Question

A model for the density\(\delta \) V of the earth's atmosphere near its surface is\(\delta = 619.09 - 0.000097\rho \)where\(\rho \)the distance from the centre of the earth) is measured in meters and\(\delta \) 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 \(6.370* {{10}^{6}}\rho 6.375* {{10}^{6}}\).Use this model to estimate the mass of the atmosphere between the ground and an altitude of\(5\;km\)

Short answer

The mass of the atmosphere between the ground and an altitude\({\rm{ Mass }} = 2.4* {10^{18}}{\rm{ Kilograms }}\)

Step-by-step solution

Define spherical coordinates.

Three coordinates in three-dimensional space that define the location of a point The length of its radius vector r, the angle between the vertical plane containing this vector and the x-axis, and the angle between this vector and the horizontal x–y plane are the three parameters.

Step 2: Now find the mass of the model.

Mass

In spherical coordinates

\(E = \left\{ {(\rho ,\theta ,\phi )\mid 6.370 * {{10}^6} \le \rho \le 6.375 * {{10}^6},\;\;\;0 \le \theta \le 2\pi ,\;\;\;0 \le \phi \le \pi } \right\}\)

\(\begin{aligned}{\rm{ Mass }} &= \int_0^\pi {\int_0^{2\pi } {\int_{6.370 * {{10}^6}}^{6.375 * {{10}^6}} {(619.09 - 0.000097\rho )} } } {\rho ^2}\sin \phi d\rho d\theta d\phi \\{\rm{ Mass }} &= \int_0^\pi {\sin } \phi d\phi \int_0^{2\pi } d \theta \int_{6.370 * {{10}^6}}^{6.375 *{{10}^6}} 6 19.09{\rho ^2} - 0.000097{\rho ^3}d\rho \\{\rm{ Mass }} &= 2 * 2\pi \left( {\frac{{619.09{\rho ^3}}}{3} - \frac{{0.000097{\rho ^4}}}{4}} \right)_{6.370 * {{10}^6}}^{6.375 * {{10}^6}}\\{\rm{ Mass }} &= 4\pi \left( {206.36{\rho ^3} - 0.00002425{\rho ^4}} \right)_{6.370 * {{10}^6}}^{6.375 * {{10}^6}}\end{aligned}\)

\( = 4\pi \left( {206.36{{\left( {6.375 * {{10}^6}} \right)}^3} - 0.00002425{{\left( {6.375 *{{10}^6}} \right)}^4}} \right) - 4\pi \left( {206.36{{\left( {6.37 * {{10}^6}} \right)}^3} - 0.00002425{{\left( {6.37 * {{10}^6}} \right)}^4}} \right)\)

\({\rm{ Mass }} = 2.4 * {10^{18}}{\rm{ Kilograms }}\)

Thus, The mass of the atmosphere between the ground and an altitude

\({\rm{ Mass }} = 2.4 * {10^{18}}{\rm{ Kilograms }}\)