Chapter 7, Problem 73GP

(a)

To determine

The recoil speed of the Earth.

Answer to Problem 73GP

The recoil speed of the Earth is $V=2.5\times {10}^{-13}\text{m}/\text{s}$ .

Explanation of Solution

Given:

The mass of the meteor is $m=1.\times {10}^8\text{ kg}$ .

The mass of the Earth is $m_E=6.0\times {10}^{24}\text{ kg}$ .

The speed of meteor is $v=15\text{km}/\text{s}$ .

Formula used:

The expression for the conservation of momentum for a completely inelastic collision is,

  $\left(m_{\text{E}}+m\right)V=mv$

Here, $m_{\text{E}}$ is the mass of the Earth, m is the mass of the meteor, V is the speed of the Earth, and v is the speed of the meteor.

Calculation:

The recoil speed of the Earth is,

  $\begin{array}{l}\left(m_{\text{E}}+m\right)V=mv\\ \left(\left(6.0\times {10}^{24}\text{ kg}\right)+\left(1.0\times {10}^8\text{ kg}\right)\right)V=\left(1.0\times {10}^8\text{ kg}\right)\left(15\text{km}/\text{s}\times \frac{1,000\text{ m}}{1\text{ km}}\right)\\ 6\times {10}^{24}V=1.5\times {10}^{12}\\ V=2.5\times {10}^{-13}\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the recoil speed of the Earth is $V=2.5\times {10}^{-13}\text{m}/\text{s}$ .

(b)

To determine

The fraction of the meteors kinetic energy was transformed to kinetic energy of the Earth.

Answer to Problem 73GP

The fraction of the meteors kinetic energy was transformed to kinetic energy of the Earth is $\frac{K_{\text{E}}}{K_{\text{m}}}=1.67\times {10}^{-17}$ .

Explanation of Solution

Given:

The mass of the meteor is $m=1.\times {10}^8\text{ kg}$ .

The mass of the Earth is $m_E=6.0\times {10}^{24}\text{ kg}$ .

The speed of meteor is $v=15\text{km}/\text{s}$ .

Formula used:

The expression for the kinetic energy is,

  $K=\frac{1}{2}m{v}^2$

Here, m is the mass and v is the velocity.

Calculation:

From part (a), the recoil speed of the Earth is $V=2.5\times {10}^{-13}\text{m}/\text{s}$ .

The kinetic energy of the meteor is,

  $\begin{array}{l}{K}_{\text{m}}=\frac{1}{2}m{v}^2\\ K_{\text{m}}=\frac{1}{2}\left(1.0\times {10}^8\text{ kg}\right){\left(15\text{km}/\text{s}\times \frac{1,000\text{ m}}{1\text{ km}}\right)}^2\\ K_{\text{m}}=1.125\times {10}^{16}\text{kg}\cdot {\text{m}}^2/{\text{s}}^2\end{array}$

The kinetic energy of the Earth is,

  $\begin{array}{l}{K}_{\text{E}}=\frac{1}{2}{m}_{\text{E}}{V}^2\\ K_{\text{E}}=\frac{1}{2}\left(6.0\times {10}^{24}\text{ kg}\right){\left(\text{2}\text{.5}\times {\text{10}}^{-13}\text{m}/\text{s}\right)}^2\\ K_{\text{E}}=0.1875\text{kg}\cdot {\text{m}}^2/{\text{s}}^2\end{array}$

The fraction of the meteors kinetic energy was transformed to kinetic energy of the Earth is,

  $\begin{array}{l}\frac{K_{\text{E}}}{K_{\text{m}}}=\frac{0.1875\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{1.125\times {10}^{16}\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}\\ \frac{K_{\text{E}}}{K_{\text{m}}}=1.67\times {10}^{-17}\end{array}$

Conclusion:

Thus, the fraction of the meteors kinetic energy was transformed to kinetic energy of the Earth is $\frac{K_{\text{E}}}{K_{\text{m}}}=1.67\times {10}^{-17}$ .

(c)

To determine

The change in Earth’s kinetic energy as a result of this collision.

Answer to Problem 73GP

The change in Earth’s kinetic energy as a result of this collision is $\text{0}\text{.19 J}$ .

Explanation of Solution

Given:

The mass of the meteor is $m=1.\times {10}^8\text{ kg}$ .

The mass of the Earth is $m_E=6.0\times {10}^{24}\text{ kg}$ .

The speed of meteor is $v=15\text{km}/\text{s}$ .

Formula used:

The expression for the change in kinetic energy is,

  $\Delta KE=\frac{1}{2}m{\left(\Delta v\right)}^2$

Here, m is the mass and v is the velocity.

Calculation:

From part (a), the recoil speed of the Earth is $V=2.5\times {10}^{-13}\text{m}/\text{s}$ .

The kinetic energy of the Earth is,

  $\begin{array}{l}\Delta KE=\frac{1}{2}{m}_{\text{E}}{V}^2\\ \Delta KE=\frac{1}{2}\left(6.0\times {10}^{24}\text{ kg}\right){\left(\text{2}\text{.5}\times {\text{10}}^{-13}\text{m}/\text{s}\right)}^2\\ \Delta KE=0.19\text{kg}\cdot {\text{m}}^2/{\text{s}}^2\times \frac{1\text{ J}}{\text{1 }\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}\\ \Delta KE=0.19\text{ J}\end{array}$

Conclusion:

Thus, the change in Earth’s kinetic energy as a result of this collision is $\Delta KE=\text{0}\text{.19 J}$ .