Loose Leaf For Explorations:  Introduction To Astronomy
Loose Leaf For Explorations: Introduction To Astronomy
9th Edition
ISBN: 9781260432145
Author: Thomas T Arny, Stephen E Schneider Professor
Publisher: McGraw-Hill Education
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Chapter 5, Problem 3P

Can the unaided human eye resolve a crater on the Moon whose angular diameter is 2 arc minutes (= 120 arc seconds)? (Take the diameter of the pupil of the eye to be about 5 millimeters and the wavelength of the light to be 500 nanometers.)

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The Mars Reconnaissance Orbiter (MRO) flies at an average altitude of 280km above the Martian Surface. If its cameras have an angular resolution of 0.2 arc seconds, what is the size of the smallest objects that the MRO can detect on the Martian surface? Use the equation: S =x × d /  206265  arcseconds / radian , where S is the true size of the object, d is the distance from the detector to the object, and x is the angular size of the object. Your answer will be in km (you can ignore the radians unit (it should appear, but the equation made a simplifying assumption that dropped it out.
The smallest detail visible through a Earth-based telescope is about 1.00 arcsecond in diameter. Use the small angle formula to determine to the right number of significant figures the size of the object in meters this would represent on Mars as it is at the closest distance from Earth. (??????? ???????? (arc−seconds)2.06×105=?????? ??????????????? ; ???−????????????−????=5.46×105 ??).
Large telescopes often have small fields of view, i.e. it can only see a very small corner of the sky. For example, the Hubble Space Telescope (HST) Advanced Camera has a field of view that is roughly square and about 0.055 degree on a side. How do I Calculate the angular area of the HST's field of view in square degrees?
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