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Chapter 4: Problem 22

On earth, two parts of a space probe weigh 11 000 N and 3400 N. These parts are separated by a center-to-center distance of 12 m and may be treated as uniform spherical objects. Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.

Short Answer

Expert verified
The gravitational force is approximately \( 1.803 \times 10^{-7} \text{ N} \).

Step by step solution

01

Identify Known Values

First, identify the given values from the problem: 1. Weight of Part 1: 11,000 N 2. Weight of Part 2: 3,400 N 3. Distance between parts: 12 m It's important to note that these weights are given as the force due to gravity on Earth, which we'll use to find the masses.
02

Convert Weights to Masses

The weight of an object is related to its mass by the equation: \[ W = m imes g \]where \( W \) is the weight, \( m \) is the mass, and \( g \) is the acceleration due to gravity (9.8 m/s虏 on Earth).For Part 1: \[ m_1 = \frac{11,000 \text{ N}}{9.8 \text{ m/s}^2} \approx 1122.45 \text{ kg} \]For Part 2:\[ m_2 = \frac{3,400 \text{ N}}{9.8 \text{ m/s}^2} \approx 346.94 \text{ kg} \]
03

Apply Gravitational Force Formula

Use Newton's Law of Universal Gravitation to find the gravitational force between the two masses:\[ F = G \frac{m_1 m_2}{r^2} \]where:- \( G = 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \) is the gravitational constant,- \( m_1 = 1122.45 \text{ kg} \) is the mass of Part 1,- \( m_2 = 346.94 \text{ kg} \) is the mass of Part 2,- \( r = 12 \text{ m} \) is the distance between the centers of the masses.
04

Calculate Gravitational Force

Substitute the masses and distance into the gravitational force formula:\[ F = 6.674 \times 10^{-11} \times \frac{1122.45 \times 346.94}{12^2} \]Calculate the force:\[ F \approx 6.674 \times 10^{-11} \times \frac{389189.383}{144} \approx 1.803 \times 10^{-7} \text{ N} \]

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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
Newton's Law of Universal Gravitation is a fundamental principle that explains the attractive force between two masses. The law states that every point mass attracts every other point mass in the universe with a force that is proportional to the product of their masses. It is inversely proportional to the square of the distance between them.

Mathematically, this can be expressed as:
  • \( F = G \frac{m_1 m_2}{r^2} \)
  • Where \( F \) is the gravitational force, \( m_1 \) and \( m_2 \) are the masses, \( r \) is the distance between the centers of the two masses, and \( G \) is the gravitational constant.
Newton's law helps us to calculate the gravitational pull in various scenarios, from small objects to massive celestial bodies.
Mass Conversion
Mass conversion is the process of determining the mass from the weight of an object.
In physics, weight is the force exerted by gravity on a mass, given by the equation:
  • \( W = m \times g \)
  • Where \( W \) is the weight, \( m \) is the mass, and \( g \) is the acceleration due to gravity.
  • On Earth, \( g \) is approximately 9.8 \( \, m/s^2 \).
To convert weight to mass:
  • Use the formula \( m = \frac{W}{g} \).
  • This allows us to calculate the mass of an object if the weight is known, as we did for the parts of the space probe.
Distance Between Masses
The distance between masses, often denoted by \( r \), is a crucial factor in calculating gravitational force.
This distance is measured from the center of one mass to the center of the other. It is essential in the formula for gravitational force:
  • The gravitational force decreases with the square of the distance, meaning if the distance doubles, the force reduces to a quarter.
This relationship emphasizes why gravitational force is weaker with greater distances, such as between planets or stars.
Gravitational Constant
The gravitational constant, symbolized as \( G \), is a key component in Newton's Law of Universal Gravitation.
It is the proportionality factor that helps calculate gravitational force.

The numerical value of \( G \) is:
  • \( G = 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \).
This constant is universal, meaning it applies everywhere in the universe. It ensures that the gravitational calculations remain consistent and applicable across different scenarios, from small-scale experiments to large celestial calculations.

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Most popular questions from this chapter

A \(5.0-\mathrm{kg}\) rock and a \(3.0 \times 10^{-4}-\mathrm{kg}\) pebble are held near the surface of the earth. (a) Determine the magnitude of the gravitational force exerted on each by the earth. (b) Calculate the magnitude of the acceleration of each object when released.

Three forces act on a moving object. One force has a magnitude of 80.0 N and is directed due north. Another has a magnitude of 60.0 N and is directed due west. What must be the magnitude and direction of the third force, such that the object continues to move with a constant velocity?

Consult Multiple-Concept Example 10 for insight into solving this type of problem. A box is sliding up an incline that makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is \(0.180 .\) The initial speed of the box at the bottom of the incline is \(1.50 \mathrm{m} / \mathrm{s}\). How far does the box travel along the incline before coming to rest?

The drawing shows a large cube (mass \(=25 \mathrm{kg}\) ) being accelerated across a horizontal frictionless surface by a horizontal force \(\overrightarrow{\mathbf{P}}\). A small cube (mass \(=4.0 \mathrm{kg}\) ) is in contact with the front surface of the large cube and will slide downward unless \(\overrightarrow{\mathbf{P}}\) is sufficiently large. The coefficient of static friction between the cubes is \(0.71 .\) What is the smallest magnitude that \(\overrightarrow{\mathbf{P}}\) can have in order to keep the small cube from sliding downward?

A fisherman is fishing from a bridge and is using a "45-N test line." In other words, the line will sustain a maximum force of \(45 \mathrm{N}\) without breaking. What is the weight of the heaviest fish that can be pulled up vertically when the line is reeled in (a) at a constant speed and (b) with an acceleration whose magnitude is \(2.0 \mathrm{m} / \mathrm{s}^{2} ?\)
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