교과서 질문
In its orbit each day, the International Space Station makes 15.65 revolutions around the earth. Assuming a circular orbit, how high is this satellite above the surface of the earth?
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Ch 13: Gravitation
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors41
- Ch 02: Motion Along a Straight Line67
- Ch 03: Motion in Two or Three Dimensions47
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws59
- Ch 06: Work & Kinetic Energy14
- Ch 07: Potential Energy & Conservation8
- Ch 08: Momentum, Impulse, and Collisions18
- Ch 09: Rotation of Rigid Bodies42
- Ch 10: Dynamics of Rotational Motion48
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics31
- Ch 13: Gravitation35
- Ch 14: Periodic Motion32
- Ch 15: Mechanical Waves25
- Ch 16: Sound & Hearing32
- Ch 17: Temperature and Heat36
- Ch 18: Thermal Properties of Matter38
- Ch 19: The First Law of Thermodynamics33
- Ch 20: The Second Law of Thermodynamics22
- Ch 21: Electric Charge and Electric Field36
- Ch 22: Gauss' Law24
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF26
- Ch 26: Direct-Current Circuits30
- Ch 27: Magnetic Field and Magnetic Forces18
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction28
- Ch 30: Inductance17
- Ch 31: Alternating Current21
- Ch 32: Electromagnetic Waves1
- Ch 33: The Nature and Propagation of Light20
- Ch 34: Geometric Optics34
- Ch 35: Interference19
- Ch 36: Diffraction25
- Ch 37: Special Relativity20
- Ch 38: Photons: Light Waves Behaving as Particles15
- Ch 39: Particles Behaving as Waves26
- Ch 40: Quantum Mechanics I: Wave Functions21
- Ch 41: Quantum Mechanics II: Atomic Structure25
- Ch 42: Molecules and Condensed Matter18
- Ch 43: Nuclear Physics16
- Ch 44: Particle Physics and Cosmology23
13장, 문제 23
Two satellites are in circular orbits around a planet that has radius 9.00 × 106 m. One satellite has mass 68.0 kg, orbital radius 7.00 × 107 m, and orbital speed 4800 m/s. The second satellite has mass 84.0 kg and orbital radius 3.00 × 107 m. What is the orbital speed of this second satellite?
검증된 단계별 안내1
Identify the relevant physics principle: The gravitational force provides the necessary centripetal force for a satellite in circular orbit. This can be expressed as: F_gravity = F_centripetal.
Write the formula for gravitational force: F_gravity = (G * M * m) / r^2, where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, and r is the orbital radius.
Write the formula for centripetal force: F_centripetal = (m * v^2) / r, where m is the mass of the satellite, v is the orbital speed, and r is the orbital radius.
Set the gravitational force equal to the centripetal force: (G * M * m) / r^2 = (m * v^2) / r. Notice that the mass of the satellite (m) cancels out from both sides of the equation.
Solve for the orbital speed (v) of the second satellite: v = sqrt((G * M) / r). Substitute the given orbital radius of the second satellite into this equation to find its orbital speed.
비슷한 문제에 대한 검증된 영상 답변:
이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
5m도움이 되었나요?
주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
Gravitational Force and Circular Motion
In circular orbits, the gravitational force provides the necessary centripetal force to keep a satellite in orbit. This relationship is expressed as F_gravity = F_centripetal, where F_gravity = G * (m1 * m2) / r^2 and F_centripetal = m * v^2 / r. Understanding this balance is crucial for determining the orbital speed of a satellite.
추천 영상:
가이드 코스
05:41
Gravitational Forces in 2D
Orbital Speed Formula
The orbital speed of a satellite in a circular orbit can be derived from the balance of gravitational and centripetal forces, resulting in v = sqrt(G * M / r), where G is the gravitational constant, M is the mass of the planet, and r is the orbital radius. This formula allows us to calculate the speed needed to maintain a stable orbit at a given radius.
추천 영상:
가이드 코스
6:03Speed and Energy of Elliptical Orbits
Kepler's Third Law
Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. For circular orbits, this implies that the orbital speed is inversely proportional to the square root of the orbital radius, which helps compare the speeds of satellites at different radii around the same planet.
추천 영상:
가이드 코스
08:32Kepler's Third Law
관련 실천
교과서 질문
For a satellite to be in a circular orbit 890 km above the surface of the earth, what orbital speed must it be given?
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교과서 질문
On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 km above the earth's surface. Assume a circular orbit. How many hours does it take this satellite to make one orbit?
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교과서 질문
A planet orbiting a distant star has radius 3.24 × 106 m. The escape speed for an object launched from this planet’s surface is 7.65 × 103 m/s. What is the acceleration due to gravity at the surface of the planet?
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교과서 질문
The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho1 Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho1 Cancri?
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교과서 질문
In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already was known to have a large satellite Charon, orbiting at 19,600 km with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto.
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