Assignment 09 - Solutions
Astronomy 1 -

Chapter 8 
 - Review Question #14 
 - Problems #7, #9

14. Explain why Mercury is never seen overhead at midnight in Earth's sky.

------------- Answer -------------------
Mercury is an inferior Planet with a semi-major axis of 0.39 AU - its maximum elongation 
	from the Sun can only be at most 28 degrees.
The Sun must be below the horizon to see Mercury - hence it will alway be less than 28 degrees
	altitude and alway near where the Sun sets or rises. Overhead = altitude of 90 degrees
	and at midnight the Sun is 90 degrees below the horizon that is an elongation of 
	Mercury of 180 degrees which is impossible.

7. Compare the gravitational tidal effect of the Sun on Mercury with the tidal effect of Earth on the Moon (see Section 7.6).

------------- Answer -------------------
The differential tidal force is proportional to the Mass of the perturbing body and inversely
	proportional to the cube of its distance away. F ~ M/r3

M(sun) = 2x1030 kg	;	M(Earth) = 6x1024 kg
rSM = r(Sun-Mercury) = 0.39AU	;	rEM = r(Earth-Moon) = 0.384/150 = 1/391 AU
	
So....		F(Mercury)/F(Moon) = [M(sun)/M(Earth)[rEM/rSM)3
			= (2x1030kg)/(6x1024kg)[1/(0.39x391)]3
			= 0.094
The tidal forces of the Sun on Mercury are 1/10 of those of the Earth on the Moon.

9. Using the rate given in the text for the formation of 10-km craters on the Moon, estimate how long would be needed for the entire Moon to be covered with new craters of this size. How much higher must the cratering rate have been in the past to cover the entire lunar surface with such craters in the 4.6 billion years since the Moon formed?

------------- Answer -------------------
Crater Formation Rate: One new 10 km dia crater once every 10 million years
				= 10-7per year
	a - find the number of craters needed on the Moon - (Area of Moon)/(Area of Crater)
	b - Divide the number needed by the rate of formation to get the time period needed.

Area of Crater ~ 10 kmx10 km = 100 km2
Radius of the Moon = 1738 km
Area of the Moon = 4(pi)R2 = 4(3.14)(1738 km)2 = 3.80x107 km2
	Number of Craters formed = 3.80x107 km2/100 km2 = 380,000 
	Time of Formation = 380,000/10-7per year = 3.8x1012 years
				= 3,800 Billion years
	The rate of formation to cover the Moon in 4.5 Billion years must be
			3800/4.5 = 840 times faster than at present.
			ie. 840 per 10 million years 
			or 84 10km dia craters per million year

Chapter 9 
 - Review Question #6, #9 
 - Problems #1, #10 

6. Name three ways in which the atmosphere of Venus differs from that of Earth.

------------- Answer -------------------
	Venus		Earth
	-------		------
1	very dense	less dense
2	Mostly CO2	Mostly N2
3	Little H20	Lot H20
4	No Oxygen	Oxygen Atmosphere
5	50 km thick	10 km thick

9. Earth and Venus are nearly alike in size and density. What primary fact caused one planet to evolve as an oasis for life, while the other became a dry and inhospitable inferno?

------------- Answer -------------------
The Solar intensity at Venus' orbit is twice that at Earth's Orbit. 
The higher temperatures evaporated the water oceans and the UV from 
the Sun lead to the destruction of the water molecule. No water did 
not allow the formation of life.

1. Using the data given in the text, calculate Venus's angular diameter, as seen by an observer on Earth, when the planet is (a) at its brightest, (b) at greatest elongation, and (c) at the most distant point in its orbit.

------------- Answer -------------------
Angular Size(radians) = Diameter/Distance
Angular size("arc) = 206,300 Diameter/Distance

Use the diameter of Venus, its orbital radius and the orbital radius of the Earth.
	diameter = 12104 km		(1 AU = 150 million km)
		= 8.07x10-5AU
	a(Venus) = 0.723 AU
	a(Earth) = 1.0 AU

Note: The distance of Venus when it is brightest can be found by using the fact that the 
elongation of Venus from the Sun at that time is 39 degrees. Likewise, at greatest elongation,
Venus has an elongation of 47 degrees and the distance can be found by trigonometry or a scale 
drawing as above.

Greatest Brillancy distance = 0.42 AU
Greatest Elongation distance = 0.69 AU

10. Calculate the orbital period of the Magellan spacecraft, moving around Venus on an elliptical orbit with a minimum altitude of 294 km and a maximum altitude of 8543 km above the planet's surface. In 1993 the spacecraft's orbit was changed to have minimum and maximum altitudes of 180 km and 541 km, respectively. What was the new period?

------------- Answer -------------------
To calculate the orbital period, we need the semi-major axis of the spacecraft.
	Radius of Venus = 6052 km
	rp = perivenus = 6052 + 294 = 6346 km
	ra = apvenus = 6052 + 8543 = 14,595 km
	a1 = (ra + rp)/2 = 10470 km
The second Orbit has
	a2 = (180 + 541 + 12104)/2 = 6412 km

Calculate the periods using
Kepler's law   			P2 ~ a3/M
	Compare with a satellite around Earth. MVenus = 0.815 MEarth
	Use the Moon as a satellite of the Earth since we know its period as 27.3 days and
		its semi-major axis is 384,000 km

	P1 = PMoon [a1/aMoon]3/2 Mvenus/Mearth
		= 27.3 [10470/384000]3/2 (0.815) = 0.100 days 
		= 2.4 hours

	P2 = PMoon [a2/aMoon]3/2 Mvenus/Mearth
		= 27.3 [6412/384000]3/2 (0.815) = 0.048 days 
		= 1.15 hours
		

Project with ECU 
 - Determine where Venus is with respect to the Sun 
 (Elongation) on the following Dates: 
 Nov 1, 1998, Dec 1, 1998, Jan 1, 1999, Feb 1, 1999, Mar 1, 1999 
 Describe where to look for Venus in those coming months and what it will look like (how bright and where). 

* Note: Venus and Jupiter pass each other at only 0.1 deg at the end of February and Mercury is at
greatest elongation east at the first of March - a great time to observe planets at Sunset!!!

Sky at Sunset Mar 1, 1999

   + Saturn



           + Venus
             + Jupiter
               + Mercury

-----------------------W------------Horizon