Question: (III) An asteroid of mass \({\bf{1}}{\bf{.0 \times 1}}{{\bf{0}}^{\bf{5}}}\;{\bf{kg}}\) traveling at a speed of \({\bf{35}}\;{{{\bf{km}}} \mathord{\left/{\vphantom {{{\bf{km}}} {\bf{s}}}} \right.} {\bf{s}}}\) relative to the Earth, hits the Earth at the equator tangentially, in the direction of Earth’s rotation, and is embedded there. Use angular momentum to estimate the percent change in the angular speed of the Earth as a result of the collision.

The angular momentum of a rotating object, about a reference point, is described as the vector product of the location of the object relative to the reference point and the object's momentum.

Given data:

The mass of the asteroid is \({m_{\rm{a}}} = 1.0 \times {10^5}\;{\rm{kg}}\).

The velocity of the asteroid is \({v_{\rm{a}}} = 35\;{{{\rm{km}}} \mathord{\left/{\vphantom {{{\rm{km}}} {\rm{s}}}} \right.} {\rm{s}}}\).

Apply the conservation of angular momentum.

\(\begin{aligned}{c}{L_{\rm{i}}} &= {L_{\rm{f}}}\\{I_{\rm{E}}}{\omega _{\rm{E}}} + {I_{\rm{a}}}{\omega _{\rm{a}}} &= \left( {{I_{\rm{E}}} + {I_{\rm{a}}}} \right){\omega _{\rm{f}}}\end{aligned}\)

Here, \({I_{\rm{E}}}\) is the moment of inertia of the Earth, \({\omega _{\rm{E}}}\) is the angular velocity of the Earth, \({I_{\rm{a}}}\) is the moment of inertial of the asteroid, \({\omega _{\rm{a}}}\) is the angular velocity of the asteroid, and \({\omega _{\rm{f}}}\) is the final angular velocity of the asteroid-Earth system.

The moment of inertia of the asteroid on the right-hand side of the above equation can be neglected as it is very small compared to that of the Earth. Therefore, you can write:

\(\begin{aligned}{c}{I_{\rm{E}}}{\omega _{\rm{E}}} + {I_{\rm{a}}}{\omega _{\rm{a}}} &= \left( {{I_{\rm{E}}}} \right){\omega _{\rm{f}}}\\{\omega _{\rm{f}}} - {\omega _{\rm{E}}} &= \frac{{{I_{\rm{a}}}{\omega _{\rm{a}}}}}{{{I_{\rm{E}}}}}\end{aligned}\)

The expression for the fractional change in the angular velocity of the Earth can be written as:

\(\begin{aligned}{c}f &= \left( {\frac{{{\omega _{\rm{f}}} - {\omega _{\rm{E}}}}}{{{\omega _{\rm{E}}}}}} \right) \times 100\% \\f &= \left( {\frac{{{I_{\rm{a}}}{\omega _{\rm{a}}}}}{{{I_{\rm{E}}}{\omega _{\rm{E}}}}}} \right) \times 100\% \\f &= \frac{{\left( {{m_{\rm{a}}}R_{\rm{E}}^2} \right)\left( {\frac{{{v_{\rm{a}}}}}{{{R_{\rm{E}}}}}} \right)}}{{\left( {\frac{2}{5}{M_{\rm{E}}}R_{\rm{E}}^2} \right)\left( {\frac{{2\pi }}{T}} \right)}} \times 100\% \\f &= \frac{{5{m_{\rm{a}}}{v_{\rm{a}}}T}}{{2{M_{\rm{E}}}2\pi {R_{\rm{E}}}}} \times 100\% \end{aligned}\)

Here, \({M_{\rm{E}}}\) is the mass of the Earth and its value is \(5.97 \times {10^{24}}\;{\rm{kg}}\), \({R_{\rm{E}}}\) is the radius of the Earth and its value is \(6.38 \times {10^6}\;{\rm{m}}\), and \(T\) is the time period of one rotation of the Earth and its value is \(1\;{\rm{day}}\).

Substitute the values in the above expression.

\(\begin{aligned}{c}f &= \frac{{5\left( {1.0 \times {{10}^5}\;{\rm{kg}}} \right)\left\{ {\left( {35\;{{{\rm{km}}} \mathord{\left/{\vphantom {{{\rm{km}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{km}}}}} \right)} \right\}\left\{ {\left( {1\;{\rm{day}}} \right)\left( {\frac{{86400\;{\rm{s}}}}{{1\;{\rm{day}}}}} \right)} \right\}}}{{2\left( {5.97 \times {{10}^{24}}\;{\rm{kg}}} \right)2\pi \left( {6.38 \times {{10}^6}\;{\rm{m}}} \right)}} \times 100\% \\f &= 3.2 \times {10^{ - 16}}\% \end{aligned}\)

Thus, the percent change in the angular speed of the Earth as a result of the collision is \(3.2 \times {10^{ - 16}}\% \).