PPS 1 Solutions CIV300/ENV346 Fall 2010 1) Toronto is found at approximately 44.5 degrees north latitude. (i) How far above the horizon does the sun achieve at solar noon on the shortest day of the year in the north hemisphere (21 Dec.)? Also what is this angle: (ii) on the equator?, (iii) on the Tropic of Capricorn? (iv) At 60 degrees north latitude? (v) At 45 degrees south?

1) The thing to recall here is the sun is directly overhead (90 from the horizon) at the Tropic of Capricorn at solar noon on the 21st of December. For every degree we move away (south or north) move the sun a degree from vertical, or a degree from the 90 at the 23.5 south. Thus, for example, at Toronto, at 44.5 degrees north, we are (23.5+44.5 = 68 degrees north of the tropic of cancer. So the sun reaches a maximum height of 90-68 = 22 degrees above the horizon).

a = a = 90 – 44.5 – 23.5 = 22

1 A simple equation to calculate the sun height at solar noon: Sun height = 90° - Latitude (L) +/- Declination (D) On the solstice, the declination is 23.5°, while to add or subtract the declination depends on whether it’s summer or winter time in the designated hemisphere:

+ D: summer time; June solstice (North) and December solstice (South) - D: winter time; December solstice (North) and June solstice (South)

Location L D + or - Sun height Toronto 44.5° N 23.5° - 22° Equator 0° 23.5° - 66.5° Tropic of Capricorn 23.5° S 23.5° + 90° (overhead) 60° North latitude 60° N 23.5° - 6.5° 45° South latitude 45° S 23.5° + 68.5°

2) (i) Calculate the average orbital speed (in m/s and km/h) of the Earth around the sun. (ii) What is its average acceleration toward the center, and what produced the required force to achieve this acceleration?

2) This is simply geometry for the speed. The acceleration is V2/R where R is the radius. This acceleration is produced by the gravitation attraction between the Earth the Sun.

2 Path length of Earth’s orbit: 2πr = 2*3.14*150 x 106 = 942 x 106 km = 942 x 109 m Duration of orbit: 365 days * 24 hours * 60 minutes * 60 seconds = 31536000 s i) Average Velocity: 942 x 109 / 31536000 ≈ 3.0 x 104 m/s ≈ 1.1 x 105 km/hr ii) Acceleration towards the centre = V2/R ≈ (3.0 x 104 m/s)2/150 x109 = 6 x10-3 m/s2 The acceleration is caused by the gravitational attraction between the Earth the Sun.

3) (i) What is the rotational velocity (in m/s and km/h) on the equator and at 44 degrees north latitude? (ii) What is its average acceleration toward the center on the equator, and what produced the required force to achieve this acceleration?

3) Similar to 2. Now we have a slight change in apparent gravity associated with the acceleration and a new force (Coriolis) starts to appear as well, since gravity and the acceleration are no longer collinear. We will deal with this more later! Radius of the Earth: R = 6350 km sR2π 2π × 6350( km ) i) At the Equator: vk== = =1662.4(m /hr )= 461.8(m /s ) tt24( hr ) At 44° N: R' =×Rkcos 44o = 6350 × 0.7193 = 4567.6(m )

sR2ππ' 2× 4567.6( km ) And thus vk' == = =1195.8(m /hr )= 332.2(m /s ) tt24( hr ) ii) Acceleration rate toward the center on the equator: v22461.8 a== =0.034(m /s2 ) (c.f., acceleration due to gravity) R 6350× 103

4) At what distance from you should a dime (with assumed diameter of 1.8 cm) be placed such that it exactly covers the Sun? How far to cover the Moon? What does this imply about the nature of a total eclipse?

ds dm dd dd

Ds Dd Observer Dm Dd Observer Dime Dime Moon Sun

3 By similar triangles:

Dd 018.0 m 11 dd ds == 9 ≈⋅ 2)105.1( mm (1.9 m, more precisely) Ds ⋅1039.1 m

Dd 018.0 m 8 dd dm == 6 ≈⋅ 2)10.4( mm (2.1 m, meaning the moon looks smaller) Dm )10(5.3 m

2 5) Assuming a solar insolation of 1370 W/m , how much total energy does the Earth receive from the sun? Approximately what fraction of the Sun’s total energy output does the Earth receive?

5) Earth’s diameter: D = 12700 km Earth’s orbit radius: R = 150 × 106 km Total energy received by the Earth: π π SI×=×××=×( D23 ) 1370 (12700 10 )2 1.74 1017 (W ) 44 Total Energy output of the sun at the Earth’s orbit = SI×(4π R2 )

π 2 SI×() D 22 D 12700 Fraction = 4 == =×=4.48 10−10 0.000000045% , SI××(4π R22 ) 16 R 16 (150× 1062 ) pretty tiny!!

6) Calculate the distance to the perceived horizon as a function of an observer’s height h. Assume a flat landscape such as the Ocean. H h R

R is Earth radius, h observer height, and H horizon dist.

2 )( −+= RhRH 22 22 2 −++= RhRhRH 22 2 2 += hRhH 2 2 ≈ 2RhH =≈ 6.32 hRhH In which the H distance is in km and h is in m.

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For h = 2 m, H ≈ 5 m; for h = 400 m (about observation deck of CN tower), H = 72 km (assuming clear sky!)

7) Consider the idea that all terrestrial energy systems are essentially extensions of the solar energy system. Discuss this idea in terms of observations one can make while watching a downpour of rain unto the surface of a lake.

Energy transformations one might discuss: • Where does the rain come from? From evaporated water, a process powered from the sun. • How did the rain come to fall here? It was transported likely mostly from the ocean by moving air, a processes powered by the sun. • Why is there a lake here at all? The lake exists in a liquid state because if is warmed by the sun, but not so much it evaporates, a balance controlled by the sun. • Why is there a sound of calling water? The acoustic energy is released by the conversion of the KE energy of the falling water, an energy that has it source from the evaporation processes powered by the sun…. • And so it continues, so one could include the vegetation, the light reflected from the water, the waves on its surface, the erosion on its shore, etc. etc.

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