/*! 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 37 What is the ratio of the sun's g... [FREE SOLUTION] | 91Ó°ÊÓ

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What is the ratio of the sun's gravitational force on you to the earth's gravitational force on you?

Short Answer

Expert verified
The ratio of the sun's gravitational force on you to the earth's gravitational force on you is very small, implying the force of gravity from the Earth on you is significantly larger than that from the Sun.

Step by step solution

01

Identify Variables

First, identify and list down the known variables. Mass of Earth \(m_e = 5.972 \times 10^{24} kg\), mass of the Sun \(m_s = 1.989 \times 10^{30} kg\), mass of human \(m_h \approx 70 kg\), distance from Earth to human \(r_e \approx 6.370 km\), distance from Sun to human \(r_s \approx 150 \times 10^6 km\). Convert all distance to the metric system (in meters).
02

Calculate Gravitational Force by Earth

Use the gravitational force formula to calculate the gravitational force exerted on the human by the Earth. \(F_e = G \frac{m_h \times m_e}{r_e^2}\)
03

Calculate Gravitational Force by Sun

Similarly, calculate the gravitational force exerted on the human by the Sun. \(F_s = G \frac{m_h \times m_s}{r_s^2}\)
04

Compute Ratio

Compute the ratio of the sun's gravitational force on you to the earth's gravitational force on you by dividing \(F_s\) by \(F_e\).

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Key Concepts

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

Gravitational Force Formula
When we talk about the universe and the celestial bodies within it, including our own planet Earth and the Sun, one of the most fascinating forces at work is the force of gravity. The gravitational force formula, which is at the heart of gravitational interactions, is crucial for understanding why objects attract each other.

The simplest form of the gravitational force formula is given by Isaac Newton’s law of universal gravitation, which states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, this is given by: \[ F = G \frac{m_1 \times m_2}{r^2} \]where F represents the gravitational force, G is the gravitational constant (approximately \(6.674 \times 10^{-11} N\cdot(m/kg)^2\)), \(m_1\) and \(m_2\) are the masses of the two objects, and \(r\) is the distance between the centers of the two masses.
Mass and Distance in Gravity
Two fundamental aspects that affect the gravitational pull between any two objects are mass and distance. Mass here refers to the amount of matter present in an object, which is often measured in kilograms. A common misconception is that weight and mass are the same, but this is not the case; mass is a measure of matter, while weight is the measure of the force of gravity on an object.

When solving problems involving gravity, we must take into account that the gravitational force increases with the mass of the objects—more massive objects exert a more substantial gravitational pull. Conversely, as the distance between the two objects increases, the gravitational force diminishes—in accordance with the inverse square law. This is due to the nature of the space through which the gravitational force operates; as space expands between objects, the intensity of the gravitational field decreases with the square of the distance. Understanding the interrelation of mass and distance is critical to solving gravitational force problems effectively.
Newton's Law of Universal Gravitation
Newton's law of universal gravitation is a cornerstone of classical physics and has monumental implications for our comprehension of the cosmos. The law reveals the underlying simplicity of gravitational interactions across the universe and provides a framework that applies to objects ranging from the smallest particles to the largest galaxies.

Newton's law teaches us that every bit of matter in the universe exerts an attractive force on every other bit, regardless of where they are. It is because of this law that planets orbit stars and moons orbit planets. The beauty of this law lies in its universality—it is as applicable to an apple falling from a tree on Earth as it is to the complex orbits of planets in far-off solar systems. The calculations done in the simplified textbook exercise demonstrate the direct application of Newton's law, showing that despite the colossal distances and differences in mass, we can quantify and compare the gravitational forces that exist between two objects. By calculating the ratios of gravitational forces, students can gain a deeper understanding of how these forces vary in different parts of our solar system.

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Most popular questions from this chapter

In \(2014,\) a space probe approached the rocky core of the comet Churyumov- Gerasimenko, which is only a few \(\mathrm{km}\) in diameter. The probe then entered orbit around the comet at a distance of \(30 \mathrm{km} .\) The comet was found to have a mass of \(1.0 \times 10^{13} \mathrm{kg} .\) What was the orbital period of the probe around the comet, in earth days?

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The ultracentrifuge is an important tool for separating and analyzing proteins in biological research. Because of the enor mous centripetal accelerations that can be achieved, the apparatus (see Figure 6.18 ) must be carefully balanced so that each sample is matched by another on the opposite side of the rotor shaft. Any difference in mass of the opposing samples will cause a net force in the horizontal plane on the shaft of the rotor; this force can actually be large enough to destroy the centrifuge. Suppose that a scientist makes a slight error in sample preparation, and one sample has a mass 10 mg greater than the opposing sample. If the samples are \(10 \mathrm{cm}\) from the axis of the rotor and the ultracentrifuge spins at 70,000 rpm, what is the magnitude of the net force on the rotor due to the unbalanced samples?

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