/*! 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 2 (I) The Sun subtends an angle of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(I) The Sun subtends an angle of about 0.5\(^{\circ}\) to us on Earth, 150 million km away. Estimate the radius of the Sun.

Short Answer

Expert verified
The radius of the Sun is approximately 696,000 km.

Step by step solution

01

Understanding the Problem

The problem gives us the angle subtended by the Sun at Earth as 0.5° and the distance from Earth to the Sun as 150 million km. We need to find the radius of the Sun using these values.
02

Using Angular Diameter Formula

We can use the formula for angular diameter: \[ \theta = \frac{2r}{D} \], where \( \theta \) is the angle in radians, \( r \) is the radius of the Sun, and \( D \) is the distance to the Sun. Convert 0.5° to radians for calculation, using \( 1° = \frac{\pi}{180} \, \text{radians} \).
03

Converting Degrees to Radians

Convert the given angle from degrees to radians. Thus, \( 0.5^{\circ} \) equals \( 0.5 \times \frac{\pi}{180} = \frac{\pi}{360} \) radians.
04

Rearranging the Formula and Substituting Values

Use the rearranged formula \( r = \frac{D \times \theta}{2} \). Substitute \( \theta = \frac{\pi}{360} \) radians and \( D = 150 \times 10^6 \) km.
05

Calculating the Radius

Calculate the radius using the formula: \[ r = \frac{150 \times 10^6 \, \text{km} \times \frac{\pi}{360}}{2} \]. Simplifying gives \[ r = \frac{150 \times 10^6 \times \pi}{720} \].
06

Final Calculation

Calculate the exact value using the provided numbers. Plug in \( \pi \approx 3.14159 \) to get: \[ r \approx \frac{150 \times 10^6 \times 3.14159}{720} \approx 696,000 \, \text{km} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Angular Diameter
The angular diameter is a measure of how large an object appears from a specific point of view. Imagine extending lines from your observation point to the edges of the object. The angle between these lines is the angular diameter. For celestial bodies, this concept helps astronomers estimate size and distance.

It's important to note that angular diameter does not measure the object's actual size. Instead, it reflects how much space the object takes up in your field of view. This is crucial in astronomy where distances are vast, making direct measurement difficult.
  • The angular diameter formula is: \( \theta = \frac{2r}{D} \), where:
  • \( \theta \) is the angular diameter in radians.
  • \( r \) is the radius of the object.
  • \( D \) is the distance to the object.
Understanding the angular diameter is helpful for estimating sizes of planets, stars, and other astronomical phenomena from Earth.
Angle Conversion Made Easy
Angles can be measured in various units, the most common being degrees and radians. Converting between these units is a key skill, especially in mathematics and astronomy.The full circle comprises 360 degrees, and to convert degrees into radians, you use the relation: \( 1\degree = \frac{\pi}{180} \text{ radians} \). This conversion is straightforward and is achieved by multiplying the number of degrees by \( \frac{\pi}{180} \).
  • For example, to convert 0.5°, you'll compute:
  • \( 0.5 \times \frac{\pi}{180} = \frac{\pi}{360} \text{ radians} \).
Remember that radians are preferred in mathematical equations and physics because they provide a direct way of relating angle and arc length. Through consistent practice, converting between degrees and radians becomes second nature.
Basics of Trigonometry in Astronomy
Trigonometry is crucial in the field of astronomy, particularly when it comes to figuring out the sizes and distances of celestial objects. It involves relationships between angles and side lengths in right-angled triangles. In the context of the original problem, a simplified form of trigonometry is used, where you consider the Sun, Earth, and the radius of the Sun as forming a right triangle.
  • The angular diameter formula is derived from the tangent function in trigonometry.
  • Trigonometry intimates the ratio between the opposite side (the radius) and the adjacent side (distance from Earth to the Sun).
By solving equations derived from trigonometric principles, astronomers can calculate unknown distances and sizes using known angles—making trigonometry indispensable for expanding our understanding of the universe.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(II) In traveling to the Moon, astronauts aboard the \(Apollo\) spacecraft put the spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis. Determine \((a)\) the angular acceleration, and \((b)\) the radial and tangential components of the linear acceleration of a point on the skin of the ship 6.0 min after it started this acceleration.

(I) An automobile engine slows down from 3500 rpm to 1200 rpm in 2.5 s. Calculate \((a)\) its angular acceleration, assumed constant, and \((b)\) the total number of revolutions the engine makes in this time.

(I) \((a)\) What is the angular momentum of a 2.8-kg uniform cylindrical grinding wheel of radius 28 cm when rotating at 1300 rpm? \((b)\) How much torque is required to stop it in 6.0 s?

(II) A small 350-gram ball on the end of a thin, light rod is rotated in a horizontal circle of radius 1.2 m. Calculate \((a)\) the moment of inertia of the ball about the center of the circle, and \((b)\) the torque needed to keep the ball rotating at constant angular velocity if air resistance exerts a force of 0.020 N on the ball. Ignore air resistance on the rod and its moment of inertia.

(II) A bicycle with tires 68 cm in diameter travels 9.2 km. How many revolutions do the wheels make?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.