/*! 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 6 Venus has a diameter of about 80... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Venus has a diameter of about 8000 miles. When it is visible as a "morning star" (or "evening star"), it is about as far away as the sun, i.e., about 93 million miles. It looks "larger than a point" to the unaided eye. Are you seeing the true size of Venus?

Short Answer

Expert verified
No, we do not see Venus in its true size due to its distance.

Step by step solution

01

Understanding the Question

The question asks whether we see Venus in its true size when it is visible in the sky. To address this, we need to consider the distance of Venus from the Earth and the way objects appear smaller when they are further away.
02

Calculate Angular Size

To determine if we are seeing Venus' true size, we need to calculate its angular size, which is a measure of how large it appears to an observer on Earth. The formula for angular size is given by \( \theta = \frac{d}{D} \), where \( d \) is the diameter of the object and \( D \) is the distance from the observer to the object. Convert the units if necessary so that both \( d \) and \( D \) are in the same units.
03

Convert Distance and Diameter

First, we convert the diameter of Venus from miles to the same unit as Distance, i.e., million miles. Venus' diameter is 8000 miles, which is \( 0.008 \) million miles. The distance to Venus is the same as the Sun, \( 93 \) million miles.
04

Compute Angular Size

Using the formula \( \theta = \frac{d}{D} \), substitute \( d = 0.008 \) million miles and \( D = 93 \) million miles into the formula to find \( \theta = \frac{0.008}{93} \approx 8.6 \times 10^{-5} \) radians. This small angular size confirms that even though Venus looks larger than a point, it is significantly smaller than its true size as perceived from Earth.
05

Concluding the Observation

Given the small angular size computed, Venus does not appear with its true physical dimensions to an observer on Earth, even if it does not look like a point. This effect is typical for distant celestial objects because of their immense distance compared to their actual size.

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.

Celestial Objects
Celestial objects are fascinating wonders of the universe. These objects include stars, planets, moons, asteroids, and comets that we can observe in the night sky.
One key example is Venus, often recognized for its bright appearance as a 'morning star' or 'evening star'. Celestial objects like Venus are a part of our solar system and are visible to us because they reflect the sun's light.
Understanding celestial objects involves observing their size, brightness, and motion through the sky. Their vast distances from us mean that we often perceive them as smaller than they actually are. In the case of Venus, its diameter is around 8,000 miles, which makes it a significant planetary body.
Observing celestial objects helps us understand the universe we live in. With advanced telescopes and technology, astronomers study these bodies to learn more about their compositions, atmospheres, and potential for supporting life.
Observational Astronomy
Observational astronomy is the practice of studying celestial objects through direct observation. This field allows astronomers and enthusiasts alike to explore the night sky, discovering various stars, planets, and other celestial phenomena.
It involves the use of equipment like telescopes to magnify these distant objects, making them appear larger and clearer than they do to the naked eye. Observers can watch these celestial bodies from different locations on Earth and note changes or movements over time.
A significant aspect of observational astronomy is measuring the angular size of an object. The angular size gives an idea of how large an object appears from a particular point of view. For example, Venus seems larger than a point to the unaided eye when viewed from Earth, but it is much smaller than its true size due to its immense distance.
By refining observational techniques, astronomers can gain insights into the physical characteristics and behaviors of these celestial objects, contributing immensely to our knowledge of the cosmos.
Distance Measurements
Distance measurements are crucial in the world of astronomy. They help us determine how far away celestial objects are from the Earth, which in turn influences how we perceive them.
Distances in astronomy are vast and often measured in units like light-years, astronomical units (AU), or million miles. For example, Venus is approximately 93 million miles away from Earth when it is visible as a 'morning star'.
Knowing the distance to celestial objects allows astronomers to calculate their angular size, a vital step in understanding how large they appear in the sky. The formula to calculate angular size is \( \theta = \frac{d}{D} \), where \( d \) is the diameter of the object, and \( D \) is its distance from the observer.
Properly measuring these distances and sizes helps us create accurate models of the universe, demonstrating the relative positions and movements of celestial bodies. This understanding aids in navigation, space exploration, and our broader comprehension of the cosmos.

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

A diffraction-limited laser beam of diameter \(1 \mathrm{~cm}\) is pointed at the moon. What is the diameter of the area illuminated on the moon? (The moon is \(240,000 \mathrm{mi}\) away.) Take the light wavelength to be \(6328 \AA\). Neglect scattering in the earth's atmosphere.

Eye-pupil size and mental activity. If someone shows you a picture of a good- looking individual of the opposite sex, your eye-pupil diameter may increase by as much as \(30 \%\), according to Eckhard \(\mathrm{H}\). Hess, Scientific American p. 46 (April, 1965 ). This large a change is very easy to detect in your own pupil by using a pinhole in a piece of aluminum foil that covers one eye, with a bright source illuminating the pinhole, as discussed in Sec. 9.7. Perhaps by just thinking, you can vary your pupil size, depending on what you think about. Have someone read to you. (Concentrate on listening, not on the pupil size.)

Derive a formula for the magnification of a pinhole magnifier. Check the formula as follows: Make two marks \(2 \mathrm{~cm}\) apart on one piece of paper; make two marks \(2 \mathrm{~mm}\) apart on another. Put the pinhole over one eye and nothing over the other. Both pieces of paper should be illuminated from behind (at least, that is easiest to look at). With both eyes open, look with one eye through the pinhole at the \(2 \mathrm{~mm}\) marks and with your other eye look at the \(2 \mathrm{~cm}\) marks. Bring the \(2 \mathrm{~mm}\) marks closer until you superpose the two sets of marks with your two eyes. Measure appropriate distances.

Suppose you cover one of two slits with a microscope slide and the other with nothing. If the slide has thickness \(1 \mathrm{~mm}\), show that monochromatic light of wavelength \(5000 \AA\) gets a retardation in one slit of about 1000 wavelengths relative to the other. If the double-slit pattern is not to wash out, the light must be fairly monochromatic. How narrow a band of wavelengths (in angstroms) is required so that the relative phase shift of the two slits varies by less than 180 deg from one edge of the wavelength band to the other? How could you use this fact to measure the bandwidth of a spectral line? (What would you measure and plot versus what, and how would you obtain the bandwidth from the plot?)

Light is emitted from an unpolarized point source. First it passes through a linear polarizer with easy transmission axis at 45 deg to the \(x\) and \(y\) axes. Then it is incident on a double slit. Each slit is covered by a linear polarizer, one slit having the polarization axis along \(\hat{x}\), the other having it along \(\hat{y}\). (a) Suppose you look at the interference pattern with the unaided eye. Do you expect to have the usual two-slit interference pattern? What do you expect? (b) Next suppose you look at the interference pattern while holding a polaroid linear polarizer in front of one eye. What do you expect to see? What happens as you rotate the polaroid in front of your eye? (c) Now suppose you look at the pattern through a circular polarizer run backward as an analyzer. What pattern do you expect to see? There are many nice variations you can make on this problem: (i) Put a righthanded circular polarizer over one slit and a left-banded circular polarizer over the other and repeat the above observations. (ii) Add a quarter- or half-wave plate just behind the slits, etc.

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.