Question

Suppose that an asteroid of mass$\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$is nearly at rest outside the solar system, far beyond Pluto. It falls toward the Sun andcrashes into the Earth at theequator, coming in at angle of$\mathbf{30}\mathbf{°}$to the vertical as shown, against the direction of rotation of the Earth (Figure 11.101; not to scale). It is so large that its motion is barely affected by the atmosphere.

(a) Calculate the impact speed. (b) Calculate the change in the length of a day due to the impact.

Short answer

The impact speed of the asteroid is$v=4.37\times {10}^{4}m/s.$

The change in length of a day due to the impact is 1 h.

Step-by-step solution

Definition of Angular Speed.

Angular speed is defined as the rate of change of angular displacement.

Angular velocity is a vector quantity that expresses both direction and magnitude, while angular speed describes magnitude only.

Mass of the asteroid,$m=2\times {10}^{21}kg$

Angle between the direction of the asteroid and the vertical, $\theta =30°$

The figure shows an asteroid fall toward the Sun and crashes into the Earth at the equator, coming in at angle of$30$degrees to the vertical.

Find the total potential energy of the asteroid.

Initially the asteroid is at rest, so the initial kinetic energy of the system is zero$(K_i=0)$

Initially the asteroid is far beyond Pluto, so the initial potential energy of the system is zero$(U_i=0)$

Finally the asteroid hits the Earth, so the potential energy of the asteroid is the sum of gravitational potential energy due to Earth and the gravitational potential energy due to Sun.

The gravitational potential energy of the asteroid due to Earth is

$V_E=-\frac{G{M}_{E}m}{R_E}......\left(1\right)$

Here,

$G\to$Universal gravitational constant$(6.67\times {10}^{-11}M.m^2/k{g}^2)$

$M_E\to$Mass of Earth $(6\times {10}^{24}kg)$

$R_E\to$Radius of Earth$(6.4\times {10}^{6}m)$

The gravitational potential energy of the asteroid due to the Earth is

$V_S=-\frac{G{M}_{S}m}{R_{SE}}......\left(2\right)$

Here,

$M_S\to$Mass of Sun$(2\times {10}^{30}kg)$

$R_{SE}\to$Average distance between Sun and Earth$(1.496\times {10}^{11}m)$

The total final potential energy of the asteroid is

$\begin{array}{rcl}{U}_f& =& V_E+V_S\\ & =& -\frac{G{M}_{E}m}{R_E}-\frac{G{M}_{S}m}{R_{SE}}\\ & & \end{array}$

Let $v$be the impact speed of the asteroid. The final kinetic energy of the asteroid is

$K_f=\frac{1}{2}m{v}^2$

Find the impact speed of the asteroid.

Apply the law of conservation of energy to the system:

$K_i+U_i=K_f+K_f$

$0+0=\frac{1}{2}m{v}^2=-\frac{G{M}_{E}m}{R_E}-\frac{G{M}_{S}m}{R_{SE}}$

$\begin{array}{rcl}\frac{1}{2}m{v}^2& =& GM\left(\frac{M_E}{R_E}+\frac{M_S}{R_{SE}}\right)\\ v& =& \sqrt{2G\left(\frac{M_E}{R_E}+\frac{M_S}{R_{SE}}\right)}......\left(3\right)\\ & & \end{array}$

On substituting the known values in the equation (3), the impact speed of the asteroid can be calculated as

$v=\sqrt{2(6.67\times {10}^{-11}N.m^2/k{g}^2)\left(\frac{(6\times {10}^{24}kg)}{(6.4\times {10}^{6}m)}+\frac{(2\times {10}^{30}kg)}{(1.496\times {10}^{11}m)}\right)}$

$=4.37\times {10}^{4}m/s$

Therefore, the impact speed of the asteroid is$v=4.37\times {10}^{4}m/s.$

Find the change in length of a day due to the impact.

(b) We assume the Earth as a solid sphere, so the moment of inertia of the Earth rotating about its axis is

$I=\frac{2}{5}{M}_E{R^2}_E......\left(4\right)$

Time period of Earth, $T_i=24hr$

$=24hr\left(\frac{60\min }{1hr}\right)\left(\frac{60s}{1\min }\right)$

Initial angular speed of Earth is

${\omega }_i=\frac{2\pi }{T_i}$

$=\frac{2\pi }{86400s}$

$=7.27\times {10}^{-5}rad/s$

Initial angular momentum of the system is

$L_i=Mv{R}_E\sin \theta +I_E{\omega }_i$

$\begin{array}{rcl}& =& Mv{R}_E\sin \theta +\left(\frac{2}{5}{M}_E{R^2}_E\right){\omega }_i\\ & =& -mv{R}_E\sin \theta +\frac{2}{5}{M}_E{R^2}_E{\omega }_i\\ & & \end{array}$

Final angular momentum of the system is

$\begin{array}{rcl}{L}_f& =& I_E{\omega }_f\\ & =& \frac{2}{5}{M}_E{R^2}_E{\omega }_f\\ & & \end{array}$

Using the law of conservation of angular momentum, we get

$$\begin{gathered} L_f=I_i \\ \frac{2}{5}{M}_E{R^2}_E{\omega }_f=-mv{R}_E\sin \theta +\frac{2}{5}{M}_E{R^2}_E{\omega }_i \end{gathered}$$

${\omega }_f=\frac{\frac{2}{5}{M}_E{R^2}_E{\omega }_i=-mv{R}_E\sin \theta }{\frac{2}{5}{M}_E{R^2}_E}$

${\omega }_f={\omega }_i-\frac{mv\sin \theta }{\frac{2}{5}{M}_E{R}_E}......\left(5\right)$

On substituting the known values in the above equation (5), the final angular speed of the Earth can be calculated as

${\omega }_f={\omega }_i-\frac{mv\sin \theta }{\frac{2}{5}{M}_E{R}_E}$

$=(7.27\times {10}^{-5}rad/s)-\frac{(2\times {10}^{21}kg)(4.37\times {10}^{4}m/s)sin30°}{\left(\left(\frac{2}{5}\right)(6.4\times {10}^{6}m)(6\times {10}^{24}kg)\right)}$

$=(7.27\times {10}^{-5}rad/s)-(2.84\times {10}^{-6}rad/s)$

$=6.98\times {10}^{-5}rad/s$

The final period of rotation of Earth can be calculated as

$$\begin{gathered} T_f=\frac{2\pi }{{\omega }_f} \\ =\frac{2\pi }{6.98\times {10}^{-5}rad/s} \\ =90016s \end{gathered}$$

Therefore, the change in length of a day due to the impact can be calculated as

$\begin{array}{rcl}\Delta T& =& T_f-T_i\\ & =& 90016s-86400s\\ & & \end{array}$

$$\begin{gathered} =3616s \\ =3616s\left(\frac{1h}{3600s}\right) \\ \Delta T=1.0h \end{gathered}$$