a) assume io is a sphere with a radius of 1820 km and is covered to a depth of 1 kilometer to cover any new craters. what volume of fresh material must be produced in cubic meters? b) if the present surface was produced by the 420 calderas, what is the volume produced by each caldera in cubic meters? c) the typical time between large meteor impacts is about 500,000 years. how much material would have to be produced by each caldera each year to cover the surface between impacts? d) what is the total resurfacing rate each year in centimeters/year if the 1 kilometer depth was generated over 500,000 years?

Question

15-11-2022
Geography

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a) assume io is a sphere with a radius of 1820 km and is covered to a depth of 1 kilometer to cover any new craters. what volume of fresh material must be produced in cubic meters? b) if the present surface was produced by the 420 calderas, what is the volume produced by each caldera in cubic meters? c) the typical time between large meteor impacts is about 500,000 years. how much material would have to be produced by each caldera each year to cover the surface between impacts? d) what is the total resurfacing rate each year in centimeters/year if the 1 kilometer depth was generated over 500,000 years?

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rk16 rk16 · Answer 1 · 2022-11-19T00:07:11+01:00

The volume of surface of IO is 4.2×[tex]10^{16}m^3[/tex], the volume produced per caldera is 1×[tex]10^{14}m^3[/tex], the material would have produced per caldera per year is 2×[tex]10^8m^3[/tex]/year, the total resurfacing generate per year is 0.2cm/year.

How to calculate the volume of surface and resurfacing generate per year?

The volume of surface can be calculated by multiplying area of IO by depth of surface.

Area IO = [tex]4\pi r^2[/tex]

= 4 × 3.14 × (1820000)^2

= 4.2×[tex]10^{13}m^3[/tex]

Volume of surface = Area IO × depth of surface

= 4.2×[tex]10^{13}m^3[/tex] × [tex]10^3m^3[/tex]

= 4.2×[tex]10^{16}m^3[/tex]

Volume produced by each caldera can be calculated by dividing volume of surface by total of calderas.

Volume produced per caldera = 4.2×[tex]10^{16}m^3[/tex] / 420

= 1×[tex]10^{14}m^3[/tex]

Material produced per caldera per year can be calculated by dividing volume produced per caldera by the year between meteor impacts.

Material produced per caldera per year = 1×[tex]10^{14}m^3[/tex] / 500,000

= 2×[tex]10^8m^3[/tex]/year

Resurfacing generated per year can be calculated by dividing the depth of surface by the year

Resurfacing generated per year = 100,000cm / 500,000

= 0.2cm/year

Thus, the volume of surface fresh material is 4.2×[tex]10^{16}m^3[/tex], the volume produced per caldera is 1×[tex]10^{14}m^3[/tex], the material would have produced per caldera per year is 2×[tex]10^8m^3[/tex]/year, the total resurfacing generate per year is 0.2cm/year.

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