MATH SOLVE

3 months ago

Q:
# A model for the density δ of the earth’s atmosphere near its surface isδ=619.09−0.000097pwhere p (the distance from the center of the earth) is measured in meters and δ is measured in kilograms per cubic meter. If we take the surface of the earth to be a sphere with radius 6370 km, then this model is a reasonable one for:6.370×106≤p≤6.375×106 Use this model to estimate the mass of the atmosphere between the ground and an altitude of 5 km.

Accepted Solution

A:

Answer: 3.751*10^18kgStep-by-step explanation:δ =619.09−0.000097p....equa1 where p (the distance from the center of the earth) is measured in meters and δ is measured in kilograms per cubic meter.Calculating the density of air at 5km above earth surfaceP = 5000m + 6370000m = 6.375*10^6mδ = 619.09 -(.000097* 6.375*10^6) δ = 0.715kg/m^3 = densitySince Mass = density*volume...equ2To calculate volume of air around the spherical earth at height 5kmV = (4/3 pai R^3) - (4/3pai r^3) ...equation 3 where R =6.375*10^6m, r = 6.37*10^6Substituting R and r in equation 2 to solve for volume of airV = 1.085*10^21 - 1.08*10^21V = 5.25*10^18m^3Substituting δ and V into equation 2 to solve for mass of airM = 0.715 * (5.25*10^18)M = 3.751*10^18kg