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.

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

4.9x10^20 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

-9.8 m/s^s because thats the earth gravity so it will lose 9.8 m/s^2 until its stop and thats because its the opposite of the force towards the earth!
Hope it helps