Combined Topic 5 Challenge — Asteroid Deflection Mission

Combined Topic 5 Challenge: The Asteroid Deflection Mission

Context: To defend Earth from a potential impact, a "kinetic impactor" spacecraft of mass $4000\text{ kg}$ is launched to intercept a stationary asteroid of mass $6.0\times {10}^{10}\text{ kg}$.

Part 1: Further Momentum (5A)

Question 1

The spacecraft is travelling at $15\text{ km s}{-}^1$ when it hits the asteroid head-on and becomes embedded in its surface.

• Calculate the final velocity of the asteroid-spacecraft system immediately after the collision.
• State the principle you used to perform this calculation and any assumptions made regarding external forces.

Worked Solution:

1. Given Data:

• Mass of spacecraft ($m_s$) = $4000\text{ kg}$
• Initial velocity of spacecraft ($u_s$) = $15\text{ km s}{-}^1=15000\text{ m s}{-}^1$
• Mass of asteroid ($m_a$) = $6.0\times {10}^{10}\text{ kg}$
• Initial velocity of asteroid ($u_a$) = $0\text{ m s}{-}^1$ (Stationary)

2. Physics Principle:

The calculation uses the Principle of Conservation of Linear Momentum, which states that total linear momentum in an isolated system remains constant before and after a collision event.

$$m_s{u}_s+m_a{u}_a=(m_s+m_a)v$$

3. Step-by-Step Calculation:

$$(4000\times 15000)+0=(4000+6.0\times {10}^{10})\times v$$

$$6.0\times {10}^7=(6.0000004\times {10}^{10})\times v$$

$$v=\frac{6.0\times {10}^7}{6.0\times {10}^{10}}=0.0010\text{ m s}{-}^1$$

Final Velocity (v) = 1.0 × 10⁻³ m s⁻¹

Assumptions: No external gravitational forces or electromagnetic field distributions alter the system parameters during the transient collision phase.

Question 2

By calculating the total kinetic energy before and after the impact, determine whether this collision is elastic or inelastic. Explain your reasoning by referring to the energy transfer that occurs during the impact.

Worked Solution:

1. Initial Kinetic Energy Calculations ($E_{\mathrm{ki}}$):

$$E_{\mathrm{ki}}=\frac{1}{2}{m}_s{u}_s^2=\frac{1}{2}\times 4000\times {\left(15000\right)}^2=4.5\times {10}^{11}\text{ J}$$

2. Final Kinetic Energy Calculations ($E_{\mathrm{kf}}$):

$$E_{\mathrm{kf}}=\frac{1}{2}(m_s+m_a)v^2=\frac{1}{2}\times (6.0\times {10}^{10})\times {\left(0.0010\right)}^2=3.0\times {10}^4\text{ J}$$

3. Evaluation Strategy:

Because the kinetic energy values show a massive loss ($E_{\mathrm{kf}}<<E_{\mathrm{ki}}$), kinetic energy is completely non-conserved.

The collision is totally inelastic.

Energy Transformation Analysis: The bulk of the initial kinetic energy ($\approx 4.5\times {10}^{11}\text{ J}$) converts dynamically into internal thermal updates, material sound waves, and structural work performed while crushing/embedding the spacecraft frame into the rocky crust matrix.

Question 3

The collision lasts for $0.20\text{ seconds}$. Calculate the average impulse exerted on the asteroid and the resulting average force experienced by the spacecraft during the impact.

Worked Solution:

1. Impulse Calculation ($I$) via Change in Momentum:

$$I=\Delta p=m_{a}v-m_a{u}_a$$

$$I=(6.0\times {10}^{10})\times 0.0010-0=6.0\times {10}^7\text{ N s}$$

2. Average Impact Force Calculation ($F$):

Using Newton's second structural law formulation ($F=\frac{\Delta p}{\Delta t}$):

$$F=\frac{I}{\Delta t}=\frac{6.0\times {10}^7}{0.20}=3.0\times {10}^8\text{ N}$$

Average Impulse = 6.0 × 10⁷ N s
Average Force = 3.0 × 10⁸ N

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Part 2: Circular Motion (5B)

Question 4

Following the collision, the asteroid is nudged into a new circular "parking orbit" around a nearby moon at a radius of $3500\text{ km}$ from the moon's centre.

• Explain why the asteroid is considered to be accelerating even if its orbital speed remains constant.
• Calculate the centripetal force required to maintain this circular orbit, given the asteroid's new velocity.

Worked Solution:

1. Mechanics Concept Explanation:

Velocity constitutes a vector quantity possessing explicit scalar speed and a specific spatial orientation. While tracking along a circular path, structural speed metrics are completely uniform, but the directional tracking vector updates dynamically at every instant. Since acceleration is fundamentally defined as the path update rate of velocity over time, any vector direction changes trigger net centripetal acceleration aimed towards the global rotational focus.

2. Centripetal Force Matrix Evaluation ($F_c$):

• Effective System Mass ($M$) = $6.0\times {10}^{10}\text{ kg}$
• Orbital Path Radius ($r$) = $3500\text{ km}=3.5\times {10}^6\text{ m}$
• Calculated System Velocity ($v$) = $0.0010\text{ m s}{-}^1$

$$F_c=\frac{M{v}^2}{r}=\frac{(6.0\times {10}^{10})\times {\left(0.0010\right)}^2}{3.5\times {10}^6}$$

$$F_c=\frac{60000}{3.5\times {10}^6}\approx 0.01714\text{ N}$$

Centripetal Force = 1.7 × 10⁻² N

Question 5

Determine the angular velocity ($\omega$) of the asteroid in this orbit in radians per second.

Worked Solution:

1. Formula Definition:

$$v=r\omega ⟹\omega =\frac{v}{r}$$

2. Mathematical Execution:

$$\omega =\frac{0.0010}{3.5\times {10}^6}\approx 2.857\times {10}^{-10}\text{ rad s}{-}^1$$

Angular Velocity (ω) = 2.9 × 10⁻¹⁰ rad s⁻¹

Question 6

A sensor on the asteroid's surface rotates as the asteroid spins. If the asteroid completes one full rotation every $24\text{ hours}$, calculate the centripetal acceleration experienced by a sensor located $50\text{ m}$ from the asteroid's centre of mass.

Worked Solution:

1. Chrono Unit Metric Conversion:

$$T=24\times 60\times 60=86400\text{ s}$$

2. Intrinsic Spin Angular Velocity Evaluation (${\omega }_{\text{spin}}$):

$${\omega }_{\text{spin}}=\frac{2\pi }{T}=\frac{2\pi }{86400}\approx 7.2722\times {10}^{-5}\text{ rad s}{-}^1$$

3. Centripetal Acceleration Calculation Loop ($a$):

Applying the standard mechanical format equation $a={\omega }^{2}r$ using the targeted radius displacement $r=50\text{ m}$:

$$a={(7.2722\times {10}^{-5})}^2\times 50$$

$$a=(5.2885\times {10}^{-9})\times 50\approx 2.644\times {10}^{-7}\text{ m s}{-}^2$$

Centripetal Acceleration = 2.6 × 10⁻⁷ m s⁻²