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Ch 03: Motion in Two or Three Dimensions
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Young & Freedman Calc 14th EditionCh 03: Motion in Two or Three Dimensions문제 23b
3장, 문제 23b

The earth has a radius of 6380 km and turns around once on its axis in 24 h. If arad at the equator is greater than g, objects will fly off the earth's surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

검증된 단계별 안내
1
First, understand that the problem involves centripetal acceleration at the Earth's equator. The centripetal acceleration \( a_{rad} \) is given by the formula \( a_{rad} = \frac{v^2}{r} \), where \( v \) is the tangential velocity and \( r \) is the radius of the Earth.
Next, calculate the tangential velocity \( v \) at the equator using the formula \( v = \frac{2\pi r}{T} \), where \( T \) is the period of rotation. Here, \( r = 6380 \) km, which needs to be converted to meters for consistency in units.
Set the centripetal acceleration \( a_{rad} \) equal to the gravitational acceleration \( g \), which is approximately \( 9.8 \text{ m/s}^2 \). This is the condition for objects to fly off the Earth's surface.
Substitute \( v = \frac{2\pi r}{T} \) into \( a_{rad} = \frac{v^2}{r} \) and set \( a_{rad} = g \). This gives \( g = \frac{(\frac{2\pi r}{T})^2}{r} \). Simplify this equation to solve for \( T \).
Rearrange the equation to find \( T \), which is the period of rotation required for \( a_{rad} \) to be greater than \( g \). The final expression will be \( T = \frac{2\pi r}{\sqrt{gr}} \). Calculate \( T \) using the known values of \( r \) and \( g \).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
3m
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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Centripetal Acceleration

Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. It is given by the formula arad = v^2/r, where v is the tangential velocity and r is the radius of the circular path. At the Earth's equator, this acceleration is due to the Earth's rotation.
추천 영상:
가이드 코스
06:48
Intro to Centripetal Forces

Gravitational Acceleration

Gravitational acceleration, denoted as g, is the acceleration due to Earth's gravity, approximately 9.81 m/s² at the surface. It acts downward towards the center of the Earth. For objects to remain on the surface, gravitational acceleration must be greater than the centripetal acceleration caused by Earth's rotation.
추천 영상:
가이드 코스
07:32
Weight Force & Gravitational Acceleration

Rotational Period

The rotational period is the time it takes for an object to complete one full rotation around its axis. For Earth, this period is 24 hours. If the rotational period decreases, the tangential velocity at the equator increases, leading to higher centripetal acceleration. The question asks for the period at which this acceleration would exceed gravitational acceleration, causing objects to fly off.
추천 영상:
가이드 코스
06:28
Satellite Period
관련 실천
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교과서 질문

A model of a helicopter rotor has four blades, each 3.40 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. What is the radial acceleration of the blade tip expressed as a multiple of g?

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교과서 질문

A model of a helicopter rotor has four blades, each 3.40 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. What is the linear speed of the blade tip, in m/s?

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교과서 질문

The earth has a radius of 6380 km and turns around once on its axis in 24 h. What is the radial acceleration of an object at the earth's equator? Give your answer in m/s2 and as a fraction of g.

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교과서 질문

A 124 kg balloon carrying a 22 kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0 kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. How high is the balloon when the rock is thrown?

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교과서 질문

A 124 kg balloon carrying a 22 kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0 kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. At the instant the rock hits the ground, how far is it from the basket?

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