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Posts Tagged ‘**basic astronomy**’

### Exercise questions

Posted January 24, 2009

on:I posted here some exercise problems that you can use to expand your knowledge in some chapter in basic astronomy :

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1.) The following statements are true or false:

- (a) Of all the natural satellites in the Solar System only the Moon always turns the same face towards its primary.
- (b) The mass of a planet in the Solar System can be determined only if it possesses one or more satellites.
- (c) The planet with the largest apparent angular diameter when nearest the Earth is Venus.
- (d) Pluto is the planet farthest from the Sun.
- (e) A lunar eclipse may occur if the Moon is new.

2.) Calculate the mean density of Jupiter from the following data, assuming the orbits of Earth and Jupiter to be circular and coplanar:

- Angular semi-diameter of Jupiter at opposition = 21”,8
- Orbital radius of Jupiter = 5,2 A.U.
- Mass of Jupiter/mass of Earth = 318
- Mean density of Earth = 5,5 kg/m^-3
- Sun’s horizontal parallax = 8”,8

3.) The two components of a binary star are approximately equal brightness. Their maximum separation is 1”,3 and the period is 50,2 years. The composite spectrum shows double lines with a maximum separation of 0,18 Angstrom at 5000 Angstrom. Assuming that the plane of the orbit contains the line of sight, calculate (i) the total mass of the system in the terms of the solar mass, (ii) the parallax of the system.

=================================================== (source : ** Astronomy, The Structure Of Universe**).

After we talk about parallax, now we will discuss about angular diameter.

**I. Definition**

The angle that the actual diameter of an object makes in the sky; also known as *angular size* or *apparent diameter*. The **angular diameter** of an object as seen from a given position is the “visual diameter” of the object measured as an angle. The visual diameter is the diameter of the perspective projection of the object on a plane through its center that is perpendicular to the viewing direction. Because of foreshortening, it may be quite different from the actual physical diameter for an object that is seen under an angle. For a disk-shaped object at a large distance, the visual and actual diameters are the same.The Moon, with an actual diameter of 3,476 kilometers, has an angular diameter of 29′ 21″ to 33′ 30″, depending on its distance from Earth. If both angular diameter and distance are known, *linear diameter* can be easily calculated.

The Sun and the Moon have angular diameters of about half a degree, as would a 10-centimeter (4-inch) diameter orange at a distance of 11.6 meters (38 feet). People with keen eyesight can distinguish objects that are about an arc minute in diameter, equivalent to distinguishing between two objects the size of a penny at a distance of 70 meters (226 feet). Modern telescopes allow astronomers to routinely distinguish objects one arc second in diameter, and less. The Hubble Space Telescope, for example, can distinguish objects as small as 0.1 arc seconds. For comparison, 1 arc second is the apparent size of a penny seen at a distance of 4 kilometers (2.5 miles).

The angular diameter is proportional to the actual diameter divided by its distance. If any two of these quantities are known, the third can be determined.

For example if an object is observed to have an apparent diameter of 1 arc second and is known to be at a distance of 5,000 light years, it can be determined that the actual diameter is 0.02 light years.

**II. Formulas**

The angular diameter of an object can be calculated using the formula:

in which δ is the angular diameter, and *d* and *D* are the visual diameter of and the distance to the object, expressed in the same units. When *D* is much larger than *d*, δ may be approximated by the formula δ = *d* / *D*, in which the result is in radians.

For a spherical object whose *actual* diameter equals *d*_{act}, the angular diameter can be found with the formula:

for practical use, the distinction between *d* and *d*_{act} only makes a difference for spherical objects that are relatively close.

**III. Estimating Angular Diameter**

This illustration shows how you can use your hand to make rough estimates of angular sizes. At arm’s length, your little finger is about 1 degree across, your fist is about 10 degrees across, etc. *Credit: NASA/CXC/M.Weiss*

**IV. Use in Astronomy**

In astronomy the sizes of objects in the sky are often given in terms of their angular diameter as seen from Earth, rather than their actual sizes.

The angular diameter of Earth’s orbit around the Sun, from a distance of one parsec, is 2″ (two arcseconds).

The angular diameter of the Sun, from a distance of one light-year, is 0.03″, and that of the Earth 0.0003″. The angular diameter 0.03″ of the Sun given above is approximately the same as that of a person at a distance of the diameter of the Earth.[1]

This table shows the angular sizes of noteworthy celestial bodies as seen from the Earth:

Sun | 31.6′ – 32.7′ |

Moon | 29.3′ – 34.1′ |

Venus | 10″ – 66″ |

Jupiter | 30″ – 49″ |

Saturn | 15″ – 20″ |

Mars | 4″ – 25″ |

Mercury | 5″ – 13″ |

Uranus | 3″ – 4″ |

Neptune | 2″ |

Ceres | 0.8″ |

Pluto | 0.1″ |

* Betelgeuse: 0.049″ – 0.060″

* Alpha Centauri A: ca. 0.007″

* Sirius: ca. 0.007″

This meaning the angular diameter of the Sun is ca. 250,000 that of Sirius (it has twice the diameter and the distance is 500,000 times as much; the Sun is 10,000,000,000 times as bright, corresponding to an angular diameter ratio of 100,000, so Sirius is roughly 6 times as bright per unit solid angle).

The angular diameter of the Sun is also ca. 250,000 that of Alpha Centauri A (it has the same diameter and the distance is 250,000 times as much; the Sun is 40,000,000,000 times as bright, corresponding to an angular diameter ratio of 200,000, so Alpha Centauri A is a little brighter per unit solid angle).

The angular diameter of the Sun is about the same as that of the Moon (the diameter is 400 times as large and the distance also; the Sun is 200,000-500,000 times as bright as the full Moon (figures vary), corresponding to an angular diameter ratio of 450-700, so a celestial body with a diameter of 2.5-4″ and the same brightness per unit solid angle would have the same brightness as the full Moon).

Even though Pluto is physically larger than Ceres, when viewed from Earth, e.g. through the Hubble Space Telescope, Ceres has a much larger apparent size.

While angular sizes measured in degrees are useful for larger patches of sky (in the constellation of Orion, for example, the three stars of the belt cover about 3 degrees of angular size), we need much finer units when talking about the angular size of galaxies, nebulae or other objects of the night sky.

Degrees, therefore, are subdivided as follows:

* 360 degrees (º) in a full circle

* 60 arc-minutes (′) in one degree

* 60 arc-seconds (′′) in one arc-minute

To put this in perspective, the full moon viewed from earth is about ½ degree, or 30 arc minutes (or 1800 arc-seconds). The moon’s motion across the sky can be measured in angular size: approximately 15 degrees every hour, or 15 arc-seconds per second. A one-mile-long line painted on the face of the moon would appear to us to be about one arc-second in length.

Source : Wikipedia and encyclopedia of science.

### Basic Astronomy (part 1)

Posted September 20, 2008

on:Before we learn further about astronomy, there are some basic knowledges that we must know and understand.

First, we will talk about measuring distance in astronomy.

Astronomical object lies in a very great distance from us. So far than our sense can perceive. That’s why our sense can’t have a 3-D visualization of the universe. Our sense can’t differ closer to farther objects. So, we need some trick to know how far an object from us. One of the simplest method used by astronomers to measure distance of some closest star is using the parallax effect.

Parallax is an optical effect seen when the observer seeing an object from two different positions. The object will be seen shifted relative to the farther background objects.

The parallax effect is one of those things you see everyday and think nothing of until it’s given some mysterious scientific-sounding name. There’s really no magic here. Consider the following simple situation.

You’re riding in a car on a highway out west. It’s a beautiful sunny day, and you can see for miles in every direction. Off to your left, in the distance, you see a snow-capped mountain. In front of that mountain, and much closer to the car, you see a lone ponderosa pine standing in a field next to the highway. I’ve diagramed this idyllic scene in the figure below:

As you drive by the field, you notice an interesting sight. When you’re in the position on the left side of the figure, the tree appears to be to the right of the mountain. You can see this in the figure by the fact that the line of sight to the tree (indicated by the green line) is rightward of the line of sight to the mountain (indicated by the blue line). A picture of what you see out the window of your car is shown below the car.

The interesting part is that as your drive on, you notice that the tree and mountain have switched positions; that is, by the time you reach the right hand position in the above figure, the tree appears to be to the left of the mountain. You can see this in the figure by noting that the line of sight to the tree (green line) is leftward of the line of sight to the mountain (blue line). A picture of what you see out the window of your car now is shown below the car.

What’s going on here? It’s pretty clear that the tree and mountain haven’t moved at all, yet the tree appears to have jumped from one side of the mountain to the other. By now, you’re probably saying *“Well, DUH, the tree is just closer to me than the mountain. What’s so remarkable about that?”* I would answer, *“There’s nothing at all remarkable about it. It’s just the effect of parallax.”* In fact, if you understand the above discussion, you already understand the parallax effect.

Now let’s talk about measuring the distance to the tree using this information. From the above information, you can see that it would be pretty easy to measure the angle between the direction to the tree and the direction to the mountain in both instances. Let’s call those angles **A** and **B**, respectively. Now, if the mountain is sufficiently distant so that the direction to the mountain from both viewpoints is the same, then the two blue lines in the figure below are parallel.

This helps a lot, because we can then show that the angle made by the two green lines (i.e., the difference in the direction to the pine tree from the two viewpoints) is equal to the sum of **A** and **B**. To see this, construct a line through the pine tree parallel to the two blue lines in the figure (this line is shown as a dotted line above). Then all of the blue lines are parallel, and each of the green lines crosses a pair of parallel lines. Reach deep back into your high school geometry (or equivalently, just stare at the above figure for a minute), and you’ll remember or realize that the angles at the pine tree labeled **A** and **B** have the same values as the angles **A** and **B** measured at the two car positions. Thus, the angle between the two green lines is the sum of **A** and **B**, which are angles we can measure from the comfort of our car.

Now, if we know the distance **D** we’ve traveled, then we have an Observer’s Triangle and we can solve for the distance to the tree using the Observer’s Triangle relation

**alpha/57.3 = D/R **where **alpha** is the angle at the tree (**A** + **B**), **D** is the distance we’ve traveled between views, and **R** is the distance from the road to the tree.** (source : Astronomy 101 Specials: Measuring Distance via the Parallax Effect).**

We will use the same method to measure the star’s distance. This method is called * trigonometric parallax* because we only use simple triangulation to find the distance. The only problem is star’s distance is so huge so the parallax effect will be so small (less than 1 arc second; 1 arc second = 1/3600 of a degree). So, that’s why this method can only measure accurately for several nearby stars. Farther star will need different, more complex, indirect method to derive its distance.

As explained before, the stars are so far away that observing a star from opposite sides of the Earth would produce a parallax angle much, much too small to detect (That’s why ancient people can’t detect this shifting to prove heliocentric view) . As a consequence, we must use large a baseline as possible. The largest one that can be easily used is the orbit of the Earth. In this case the baseline is the mean distance between the Earth and the Sun—an **astronomical unit** (AU) or 149.6 million kilometers! A picture of a nearby star is taken against the background of stars from opposite sides of the Earth’s orbit (six months apart). The parallax angle *p* is one-half of the total angular shift.

However, even with this large baseline, the distances to the stars in units of astronomical units are huge, so a more convenient unit of distance called a **parsec** is used (abbreviated with “pc”). A parsec is the distance of a star that has a parallax of one arc second using a baseline of 1 astronomical unit. Therefore, one parsec = 206,265 astronomical units. The nearest star is about 1.3 parsecs from the solar system. In order to convert parsecs into standard units like kilometers or meters, you must know the numerical value for the astronomical unit—it sets the scale for the rest of the universe. Its value was not know accurately until the early 20th century. In terms of light years, one parsec = 3.26 light years.

Using a parsec for the distance unit and an arc second for the angle, our simple angle formula above becomes extremely simple for measurements from Earth:

*p* = 1/*d*

Parallax angles as small as 1/50 arc second can be measured from the *surface* of the Earth. This means distances *from the ground* can be determined for stars that are up to 50 parsecs away. If a star is further away than that, its parallax angle *p* is too small to measure and you have to use more indirect methods to determine its distance. Stars are about a parsec apart from each other on average, so the method of trigonometric parallax works for just a few thousand nearby stars. The Hipparcos mission greatly extended the database of trigonometric parallax distances by getting above the blurring effect of the atmosphere. It measured the parallaxes of 118,000 stars to an astonishing precision of 1/1000 arc second (about 20 times better than from the ground)! It measured the parallaxes of 1 million other stars to a precision of about 1/20 arc seconds. Selecting the Hipparcos link will take you to the Hipparcos homepage and the catalogs.

The actual stellar parallax triangles are much longer and skinnier than the ones typically shown in astronomy textbooks. They are so long and skinny that you do not need to worry about which distance you actually determine: the distance between the Sun and the star or the distance between the Earth and the star. Taking a look at the skinny star parallax triangle above and realizing that the triangle should be over 4,500 times longer (!), you can see that it does not make any significant difference which distance you want to talk about. If Pluto’s entire orbit was fit within a quarter (2.4 centimeters across), the nearest star would be 80 meters away! But if you are stubborn, consider these figures for the planet-Sun-star star parallax triangle setup above (where the planet-star side is the hypotenuse of the triangle):

the Sun — nearest star distance = 267,068.23022*0* AU = 1.2948 pc

the Earth–nearest star distance = 267,068.23022*2* AU = 1.2948 pc

Pluto–nearest star distance = 267,068.23*3146* AU = 1.2948 pc !

If you are super-picky, then yes, there is a slight difference but no one would complain if you ignored the difference. For the more general case of parallaxes observed from any planet, the distance to the star in parsecs *d = ab/p*, where *p* is the parallax in arc seconds, and *ab* is the distance between the planet and the Sun in AU.

Formula (1) relates the planet-Sun baseline distance to the size of parallax measured. Formula (2) shows how the star-Sun distance *d* depends on the planet-Sun baseline and the parallax. In the case of Earth observations, the planet-Sun distance *ab* = 1 A.U. so *d = 1/p*. From Earth you simply flip the parallax angle over to get the distance! (Parallax of 1/2 arc seconds means a distance of 2 parsecs, parallax of 1/10 arc seconds means a distance of 10 parsecs, etc.)

A nice visualization of the parallax effect is the Distances to Nearby Stars and Their Motions lab (link will appear in a new window) created for the University of Washington’s introductory astronomy course. With this java-based lab, you can adjust the inclination of the star to the planet orbit, change the distance to the star, change the size of the planet orbit, and even add in the effect of proper motion. (**source : www.astronomynotes.com**)

**Units in Distance**

- Astronomical Unit (A.U). It is defined as the mean distance of the Sun from the Earth. Its value is about 149,6 million km. This unit is conveniently used to express distance to the object in solar system because we can directly compared the distance to Earth-Sun distance.
- One light year is defined as the distance that light has traveled in light years. Light has velocity about 300.000 km/s. So, one light year equals to 9,46 x 10^12 km. This unit is mostly used to express the distance of extragalactic object. Remember that light’s speed is finite so distant objects are seen as they are in the past. For example the Sun. The Sun that we see at this moment is the Sun as it was 8 minutes ago. Light needs about 8 minutes to travel the Earth-Sun distance. So, looking farther objects mean we’re looking even further to the past. That’s why light years is more commonly used to express distant object’s distance. When we say that a cluster’s distance is 8 billion light years, it means that the cluster that we seen right now is the way it looks 8 billion years ago !
- Parsec (
*Parallax second*). Star that have parallax 1 arc second have distance about 3,26 light years or 206.265 A.U (astronomical unit). Astronomer use this distance as a unit to express distance of the star. It is called a*parsec*. This unit is favorable to express star’s distance because it is closely related to star’s parallax (p). (remember that parallax = 1/distance, while the observer is on Earth, parallax is expressed in arc second and distance is expressed in parsec).

**So, for reviewing our understanding about the parallax, try to answer these questions:**

- If a star has parallax 0″,711, determine its distance (
*in light years*) from us! - Assume we can measure parallax from Mars (with the same technology that we used here on Earth). Assume that we can measure accurately using parallax method until 200 parsec from the Earth (distance limit). Determine the distance limit if we conduct the measurement of star’s distance using parallax method. Given that the distance of Mars from the Sun is about 1,52 AU.
- You observe an asteroid approaching the Earth. You have two observatories 3200 km apart, so you can measure the parallax shift of the incoming asteroid. You observe the parallax shift to be 0,022 degrees.Determine : (a) the parallax expressed in radians (b) the asteroid’s distance from Earth.
- If you measure the parallax of a star to be 0,1 arc seconds on Earth, how big would the parallax of the same star for an observer on Mars?
- If you measure the parallax of a star to be 0,5 arc seconds on Earth and an observer in a space station in the orbit around the Sun measures a parallax for the same star to be 1 arc seconds, how far is the space station from the Sun ?

You can share your solution of the above questions in the comment column.

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