Physics, 11.07.2019 14:00 royalkk

Approximating venus's atmosphere as a layer of gas 50 km thick, with uniform density 21 kg/m3, calculate the total mass of the atmosphere.

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The diameter of Venus (from tables) is 12,100 km.
Therefore its radius is
r = 12100/2 = 6050 km = 6050 x 10³ m = 6.05 x 10⁶ m

The layer of gas is 50 km thick.
The outer radius of the surrounding layer of gas is
R = r + 50 x 10³ m
= 6.05 x 10⁶ + 50 x 10³ m
= 6.1 x 10⁶ m

The volume of the layer of gas is
V = [(4π)*3] * [(6.1 x 10⁶)³ - (6.05 x 10⁶)³] m³
= [(4π)/3] * 5.535875 x 10¹⁸ m³
= 2.318862 x 10¹⁹ m³

Because the density of the gas layer is 21 kg/m³, therefore the mass of the gas layer is
m = (21 kg/m³) * (2.318862 x 10¹⁹ m³)
= 4.8696 x 10²⁰ kg

$4.8696 \times 10^{20} \, kg$

Answer from: lehbug01

4.9x10^20 kg Venus has a radius of 6051.8 km, so we'll assume the total volume of the atmosphere is the volume of a sphere with a radius of (6051.8 + 50) km minus the volume of Venus itself. So V = 4/3 pi r^3 6051.8 + 50 = 6101.8 The overall expression is V = 4/3 pi 6101.8^3 - 4/3 pi 6051.8^3 V = 4/3 pi (6101.8^3 - 6051.8^3) V = 4/3 pi 5539155986 V = 23202362337 V = 2.32x10^10 km^3 V = 2.32x10^19 m^3 Now multiply the volume by the density. So V = 2.32x10^19 m^3 * 21 kg/m^3 = 4.87x10^20 kg So the atmosphere of Venus has a mass of about 4.9x10^20 kg

Answer from: Quest

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.

Answer from: Quest

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Approximating venus's atmosphere as a layer of gas 50 km thick, with uniform density 21 kg/m3, calcu...