Chapter 8, Problem 71P

(a)

To determine

The direction of impact with the meteorite.

Explanation of Solution

Introduction:

The meteorites are rock-like shapes that move in the space together with earth and the other planets. These meteorites are large in shapes, have a rigid structure that does not follow any desired path. This meteorite is a solid shaped body that originated from a comet or an asteroid.

The striking of the meteorites with the earth is exactly in the opposite direction to the orbital velocity with which the earth is travelling.

Conclusion:

The direction of striking meteorite is exactly opposite to the direction of the orbital velocity of the earth.

(b)

To determine

The maximum percentage change in earth’s orbital speed as a result of collision.

Answer to Problem 71P

The maximum percentage change in earth’s orbital speed as a result of collision is $2.71\times {10}^{-15}\%$ .

Explanation of Solution

Given:

The mass of the meteorite is $m_m=2.72\times {10}^5\text{tonnes}$ .

The speed of the meteorite is $v_m=17.9\text{km/s}$ .

The earth’s orbital speed is about $v_e=30\text{km}/\text{s}$ .

Formula used:

The expression for conservation of momentum is given as,

  $\Delta \overrightarrow{P}={\overrightarrow{P}}_f-{\overrightarrow{P}}_i$

Here, ${\overrightarrow{P}}_f$ is the final momentum of the body and ${\overrightarrow{P}}_i$ is the initial momentum of the body.

The expression for the momentum is given as,

  $P=m\times v$

The expression for the percentage change in earth’s orbital speed is given as,

  $\left|\frac{\Delta v}{v_e}\right|=\left|\frac{v_e-v_f}{v_e}\right|$

Here, $v_f$ is the final velocity of the system after collision.

Calculation:

Applying the linear conservation of momentum into the system as,

  $\begin{array}{c}\Delta \overrightarrow{P}=0\\ {\overrightarrow{P}}_f-{\overrightarrow{P}}_i=0\end{array}$

As after the collision the meteorite and earth moves horizontally. Therefore,

  $\begin{array}{c}{P}_{f,x}-P_{i,x}=0\\ \left(m_e+m_m\right)v_f-\left(m_e{v}_e-m_m{v}_m\right)=0\\ v_f=\frac{\left(m_e{v}_e-m_m{v}_m\right)}{\left(m_e+m_m\right)}\\ =\frac{m_e{v}_e}{\left(m_e+m_m\right)}-\frac{m_m{v}_m}{\left(m_e+m_m\right)}\end{array}$

On further solving as,

  $v_f=\frac{v_e}{\left(1+\frac{m_m}{m_e}\right)}-\frac{\frac{m_m}{m_e}{v}_m}{\left(1+\frac{m_m}{m_e}\right)}$

Since the mass of earth is very large as compared to the mass of the meteorite.

  $m_m<<m_e$

Therefore, The term $\frac{m_m}{m_e}$ is neglected.

Thus, the final velocity is expressed as,

  $\begin{array}{c}{v}_f=v_e-\frac{\frac{m_m}{m_e}{v}_m}{\left(1+\frac{m_m}{m_e}\right)}\\ =v_e-\left(\frac{m_m}{m_e}{v}_m\right){\left(1+\frac{m_m}{m_e}\right)}^{-1}\\ =v_e-\left(\frac{m_m}{m_e}{v}_m\right)\left(1-\frac{m_m}{m_e}\right)\\ =v_e-\left(\frac{m_m}{m_e}{v}_m\right)\end{array}$

The expression for the percentage change in earth’s orbital speed is given as,

  $\begin{array}{c}\left|\frac{\Delta v}{v_e}\right|=\left|\frac{v_e-v_f}{v_e}\right|\\ =\left|1-\frac{v_f}{v_e}\right|\end{array}$

The percentage can be calculated as,

  $\begin{array}{c}\left|\frac{\Delta v}{v_e}\right|=\left|1-\frac{v_f}{v_e}\right|\\ =\left|1-\frac{v_e-\left(\frac{m_m}{m_e}{v}_m\right)}{v_e}\right|\\ =\left|\frac{\left(\frac{m_m}{m_e}{v}_m\right)}{v_e}\right|\\ =\left|\frac{\left(\frac{2.72\times {10}^5\text{tonne}\left(\frac{{10}^3\text{kg}}{\text{tonne}}\right)}{5.98\times {10}^{24}\text{kg}}\times \left(17.9\text{km}/\text{s}\right)\right)}{30\text{km}/\text{s}}\right|\end{array}$

On further solving as,

  $\begin{array}{c}\left|\frac{\Delta v}{v_e}\right|=2.71\times {10}^{-17}\\ =2.71\times {10}^{-15}\%\end{array}$

Conclusion:

Therefore, the maximum percentage change in earth’s orbital speed as a result of collision is $2.71\times {10}^{-15}\%$ .

(c)

To determine

The mass of an asteroid to change the earth’s orbital speed by $1\%$ .

Answer to Problem 71P

The mass of an asteroid to change the earth’s orbital speed by $1\%$ is $1\times {10}^{23}\text{kg}$ .

Explanation of Solution

Given:

The weight of the meteorite is $m_m=2.72\times {10}^5\text{tonnes}$ .

The speed of the meteorite is $v_m=17.9\text{km/s}$ .

The earth’s orbital speed is about $v_e=30\text{km}/\text{s}$ .

The percentage change in the earth’s orbital speed is $\left|\frac{\Delta v}{v_e}\right|=1\%$ .

Formula used:

The expression for percentage change in earth’s orbital speed is given as,

  $\left|\frac{\Delta v}{v_e}\right|=\left|\frac{\left(\frac{m_m}{m_e}{v}_m\right)}{v_e}\right|$

Calculation:

The mass of the meteorite can be calculated as,

  $\begin{array}{c}\left|\frac{\Delta v}{v_e}\right|=\left|\frac{\left(\frac{m_m}{m_e}{v}_m\right)}{v_e}\right|\\ \frac{1}{100}=\frac{\left(\frac{m_m}{m_e}{v}_m\right)}{v_e}\\ m_m=\frac{v_e{m}_e}{100\times v_m}\\ =\frac{30\text{km}/\text{s}\times \left(5.98\times {10}^{24}\text{kg}\right)}{100\times \left(17.9\text{km}/\text{s}\right)}\end{array}$

On further solving as,

  $m_m=1\times {10}^{23}\text{kg}$

Conclusion:

Therefore,the mass of an asteroid to change the earth’s orbital speed by $1\%$ is $1\times {10}^{23}\text{kg}$ .