/*! 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 car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of \(+11 \mathrm{~m} / \mathrm{s}\) in a time of \(28 \mathrm{~s}\). The combined mass of the boat and trailer is 410 kg. The frictional force acting on the trailer can be ignored. What is the tension in the hitch that connects the trailer to the car?

Problem 77 The drawing shows Robin Hood (mass \(=77.0 \mathrm{~kg}\) ) about to escape from a dangerous situation. With one hand, he is gripping the rope that holds up a chandelier (mass \(=195\) kg). When he cuts the rope where it is tied to the floor, the chandelier will fall, and he will be pulled up toward a balcony above. Ignore the friction between the rope and the beams over which it slides, and find (a) the acceleration with which Robin is pulled upward and (b) the tension in the rope while Robin escapes.

Consult Interactive LearningWare 4.2 at before beginning this problem. A truck is traveling at a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) along a level road. A crate is resting on the bed of the truck, and the coefficient of static friction between the crate and the truck bed is \(0.650 .\) Determine the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck.

A 350 -kg sailboat has an acceleration of \(0.62 \mathrm{~m} / \mathrm{s}^{2}\) at an angle of \(64^{\circ}\) north of east. Finc the magnitude and direction of the net force that acts on the sailboat.

Concept Questions Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The woman pushes on the man with a certain force that is parallel to the ground. (a) Must the man accelerate under the action of this force? If so, what three factors determine the magnitude and direction of his acceleration? (b) Is there a corresponding force exerted on the woman? If so, where does it originate? Is this force related to the magnitude and direction of the force the woman exerts on the man? If so, how? Problem The mass of the man is \(82 \mathrm{~kg}\) and that of the woman is \(48 \mathrm{~kg} .\) The woman pushes on the man with a force of \(45 \mathrm{~N}\) due east. Determine the acceleration (magnitude and direction) of (a) the man and (b) the woman.

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