1) Write the following using powers-of-ten notation:
a) Ten million: 107
b) Sixty thousand: 6 x 104
c) Four one-thousandths: 4-3
d) 20 years= 240 months= 2.4 x 102 months
2) The speed of light is 3.00 X 108 m/s. How long does it take light to travel from the Sun to
Earth? Give your answer in seconds, using powers-of-tens notation.
The average distance from Sun to Earth is around 1.496 X 108 kms. We know that 1 km
has 1000 metres. So, light’s speed in kms would be roughly 3 X 105 km/s. Now, we know
that time t= distance/speed; filling in the values we get:
1.496 X 108 kms / 3 X 10 5 kms. After solving the powers we get, 1.496 X 103 / 3 which is
equal to 498.67 seconds. In minutes, it comes to about 8.31 minutes.
3) Explain where the 206, 265 in the Small-Angle formula came from.
The small-angle formula is: Linear size D= ad/206,265
Where D is the linear size of the object; a= angular size of the object, and d= distance to the
object. We get 206,265 when we divide the number of arcseconds in a circle (i.e. 360 X 60 X 60;
because a circle has 360 degrees, and each degree has 60 arcminutes with each arcminutes
having 60 arcseconds) with 2π which is the ratio of circumference of a circle to its radius.
4) On April 18, 2006, the planet Venus was at a distance of 0.869 AU from Earth. The diameter of
Venus is 12,104 km. What was the angular size of Venus as seen from Earth on April 18, 2006?
Give answer in arcminutes.
We know from above that d is 0.869 AU; D= 12,104 kms; we have to find a (angular size) of
Venus that day. First, we have to convert 0.869 into kms. We know that 1 AU= 1.496 X 108 kms,
so 0.869 AU= 1.496 x 108 kms/ 1 AU X 0.869 AU which gives us roughly 1.3 X 108 kms. Now,
plugging in the formula we get:
12,104 kms= a X (1.3 X 108 kms)/206,625
a= 206,265 X 12,104 kms/1.3 X 108 kms = around 19.15 arcseconds. We know that 1
arcsecond= 1/60 arcminutes. So, 19.15/60= around 0.32 arcminutes
5) Is there any place on Earth where you could see the north celestial pole on the northern
horizon? If so, where? Is there any place on Earth where you could see the north celestial pole
on the western horizon? If so, where? Explain your answers.
On the northern horizon, the NCP is always above horizon. NCP is the Earth’s North Pole
extended in the outer space; we know that the stars move parallel to the horizon so, a person
on northern horizon will be able to see the North Star, which is less than 1 degree away from
the NCP. At the equator, both the poles lie at the horizon.
6) Why is the ecliptic tilted with respect to the celestial equator? Does the Sun appear to move
along the ecliptic, the celestial equator, or neither? By about how many degrees does the Sun
appear to move on the celestial sphere each day?
The celestial equator is just the projection of Earth’s equator into the space, and ecliptic is the path
the Sun appears to move along as the Earth orbits around it. Because the Earth is tilted by 23.5
degrees on its axis, we can imagine that celestial equator is tilted too. But, the ecliptic is not tilted as
it is based on Sun’s apparent movement. Therefore, the ecliptic is tilted with respect to the celestial
equator. Since there are 365.25 days in a year, and there are 360 degrees in a circle, the Sun appears
to move about 1 degree/day.
7) Where on Earth do you have to be to see the NCP directly overhead? What is the maximum
possible elevation of the Sun above the horizon at that location? On what date can this
maximum elevation be observed?
At the North Pole, the NCP is directly overhead. The maximum possible elevation of the Sun
above horizon is 23.5 degrees, and it can observed at the summer solstice i.e. around June 21st.
8) At local noon on March 21, when the Sun is at the Vernal Equinox, a sidereal clock will say it is
midnight. Explain why.
Midnight sidereal time is defined as when vernal equinox crosses the upper meridian. 1 sidereal
day is around 4 minutes shorter than a solar day. This difference increases on March 21 when the
Sun is at the Vernal Equinox. Alternatively, Sun takes half a year to move from Autumnal equinox
to Vernal equinox. That’s 365.25/2 days; each sidereal day is shorter by 4 minutes. That means
365.25/2 X 4 minutes= around 735 minutes or around 12 hours difference. So, at 12 noon, it’ll be
midnight on a sidereal clock.
9) How would the sequence and timing of lunar phases be affected if the Moon moved around its
orbit (a) in the same direction, but twice the speed (b) at the same speed, but in the opposite
direction? Explain your answers.
I think that if moon moved around the earth at twice its current speed, it’ll complete its revolution
around Earth in half of the time that it currently takes i.e. 29.5/2= 14.75 days. But it will not have
any effect on the phases of the moon as it is moving in the same direction. On the other hand, if
it moved in the opposite direction, the phases will still be the same—starting from New Moon,
then Waxing crescent (waxing mean increasing) albeit, we’ll see the other side of the Moon always
visible to us.
10) What is the difference between a sidereal month and a synodic month? Which is longer?
Why?
The sidereal month means the time Moon takes to complete a circle around Earth. Synodic
month is the time taken by Moon to move from one new moon to the next. Since, Earth is
constantly moving around Sun too, Moon has to move slightly more than a full 360 degree circle
to get from one New Moon to the next. Hence, synodic month is longer than a sidereal month.
11) What is the phase of the Moon if it rises at (a) midnight; (b) sunrise; (c) halfway between
sunset and midnight; (d) halfway between noon and sunset? Explain your answers.
(a) It’ll be at 3rd quarter phase
(b) It’ll be at New Moon phase
(c) Waning Gibbous
(d) Waxing Gibbous
12)
13)
14)
15)
The reason being that Moon takes about 27.5 days to complete one complete circle around
the Earth, for it to be in the same position it has to move around 360 degrees/27.5 days=
around 13.1 degrees/day. We all know that Earth takes 24 hours to complete 360 degrees
around its axis. So, 13.1 degrees/360 degrees X 24 hours= 0.87 hours or roughly 52.4
minutes is the time Earth takes so that the Moon appears at the same position in the sky.
So, if during New Moon phase it rises with Sun (it is almost invisible to us), the Waning
Gibbous happens around 10 or so days later i.e. 10 X 52.4 minutes= 524 minutes/60
minutes= around 8.7 hours later than New Moon (which will be 6 am + 8.7 hours = 14.7
hours or 2:42 PM) and so on.
How is annular eclipse of the Sun different than total eclipse of the Sun? What causes this
difference?
During total eclipse of the Sun, the Moon is between Sun and Earth. Since the Moon and Sun’s
diameters have almost the same angular size (angular size of an object appears bigger the closer
it is to us), Moon can completely block the Sun for few moments on a certain part of the Earth
(called its Umbra). But sometimes, the moon is way further from Earth, and cannot block the
Sun completely; resulting in an annular eclipse.
If you are standing directly on the Canada-U.S. border near Northgate, Saskatchewan (where
the ground can be assumed to be absolutely flat), a) what is your latitude? b) What is the
angle between the northern horizon and the north celestial pole? c) What will be the
elevation angle of the Sun at local noon on June 21, September 21, and December 21? Recall
that elevation angle is defined as the angle between the southern horizon and the object (the
Sun, in this case).
An internet search reveals the latitude of Northgate, SK is 49 degrees north. The angle between
the northern horizon and the NCP is also 49 degrees i.e. NCP is 49 degrees to us. The Sun is 23.5
degrees above the Celestial Equator on June 21 so that means the Sun is (90 degrees - 49
degrees) = 41 degrees + 23.5 degrees = 64.5 degrees. On September 21, it’ll be 41 degrees – 0
degrees = 41 degrees and on December 21 it’ll be 41 degrees – 23.5 degrees = 17.5 degrees.
Repeat the above question but use the city of Mayag¨uez in Puerto Rico as your position.
Assume again that the ground is flat. A quick internet search will help you find Mayagu¨ez’s
latitude - provide it to the nearest tenth of a degree.
The latitude of Mayag¨uez is 18.20 degrees north. Meaning the angle between the northern
horizon and the NCP is 18.20 degrees. Hence, the Celestial Equator will be (90 degrees – 18.20
degrees) 71.80 degrees above the southern horizon. At noon June 21, Sun will be 71.80 degrees
+ 23.5 degrees = 95.30 degrees; on September 21: 71.80 + 0 = 71.80 degrees, and on December
21, it’ll be 71.80 – 23.5 degrees = 48.30 degrees to the southern horizon.
What is your zodiac sign? Using any resource available to you, determine the celestial
coordinates of the brightest star in that constellation, and tell me what resource you used to
find this information. Based on those coordinates, in what month would that star transit
(cross the meridian) around midnight (hint: where is the Sun at midnight)? What would be
“your” star’s approximate elevation when it transits?
My zodiac sign is Cancer. According to the website www.space.com, β Cnc (Altarf) is the
brightest star in the Cancer constellation.
Right ascension: 9 hours; Declination: 20 degrees; Best seen between latitudes 90 degrees and
minus 60 degrees and visibility is best during March at 9 p.m.