/*! 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 18 On earth, two parts of a space p... [FREE SOLUTION] | 91Ó°ÊÓ

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On earth, two parts of a space probe weigh \(11000 \mathrm{~N}\) and \(3400 \mathrm{~N}\). These parts are separated by a center-to-center distance of \(12 \mathrm{~m}\) 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 \(1.811 \times 10^{-6} \, \text{N}\).

Step by step solution

01

Understand Gravitational Force Formula

The gravitational force between two objects is given by Newton's law of universal gravitation:\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]where \( F \) is the gravitational force, \( G \) is the gravitational constant \(6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\), \( m_1 \) and \( m_2 \) are the masses of the objects, and \( r \) is the distance between the centers of the two objects.
02

Convert Weights to Masses

The weight \( W \) of an object is given by \( W = mg \), where \( m \) is the mass and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity on Earth.For the first object, \( m_1 = \frac{11000}{9.81} = 1121.31 \, \text{kg} \).For the second object, \( m_2 = \frac{3400}{9.81} = 346.58 \, \text{kg} \).
03

Calculate Gravitational Force

Now that we have \( m_1 = 1121.31 \, \text{kg} \), \( m_2 = 346.58 \, \text{kg} \), and the distance \( r = 12 \, \text{m} \), we can substitute these values into the formula for gravitational force:\[ F = \frac{(6.674 \times 10^{-11}) \cdot 1121.31 \cdot 346.58}{12^2} \]Calculate and simplify to find:\[ F \approx 1.811 \times 10^{-6} \, \text{N} \].

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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 describes the attractive force between any two masses in the universe. According to this law, every point mass attracts every other point mass by a force acting along the line intersecting both points. The magnitude of this gravitational force (\( F \)) is expressed by the formula:\[F = \frac{G \cdot m_1 \cdot m_2}{r^2}\]where:
  • \( F \) is the gravitational force.
  • \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).
  • \( m_1 \) and \( m_2 \) are the masses of the two objects.
  • \( r \) is the distance between the centers of the two masses.
This law helps us understand how massive bodies like planets, stars, and other celestial objects exert attractive forces on each other.
mass conversion
In physics, to calculate gravitational force, we often need to convert weights into masses. This conversion is necessary because weight is the force exerted by gravity on an object, which varies depending on the gravitational field. However, mass is a measure of the amount of matter in an object and remains constant.To convert weight (\( W \)) to mass (\( m \)), we use the formula:\[m = \frac{W}{g}\]where:
  • \( W \) is the weight of the object.
  • \( g \) is the acceleration due to gravity. On Earth, \( g \approx 9.81 \, \text{m/s}^2 \).
For example, if an object weighs \( 11000 \, \text{N} \) on Earth, its mass is calculated by dividing its weight by \( 9.81 \, \text{m/s}^2 \).Applying mass conversion is essential for accurately calculating gravitational interactions in space, where the gravitational field might differ from Earth's.
distance between objects
The distance between the centers of two objects is crucial when calculating gravitational force. In the formula for gravitational force, the distance (\( r \)) is squared, which means even small changes can have a large impact on the force calculated.In space, where distances can be vast, accurately determining this value is essential. For our exercise, the space probe parts are separated by a center-to-center distance of \( 12 \, \text{m} \). Keeping this distance precise is important, especially when applying Newton's law of universal gravitation.Using the exact distance ensures that the calculated force accurately reflects the interaction between the objects. This accuracy is particularly important in real-world applications, such as satellite operation and space exploration.
calculation of gravitational constant
The gravitational constant (\( G \)) is an essential part of calculating gravitational forces. Known as the "universal constant," \( G \) provides the proportionality factor needed to apply Newton's law of universal gravitation. Its value is approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).Before its discovery, predicting the gravitational forces between celestial bodies was nearly impossible. When calculating the gravitational force in any given scenario, ensuring the proper value of \( G \) is used provides consistency and ensures accurate results. It serves as a reminder of the delicate balance and the universal nature of gravity, underpinning countless calculations and scientific studies in astronomy and physics.
acceleration due to gravity
Acceleration due to gravity (\( g \)) is the acceleration that the Earth imparts to objects on or near its surface. On Earth, \( g \) is approximately \( 9.81 \, \text{m/s}^2 \). This value is critical when distinguishing between weight and mass of objects.On other celestial bodies, \( g \) can vary significantly due to differences in size and mass. Despite these variations, this acceleration is always directed towards the center of the body and plays a central role in everything from determining an object's weight to influencing the trajectories of orbiting satellites.Understanding acceleration due to gravity is crucial for comprehending how gravitational attraction works in different environments, aiding in scientific explorations and practical calculations across various fields, like engineering and space travel.

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Most popular questions from this chapter

A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs \(31 \mathrm{~N}\). The coefficient of static friction between his hands and the book is \(0.40 .\) To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

Interactive LearningWare 4.1 at reviews the approach taken in problems such as this one. A 1580 -kg car is traveling with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of \(50.0 \mathrm{~m} ?\)

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. (a) 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 \(2.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

A large rock and a small pebble are held at the same height above the ground. (a) Is the gravitational force exerted on the rock greater than, less than, or equal to that exerted on the pebble? Justify your answer. (b) When the rock and the pebble are released, is the downward acceleration of the rock greater than, less than, or equal to that of the pebble? Why? Problem 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 acceleration of each object when released. Verify that your answers are consistent with your answers to the Concept Questions.

The alarm at a fire station rings and an 86 -kg fireman, starting from rest, slides down a pole to the floor below (a distance of \(4.0 \mathrm{~m}\) ). Just before landing, his speed is \(1.4 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

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