm{kg} .\) The distances shown in the drawing are \(r_{\mathrm{SM}}=1.50 \times 10^{11} \mathrm{m}\) and \(r_{\mathrm{EM}}=3.85 \times 10^{8} \mathrm{m} .\) Determine the magnitude of the net gravitational force on the moon.

Short Answer

Expert verified
The net gravitational force on the moon is approximately \( 4.82 \times 10^{20} \text{N} \).

Step by step solution

01

Calculate the Gravitational Force between Sun and Moon

According to Newton's Law of Universal Gravitation, the force between two masses is calculated by the formula \( F = \frac{G \, m_1 \, m_2}{r^2} \), where \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) is the gravitational constant. For the sun and moon: \( F_{\text{sM}} = \frac{(6.67 \times 10^{-11}) \times (1.99 \times 10^{30}) \times (7.35 \times 10^{22})}{(1.50 \times 10^{11})^2} \approx 4.36 \times 10^{20} \text{N} \).
02

Calculate the Gravitational Force between Earth and Moon

Using the same gravitational formula, calculate the force between Earth and Moon. \( F_{\text{EM}} = \frac{(6.67 \times 10^{-11}) \times (5.98 \times 10^{24}) \times (7.35 \times 10^{22})}{(3.85 \times 10^{8})^2} \approx 1.98 \times 10^{20} \text{N} \).
03

Determine the Net Gravitational Force on the Moon

The net gravitational force is the vector sum of \( F_{\text{sM}} \) and \( F_{\text{EM}} \). Since these forces are perpendicular, use the Pythagorean theorem to find the net force: \( F_{\text{net}} = \sqrt{(F_{\text{sM}})^2 + (F_{\text{EM}})^2} \). Substitute the values: \( F_{\text{net}} = \sqrt{(4.36 \times 10^{20})^2 + (1.98 \times 10^{20})^2} \approx 4.82 \times 10^{20} \text{N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force Calculation
Gravitational force calculation plays a pivotal role in physics, especially when dealing with celestial bodies like the sun, earth, and moon. Newton's Law of Universal Gravitation provides a straightforward formula: \[ F = \frac{G \, m_1 \, m_2}{r^2} \]Here, the formula tells us that the force (\( F \)) between two objects depends on:
  • The product of their masses (\( m_1 \) and \( m_2 \))
  • Divided by the square of the distance (\( r^2 \)) between them
  • Multiplied by the universal gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
In the problem, we first calculate the gravitational force between the sun and the moon:- Mass of the sun: \(1.99 \times 10^{30} \, \text{kg} \)- Mass of the moon: \(7.35 \times 10^{22} \, \text{kg} \)- Distance (\( r_{\text{SM}} = 1.50 \times 10^{11} \, \text{m} \))The calculated gravitational force is approximately \(4.36 \times 10^{20} \, \text{N} \).
Similarly, we calculate the force between Earth and the moon using the mass and distance provided, resulting in approximately \(1.98 \times 10^{20} \, \text{N} \). These calculations highlight how variations in mass and distance influence gravitational attraction.
Vector Sum of Forces
When dealing with multiple forces acting on a single object, like the moon being pulled by both the sun and Earth simultaneously, we must determine the vector sum of forces. Vectors represent forces with both magnitude and direction, making it crucial to understand their combination.In our specific case, the gravitational forces from the sun (\(F_{\text{sM}}\)) and the Earth (\(F_{\text{EM}}\)) are perpendicular to each other. This simplifies the process of finding their net effect. To calculate the vector sum:- First, appreciate that forces at right angles can be resolved using the Pythagorean theorem:\[ F_{\text{net}} = \sqrt{(F_{\text{sM}})^2 + (F_{\text{EM}})^2} \]- Given our earlier calculations, by substituting the values, the net gravitational force on the moon becomes approximately \(4.82 \times 10^{20} \text{N} \).
This method allows us to determine the resulting force's magnitude exerted on the moon. As such, understanding how to work with vectors is essential for solving problems involving the combination of different force directions.
Physics Problem Solving
Physics problem solving often involves multiple steps, requiring a clear understanding of the principles involved. When faced with a complex problem, like determining the gravitational force on the moon, employing a systematic approach is crucial. Here's a general approach: - **Identify what you know:** Gather all given values, such as mass and distance in gravitational calculations. - **Use relevant formulas:** Newton's Law of Universal Gravitation, vector addition principles, etc. - **Calculate step by step:** Start with known formulas for individual forces, then progress to find vector sums. - **Reassess and verify:** Double-check calculations and logic to ensure correctness. By following these steps, complex problems become manageable. The key is to maintain clarity throughout.
Breaking down each aspect and applying foundational laws ensures accuracy. This methodological way of problem-solving in physics allows students to tackle a variety of scenarios beyond this exercise, from astronomy to engineering.

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Most popular questions from this chapter

A 205-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at \(30.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.900 , and the log has an acceleration of magnitude \(0.800 \mathrm{m} / \mathrm{s}^{2}\). Find the tension in the rope.

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