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Solutions to problem set 4 of the astro/eps c12 course, covering topics such as solar eclipses, equilibrium temperatures on planets, and the energy released during impacts. The solutions explain how the moon's position in the earth's orbital plane affects solar eclipses, calculate the solar flux at saturn's distance and the resulting temperatures on the leading and trailing hemispheres, and determine the ratio of impact energies for two different masses and velocities.
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Astro/EPS C12 (2008) – Mike Wong Problem Set 4 Solution Guide
Problem 1A Solar eclipses only happen when the Moon is in the Earth’s orbital plane, so that the Sun, Moon, and Earth are lined up and the Moon’s shadow falls on the Earth. But because there is a 5◦^ tilt between the Moon’s orbital plane and the Earth’s orbital plane, the Moon is not usually in the Earth’s orbital plane. It crosses the Earth’s orbital plane twice during its monthly orbit around the Earth, but a solar eclipse occurs only when that crossing falls exactly during the new moon.
Problem 1B A total solar eclipse occurs when the Sun is completely covered by the Moon. If the Moon were twice its current angular size, events which are now total solar eclipses would remain total solar eclipses, and some near-miss events which are now partial solar eclipses would become total solar eclipses. As for partial solar eclipses, a few would be reclassified as total solar eclipses, but even more partial solar eclipses would occur that would not have due to the moon’s larger angular size (and hence the probability of its penumbra falling on the Earth).
The diagram below shows what the sky looks like from Earth. The new Moon is shown drifting across the Sun. The range of paths falling within the “total” zone increases by a lot (it more than doubles). The range of paths falling within the “partial” zone increases by only a little bit (about 5%).
TOTAL PARTIAL
PARTIAL
TOTAL
PARTIAL
Moon
Sun
PARTIAL
Problem 2A Solar flux at Saturn’s distance of 9.54 AU:
FSat = F 0 (
r 02 a^2
) = 1370 W m−^2 (
) = 15.1 W m−^2
Problem 2B The darker leading hemisphere will absorb a lot more sunlight. For a 1 m^2 patch of ground,
Fleading = (1 − Aleading)FSat = (1 − 0 .03) 15.1 W m−^2 = 14.6 W m−^2 Ftrailing = (1 − Atrailing)FSat = (1 − 0 .6) 15.1 W m−^2 = 6.0 W m−^2
Problem 2C First balance the incoming and outgoing energy per unit area, and solve for the temperature:
F = σT 4 T 4 = F/σ
T = 4
F/σ
If your calculator can’t do fourth roots, you can just take the square root of the square root:
F/σ = (F/σ)^0.^25 = (F/σ)^0.^5 ×^0.^5 = ((F/σ)^0.^5 )^0.^5 =
F/σ
Using this formula, calculate the leading and trailing side temperatures:
Tleading = 4
Fleading/σ = 4
14 .6 W m−^2
Ttrailing = 4
Ftrailing/σ = 4
6 .0 W m−^2