Chapter 3, Problem 59P

To determine

The size of the antimatter meteorite and the energy involved in the annihilation process.

Answer to Problem 59P

The size of the antimatter meteorite is $3.68\text{km}$ and the energy released is equal to the energy of $9\times {10}^{12}$ nuclear arsenals in the space.

Explanation of Solution

Write the expression for the energy obtained during annihilation.

  $E=2m_0{c}^2$        (I)

Here, $E$ is the energy from annihilation, $m_0$ is the rest mass of photon and $c$ is the speed of light.

Write the expression for the gravitational energy of the Earth.

  $U=\frac{\left(0.5\right)G{M}_E^2}{R_E}$        (II)

Here, $G$ is the gravitational constant, $M_E$ is the mass of the Earth and $R_E$ is the radius of the Earth.

Since the Gravitational energy of the Earth will become the energy of annihilation of the matter-antimatter.

Write the expression for the energy during the process.

  $E=U$

Substitute $2m_0{c}^2$ for $E$ and $\frac{\left(0.5\right)G{M}_E^2}{R_E}$ for $U$ in above expression.

  $2m_0{c}^2=\frac{\left(0.5\right)G{M}_E^2}{R_E}$

Simplify the above expression for $m_0$.

  $m_0=\frac{0.5\left(G{M}_E^2\right)}{2R_E{c}^2}$        (III)

The volume of the spherical asteroid is $\frac{4}{3}\pi r^3$ .

Write the expression for the mass of the antimatter asteroid in terms of the volume.

  $\frac{4}{3}\pi r^3=\frac{m_0}{\rho }$        (IV)

Here, $r$ is the radius of the asteroid and $\rho$ is the density of the material of asteroid.

Write the expression to compare the energy released by the meteorite to total energy released in all nuclear arsenals.

  $n=\frac{U}{E^{\prime }}$        (V)

Here, $n$ is the total number of nuclear arsenals and $E^{\prime }$ is the energy of one arsenal.

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $5.97\times {10}^{24}\text{kg}$ for $M_E$, $6378\times {10}^3\text{m}$ for $R_E$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (III).

  $\begin{array}{c}{m}_0=\frac{0.5\left(\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(5.97\times {10}^{24}\text{kg}\right)\right)}{2\left(6378\times {10}^3\text{m}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2}\\ =\frac{1.188\times {10}^{39}}{1.14\times {10}^{24}}\text{kg}\\ =1.04\times {10}^{15}\text{kg}\end{array}$

Substitute $1.04\times {10}^{15}\text{kg}$ for $m_0$ and $5000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in equation (IV).

  $\begin{array}{c}\frac{4}{3}\pi r^3=\frac{1.04\times {10}^{15}\text{kg}}{5000\text{kg}/{\text{m}}^{\text{3}}}\\ r={\left(\frac{3\left(1.04\times {10}^{15}\text{kg}\right)}{4\pi \left(5000\text{kg}/{\text{m}}^{\text{3}}\right)}\right)}^{\frac{1}{3}}\\ =3.68\text{km}\end{array}$

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $5.97\times {10}^{24}\text{kg}$ for $M_E$ and $6378\times {10}^3\text{m}$ for $R_E$ in equation (II).

  $\begin{array}{c}U=\frac{0.5\left(\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right){\left(5.97\times {10}^{24}\text{kg}\right)}^2\right)}{6378\times {10}^3\text{m}}\\ =1.87\times {10}^{32}\text{J}\end{array}$

Substitute $1.87\times {10}^{32}\text{J}$ for $U$ and $21\times {10}^{18}\text{J}$ for $E^{\prime }$ in equation (V).

  $\begin{array}{c}n=\frac{1.87\times {10}^{32}\text{J}}{21\times {10}^{18}\text{J}}\\ =9.0\times {10}^{12}\text{arsenals}\end{array}$

Thus, the size of the antimatter meteorite is $3.68\text{km}$ and the energy released is equal to the energy of $9\times {10}^{12}$ nuclear arsenals in the space.