Chapter 7, Problem 70P

(a)

To determine

The mass of the dust.

Answer to Problem 70P

  $4.99\times {10}^{13}\text{kg}$

Explanation of Solution

Given Data:

  $G$ is gravitational constant $\left(6.67\times {10}^{-11}\right){\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}$ , mass of Earth is

  $M_E=5.98\times {10}^{24}\text{kg},1\times {10}^{14}\text{kg}$ is the mass of comet $m_c$ , R is the radius of the Earth 6378.1 km.

Formula Used:

The Potential energy of the comet at the surface of earth is given as:

  $U_c=-\frac{G{M}_E{m}_c}{R}$

Mass of the dust

  $m_d=\frac{R{U}_d}{G{M}_E}$

Calculation:

Gravitational Potential energy of comet:

  $U_c=-\frac{G{M}_E{m}_c}{R}$

  $U_c=-\frac{\left(6.67\times {10}^{-11}\right)\times \left(5.98\times {10}^{24}\right)\times \left(1\times {10}^{14}\right)}{6378.1\times {10}^3}$

  $U_c=-6.25\times {10}^{21}\text{Joules}$

The half of the gravitational potential energy is used to hold the dust particles:

  $\begin{array}{l}{U}_d=\frac{U_c}{2}\\ =\frac{6.25\times {10}^{21}}{2}\\ =3.125\times {10}^{21}\text{J}\end{array}$

Mass of the dust is calculated as below:

  $U_d=-\frac{G{M}_E{m}_d}{R}$

  $m_d=\frac{R{U}_d}{G{M}_E}$

  $m_d=\frac{\left(6378.1\times {10}^3\right)\times 3.125\times {10}^{21}}{\left(6.67\times {10}^{-11}\right)\times \left(5.98\times {10}^{24}\right)}$

  $m_d=4.99\times {10}^{13}\text{kg}$

Conclusion:

Mass of the dust is $4.99\times {10}^{13}\text{kg}$ .

(b)

To determine

The mass density of the dust.

Answer to Problem 70P

  $4.99\times {10}^{-6}{\text{kg/m}}^{\text{3}}$

Explanation of Solution

Introduction:

Density is defined as mass per unit volume: $\rho =\frac{mass}{volume}$

The dust lies from earth’s surface to 20 Km in the atmosphere. Therefore, the distance is from R to R+20 Km.

The Volume between thesedistances is given as below:

  $V=\frac{4}{3}\pi \left(R_a{}^3-R_E{}^3\right)$

  $R_a$ is the radius up to dust is present and $R_E$ is the radius of the Earth.

  $V=\frac{4}{3}\pi \left(R_a^3-R_E^3\right)$

  $V=\frac{4}{3}\pi \left(\left(R_E+20000{)}^3\right)-R_E^3\right)$

  $V=\frac{4}{3}\pi \left[\left({\left(6.3\times {10}^6+20000\right)}^3\right)-{\left(6.3\times {10}^6\right)}^3\right]$

  $V=1.00\times {10}^{19}{\text{m}}^{\text{3}}$

Density of dust:

  $\rho =\frac{mass}{volume}$

  $\begin{array}{l}\rho =\frac{4.99\times {10}^{13}\text{kg}}{1\times {10}^{19}{\text{m}}^3}\\ \rho =4.99\times {10}^{-6}{\text{kg/m}}^{\text{3}}\end{array}$

Conclusion:

Density of the dust is $4.99\times {10}^{-6}{\text{kg/m}}^{\text{3}}$ .