Textbook Question
The International Space Station orbits 300 km above the surface of the earth. What is the gravitational force on a 1.0 kg sphere inside the International Space Station?
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Ch 13: Newton's Theory of Gravity
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 13, Problem 8
What is the free-fall acceleration at the surface of (a) the moon and (b) Jupiter?
Verified step by step guidance1
Identify the formula for gravitational acceleration: , where is the gravitational constant, is the mass of the celestial body, and is the radius of the celestial body.
For part (a), substitute the values for the moon: and . Plug these into the formula to calculate the free-fall acceleration on the moon.
For part (b), substitute the values for Jupiter: and . Plug these into the formula to calculate the free-fall acceleration on Jupiter.
Simplify the expression for each case by performing the operations in the numerator and denominator separately, then divide the results. Ensure the units are consistent throughout the calculation.
Interpret the results: The calculated values represent the free-fall acceleration at the surface of the moon and Jupiter, respectively. These values indicate how quickly an object would accelerate due to gravity if dropped from rest on each celestial body.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Free-Fall Acceleration
Free-fall acceleration refers to the acceleration of an object due solely to the influence of gravity, without any other forces acting on it. On Earth, this acceleration is approximately 9.81 m/s². However, this value varies depending on the mass and radius of the celestial body in question, as gravitational force is directly proportional to mass and inversely proportional to the square of the distance from the center of the mass.
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Vertical Motion & Free Fall
Gravitational Force
Gravitational force is the attractive force that exists between any two masses. It is described by Newton's law of universal gravitation, which states that the force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers. This principle helps calculate the gravitational acceleration on different celestial bodies, such as the Moon and Jupiter.
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Celestial Body Characteristics
The characteristics of a celestial body, such as its mass and radius, significantly influence its gravitational pull. For instance, the Moon has a smaller mass and radius compared to Earth, resulting in a lower free-fall acceleration of about 1.62 m/s². In contrast, Jupiter, being the largest planet in the solar system, has a much greater mass, leading to a free-fall acceleration of approximately 24.79 m/s² at its cloud tops.
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Related Practice
Textbook Question
A recently discovered extrasolar planet appears to be rockier and denser than earth. It is 16 times as massive as earth, but its diameter is only twice that of earth. What is the free-fall acceleration on the surface of this planet?
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Textbook Question
Two 65 kg astronauts leave earth in a spacecraft, sitting 1.0 m apart. How far are they from the center of the earth when the gravitational force between them is as strong as the gravitational force of the earth on one of the astronauts?
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Textbook Question
Suppose we could shrink the earth without changing its mass. At what fraction of its current radius would the free-fall acceleration at the surface be three times its present value?
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Textbook Question
Asteroid 253 Mathilde is one of several that have been visited by space probes. This asteroid is roughly spherical with a diameter of 53 km. The free-fall acceleration at the surface is 9.9 ✕ 10-3 m/s2. What is the asteroid's mass?
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Textbook Question
What is the force of attraction between a 50 kg woman and a 70 kg man sitting 1.0 m apart?
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