91Ó°ÊÓ

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 \(3.2 \times {10^{ - 16}}\% \).