1. The KT impact. [11 marks total] The size of the KT impact can be estimated fr

Question

1. The KT impact. [11 marks total] The size of the KT impact can be estimated from an Irridium (Ir) anomaly in sediments associated with the impact. Irridium is a siderophile (iron loving) element. Earth was originally homogeneous, and formed by collisional aggregation of early solar system material. In the first 30 million years of Earth history, segregation into the iron core and silicate mantle. Ir partitioned into the iron and is thus in the core, leaving the bulk silicate Earth depleted in Ir. A positive Ir anomaly in sediments suggests an extra-terrestrial source.

(a) The end-Cretaceous is marked by a clay layer with a strong positive Ir anomaly. This layer is, on average, 4 cm thick with a density of 2.5 g cm3 and iridium content of 10 parts per billion by weight. Assuming the Earth is spherical, with radius 6370 km, calculate how much iridium was delivered. [3 mark(s)]

(b) Assuming that the meteor was spherical with a density of 3 g cm3 and an iridium content of 0.6 parts per million by weight, estimate the mass of the meteor and hence its diameter. [4 mark(s)]

(c) i. Assuming an impact velocity of 20 km s1 , what energy was delivered by the impact? [2 mark(s)] ii. The BC Hydro system has a generation rate of 11 GW. How many years would it take for the BC hydro system to generate equivalent energy? [2 mark(s)]

2. Energy balance in the solar system [10 marks total]

(a) The total amount of energy that the Sun emits is the Solar Luminosity, L = 3.85 × 1026 W. The flux of energy per unit area normal to (i.e. at right angles to) the direction of the Sun’s rays is given by an inverse square law, F = L 4d2 where d is the distance from the Sun. The Earth–Sun distance (1 Astronomical Unit, AU) is 1.496×1011 m. Venus is located at 0.72 AU and Mars at 1.52 AU. Calculate the solar flux at the orbit of each planet (for Earth, this is called the Solar Constant). [1 mark(s)]

(b) i. Calculate the effective temperature for each planet. Use planetary albedos of p,Venus = 0.9, p,Earth = 0.3, p,Mars = 0.25 [2 mark(s)] ii. Surface temperatures of the planets are Ts,Venus = 740 K, Ts,Earth = 289 K and Ts,Mars = 214 K. What does the effective temperature tell us about the surface temperature? [1 mark(s)]

(c) In class, we used a conceptual climate model dTs dt = A cm S 4 (1 ) 1 + 3 4 Ts 4 in which S is solar constant, is albedo, Ts is surface temperature, is Stefan’s constant, t is time, m is the mass of the atmosphere–ocean and c is the specific heat capacity of the atmosphere–ocean, A is the surface area of the planet and thermal optical depth parameterizes the greenhouse effect..

i. Assume steady state conditions, then solve to give , and hence calculate the optical depth for each planet. [3 mark(s)]

ii. Calculate the surface temperature that Earth would have if it had a greenhouse effect of equal strength to (i) Venus and (ii) Mars (but use Earth’s albedo of 0.3). [2 mark(s)]

iii. If temperature was all that mattered for habitability, would Earth likely be habitable for humans in either of these cases? Briefly explain your answer. [1 mark(s)]

Explanation / Answer

Answer: Only question 1 has been answered according to the Chegg rule

(a) Earth radius 6370 km. surface area= 4*phi* square(6370) sq-km

4 cm thick clay layer has 2.5 g/cube-m and Ir is 10 parts per billion by weight.

Ir weight delivered = 10ppb*Clay weight= 4*phi* square(6370) * 106 * 0.04 *2.5* 10-11= 509.8 g (Answer (a))

(b) Its radius = r cm, say

Ir Content = Mass * 0.6ppm = vol*3g/ccm * 0.6ppm = 4/3*phi*r3 *3* 0.6*10-6

Mass = 509.8/0.6 ppm = 509.8*106 / 0.6 = 849,666,666.7 g = 849,666.6667 kg (answer)

Then r = cube-root[849666666.7/ (4/3)phi*3] = 407.42 cm = 4.0742 m

Then diameter= 8.1484 m (Answer)

(c) Energy imparted on surface = 0.5 m v2 = 0.5* 849,666.6667 * 20,0002 = 1.6993*10e14 J

BC hydro produce 11GW= 11*10e9 J/s, then, required time = 1.6993*10e14 /11*10e9 = 15448.48485 s= 0.00045 years

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