Question

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer 1

Answer:

$\frac{1}{10}$M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒$I_{1} \times W_{1} =I_{2} \times W_{2}$

• $I_{1}$ is the moment of interia of earth before impact
• $W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
• $I_{2}$ is moment of interia of earth and asteroid system
• $W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let  $W_{1}=W$

since $\text{Time period of rotation}∝$$\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$  times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$(moment of inertia of solid sphere)

where M is mass of earth

           R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$$+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$= $(\frac{2}{5} \times M\times R^{2}$+ $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒$M_{1}=}$$\frac{1}{10} \times M$

Answer 2

For the day to become 25.0% longer than it presently is, the mass of this asteroid must be $m_1=\frac{1}{10} m$.

How to calculate the mass of this asteroid.

Since the asteroid is travelling straight (directly) towards the center of the Earth, the angular momentum of Earth is conserved and this is given by this formula:

$I_1\omega_1 = I_2\omega_2$

Also, the time period of oscillation is indirectly proportional to the angular velocity ($T\; \alpha \;\frac{1}{\omega}$)

If the time period increases by 25%, it becomes 125% ($\frac{5}{4}$), the angular velocity would decrease by $\frac{4}{5}$

Before impact, the angular velocity of Earth about an axis that passes through the center of Earth is given by:

$\omega_1 = \frac{4\omega_1}{5}$

For the solid sphere:

Mathematically, the moment of inertia of a solid sphere is given by this formula:

$I=\frac{2}{5} mr^2$

Where:

• I is the moment of inertia.

• m is the mass.

• r is the radius.

For the final moment of inertia:

$I_2=\frac{2}{5} mr^2+m_1r^2$

Note: We assume that the asteroid has a very small mass compared to the Earth and that the Earth is uniform throughout.

$I_1\omega_1 = I_2\omega_2\\\\\frac{2}{5} mr^2(\omega_1)=(\frac{2}{5} mr^2+m_1r^2) \frac{4\omega_1}{5}\\\\\frac{1}{2} mr^2=\frac{2}{5} mr^2+m_1r^2\\\\m_1r^2=\frac{1}{10} mr^2\\\\m_1=\frac{1}{10} m$

Read more on inertia here: https://brainly.com/question/3406242