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Chapter 14: Problem 1

What must the separation be between a \(5.2 \mathrm{~kg}\) particle and a \(2.4 \mathrm{~kg}\) particle for their gravitational attraction to have a magnitude of \(2.3 \times 10^{-12} \mathrm{~N} ?\)

Short Answer

Expert verified
The separation must be approximately 19.05 meters.

Step by step solution

01

Understand the Gravitational Force Formula

The gravitational force between two masses can be calculated using Newton's Law of Universal Gravitation: \[ F = \frac{G \times m_1 \times m_2}{r^2} \] where \( F \) is the gravitational force, \( G \) is the gravitational constant \( (6.67430 \times 10^{-11} \text{m}^3 \text{kg}^{-1} \text{s}^{-2}) \), \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the separation between the masses.
02

Substitute Known Values

We are given \( m_1 = 5.2 \text{kg} \), \( m_2 = 2.4 \text{kg} \), and \( F = 2.3 \times 10^{-12} \text{N} \). Substitute these values into the formula:\[ 2.3 \times 10^{-12} = \frac{(6.67430 \times 10^{-11}) \times 5.2 \times 2.4}{r^2} \]
03

Solve for \( r^2 \)

First, compute the numerator:\[ 6.67430 \times 10^{-11} \times 5.2 \times 2.4 = 8.3281296 \times 10^{-10} \]Now, rearrange the equation to solve for \( r^2 \):\[ r^2 = \frac{8.3281296 \times 10^{-10}}{2.3 \times 10^{-12}} \]Then,\[ r^2 \text{=}\frac{8.3281296 \times 10^{-10}}{2.3 \times 10^{-12}} = 362.963 \]
04

Calculate \( r \)

Finally, take the square root of both sides to find \( r \):\[ r = \text{鈭殅362.963 \text{鈮坿 19.05 \text{m} \]

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Key Concepts

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

Newton's Law of Universal Gravitation
The foundation of calculating gravitational force is rooted in Newton's Law of Universal Gravitation. Sir Isaac Newton proposed that every mass in the universe attracts every other mass with a force that is directly proportional to the product of their masses. However, this force is inversely proportional to the square of the distance between their centers. This relationship can be expressed with the formula: \[ F = \frac{G \times m_1 \times m_2}{r^2} \] Here,
  • \( F \) is the gravitational force between the masses, measured in Newtons (N).
  • G is the gravitational constant, a value that quantifies the strength of gravity (more on this below).
  • \( m_1 \) and \( m_2 \) are the masses of the two objects.
  • \( r \) is the separation between the centers of the two masses.
This equation helps us understand how gravitational forces operate between any two objects, whether they are planets in space or small particles right here on Earth.
Gravitational Constant
The gravitational constant, often represented by the letter G, is a proportionality factor used in the equation for Newton's Law of Universal Gravitation. Its value is approximately \( 6.67430 \times 10^{-11} \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \). This constant allows us to calculate the gravitational force between two masses in the correct units and with the right magnitude. The small value indicates that gravitational forces are relatively weak compared to other fundamental forces, such as electromagnetic forces. However, due to the vast masses involved in astronomical systems, these forces can become enormously significant.
Mass Separation
In the provided exercise, mass separation is the key unknown variable we are solving for. The separation, denoted by \( r \), is the distance between the centers of the two objects. According to Newton's Law of Universal Gravitation, the force of gravity decreases with the square of this distance. To find \( r \), after substituting the values of the given masses, gravitational force, and the gravitational constant into the equation, we solve step by step. Firstly, we rearrange the formula to solve for \( r^2 \). Then, by taking the square root of both sides, we obtain the value of \( r \). This example shows not only the mathematical manipulation needed but also emphasizes the conceptual understanding that gravitational force diminishes as the separation between two masses increases.

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Most popular questions from this chapter

One of the Echo satellites consisted of an inflated spherical aluminum balloon \(30 \mathrm{~m}\) in diameter and of mass \(20 \mathrm{~kg}\). Suppose a meteor having a mass of \(7.0 \mathrm{~kg}\) passes within \(3.0 \mathrm{~m}\) of the surface of the satellite. What is the magnitude of the gravita- tional force on the meteor from the satellite at the closest approach?

A solid sphere of uniform density has a 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 \(m\) 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.

Split into Two \(A\) mass \(M\) is split into two parts, \(m\) and \(M-m\), which are then separated by a certain distance. What ratio \(m / M\) maximizes the magnitude of the gravitational force between the parts?

Some believe that the positions of the planets at the time of birth influence the newborn. Others deride this belief and claim that the gravitational force exerted on a baby by the obstetrician is greater than that exerted by the planets. To check this claim, calculate and compare the magnitude of the gravitational force exerted on a \(3 \mathrm{~kg}\) baby (a) by a \(70 \mathrm{~kg}\) obstetrician who is \(1 \mathrm{~m}\) away and roughly approximated as a point mass, (b) by the massive planet Jupiter \(\left(m=2 \times 10^{27} \mathrm{~kg}\right)\) at its closest approach to Earth \(\left(=6 \times 10^{11} \mathrm{~m}\right)\), and \((\mathrm{c})\) by Jupiter at its greatest distance from Earth \(\left(=9 \times 10^{11} \mathrm{~m}\right)\). (d) Is the claim correct?

Planet Roton, with a mass of \(7.0 \times 10^{24} \mathrm{~kg}\) and a radius of \(1600 \mathrm{~km}\), gravitationally attracts a meteorite that is initially at rest relative to the planet, at a great enough distance to take as infinite. The meteorite falls toward the planet. Assuming the planet is airless. find the speed of the meteorite when it reaches the planet's surface.
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