/*! 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 23 Two particles are separated by \... [FREE SOLUTION] | 91Ó°ÊÓ

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Two particles are separated by \(19 \mathrm{~m}\) when their gravitational attraction has a magnitude of \(4.0 \times 10^{-12} \mathrm{~N}\). If the mass of particle 1 is \(5.2 \mathrm{~kg}\), what is the mass of particle \(2 ?\)

Short Answer

Expert verified
The mass of particle 2 is approximately 4.30 kg.

Step by step solution

01

Identify the known values

We know the following: the gravitational force between the particles is given as \( F = 4.0 \times 10^{-12} \, \text{N} \). The distance between the particles is \( r = 19 \, \text{m} \). The mass of the first particle is \( m_1 = 5.2 \, \text{kg} \).
02

Recall the formula for gravitational force

The gravitational force between two masses is given by Newton's law of universal gravitation: \( F = \frac{G m_1 m_2}{r^2} \), where \( G = 6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2 \) is the gravitational constant.
03

Rearrange the formula to solve for mass

To find the mass of the second particle \( m_2 \), rearrange the equation: \( m_2 = \frac{F r^2}{G m_1} \).
04

Substitute the known values into the equation

Plug in the given values: \( m_2 = \frac{(4.0 \times 10^{-12} \, \text{N}) (19 \, \text{m})^2}{(6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2) (5.2 \, \text{kg})} \).
05

Calculate the value of \( m_2 \)

First calculate \( r^2 = 19^2 = 361 \). Then calculate the numerator: \( 4.0 \times 10^{-12} \times 361 = 1.444 \times 10^{-9} \). Now calculate \( m_2 = \frac{1.444 \times 10^{-9}}{6.674 \times 10^{-11} \times 5.2} = 4.30 \, \text{kg} \).

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Key Concepts

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

Gravitational Force
Gravitational force is the attraction that exists between any two masses. This fundamental force is described by Newton's Law of Universal Gravitation, which can be summed up in the formula \[F = \frac{G m_1 m_2}{r^2}\]where:
  • \(F\) is the gravitational force.
  • \(G\) is the gravitational constant.
  • \(m_1\) and \(m_2\) are the masses of the two objects.
  • \(r\) is the distance between the centers of the two masses.
People often find gravitational force in many everyday contexts, although it is most noticeable when involving large masses like the Earth and the Moon. The force is always attractive, pulling objects toward each other.
One crucial aspect to remember is that the gravitational force is dependent on both the masses involved and the distance between them. Therefore, changing either of these variables will directly affect the gravitational pull between the objects.
Mass Calculation
Calculating mass using gravitational force involves rearranging Newton's formula to solve for the unknown mass. Specifically, if you want to find one of the masses, you can rearrange the equation like this:\[m_2 = \frac{F r^2}{G m_1}\]This formula emerged from isolating \(m_2\) on one side of the gravitational force equation. Once you have this formula ready, you substitute the known values for \(F\), \(r\), \(G\), and \(m_1\). Upon performing these operations, you successfully calculate the unknown mass \(m_2\).
Let’s break down the calculation step by step:
  • Square the distance \(r\) between the objects.
  • Calculate \(F r^2\), which is the product of the gravitational force and the squared distance.
  • Divide this product by \(G m_1\), where \(G\) is the gravitational constant and \(m_1\) is the known mass.
By following these steps, the unknown mass can be easily found, as demonstrated in the original exercise.
Gravitational Constant
The gravitational constant, denoted as \(G\), is a key component in the law of universal gravitation. It represents the constant of proportionality in the gravitational force equation and is an essential factor in our calculations.
\(G\) has a standard value of \[6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2\]This small numerical value indicates why we don't notice gravitational forces in our daily lives unless massive bodies or large distances are involved, such as with planets or stars.
The gravitational constant plays an instrumental role in predicting gravitational attraction accurately. Without \(G\), the interaction between masses in space could not be mathematically described with precision. Understanding \(G\) helps explain why celestial bodies move the way they do and allows scientists to determine fundamental properties of planets and other astronomical entities.

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Most popular questions from this chapter

Three identical stars of mass \(M\) form an equilateral triangle that rotates around the triangle's center as the stars move in a common circle about that center. The triangle has edge length \(L\). What is the speed of the stars?

A solid sphere has a uniformly distributed mass of \(1.0 \times 10^{4}\) \(\mathrm{kg}\) and a radius of \(1.0 \mathrm{~m}\). What is the magnitude of the gravitational force due to the sphere on a particle of mass \(m=2.0 \mathrm{~kg}\) when the particle is located at a distance of (a) \(1.5 \mathrm{~m}\) and (b) \(0.50 \mathrm{~m}\) from the center of the sphere? (c) Write a general expression for the magnitude of the gravitational force on the particle at a distance \(r \leq 1.0 \mathrm{~m}\) from the center of the sphere.

(a) What is the escape speed on a spherical asteroid whose radius is \(700 \mathrm{~km}\) and whose gravitational acceleration at the surface is \(4.5 \mathrm{~m} / \mathrm{s}^{2} ?\) (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of \(1000 \mathrm{~m} / \mathrm{s}\) ? (c) With what speed will an object hit the asteroid if it is dropped from \(1000 \mathrm{~km}\) above the surface?

In 1993 the spacecraft Galileo sent an image (Fig. 13-32) of asteroid 243 Ida and a tiny orbiting moon (now known as Dactyl), the first confirmed example of an asteroid-moon system. In the image, the moon, which is \(1.5 \mathrm{~km}\) wide, is \(100 \mathrm{~km}\) from the center of the asteroid, which is \(55 \mathrm{~km}\) long. Assume the moon's orbit is circular with a period of \(27 \mathrm{~h}\). (a) What is the mass of the asteroid? (b) The volume of the asteroid, measured from the Galileo images, is \(14100 \mathrm{~km}^{3}\). What is the density (mass per unit volume) of the asteroid?

Assume a planet is a uniform sphere of radius \(R\) that (somehow) has a narrow radial tunnel through its center (Fig. 13-7). Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let \(F_{R}\) be the magnitude of the gravitational force on the apple when it is located at the planet's surface. How far from the surface is there a point where the magnitude is \(\frac{1}{3} F_{R}\) if we move the apple (a) away from the planet and (b) into the tunnel?

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