A 1000 kg object is found heading towards the Earth. At the moment it is 4 Earth radii away from...

Question:

A 1000 kg object is found heading towards the Earth. At the moment it is 4 Earth radii away from the surface of our planet, the asteroid is moving at 100 m/s. If it collides with our planet, how fast will it be moving right before it hits the surface? How much kinetic energy is going to be released in this collision?

Energy Conservation:

Energy conservation means that the energy of a system can only be transformed from one type to another, but never be destroyed or created. Thus, if there is no external force acted on the system (closed system), the total energy is always conserved. The total mechanical energy (kinetic energy + potential energy) is also conserved if there is no dissipation of energy through the heat, light, sound, and so on.

Answer and Explanation:

Total mechanical energy is conserved in the system, so

{eq}KE_i + PE_i = KE_f + PE_f {/eq}

Here,

  • KE and PE are kinetic energy and potential energy, respectively,
  • i, f refer the initial and final states, respectively.

Let m and M be the masses of the object (1000 kg) and the Earth (={eq}5.972 \times 10^{24} {/eq} kg), respectively.

Let {eq}v_i, v_f {/eq} be the speed of the object of initial (=100 m/s) and the final (at the impact time) states, respectively.

Let R and h be the radius of the Earth and the initial height of the object from the surface of the Earth, respectively, so

{eq}R = 6.387 \times 10^6 \ \textrm m\\ h = 4R = 2.551 \times 10^7 \ \textrm m {/eq}

Then,

{eq}\begin{align} KE_i &= \frac{1}{2}mv_i^2\\ &= 5.0 \times 10^6 \ \textrm J\\ PE_i &= - \frac{GMm}{R+h}\\ &= -\dfrac{6.67\times 10^{-11} \ \textrm{N}\cdot \textrm{m}^2\textrm{/kg}^2 \times 5.972 \times 10^{24} \ \textrm{kg} \times 1000 \ \textrm{kg}}{6.387 \times 10^6 \ \textrm m+2.551 \times 10^7 \ \textrm m}\\ &=-1.25 \times 10^{10} \ \textrm J\\ KE_f &= \frac{1}{2}mv_f^2\\ &= 500v_f^2\\ PE_f &= - \frac{GMm}{R}\\ &= -\dfrac{6.67\times 10^{-11} \ \textrm{N}\cdot \textrm{m}^2\textrm{/kg}^2 \times 5.972 \times 10^{24} \ \textrm{kg} \times 100 \ \textrm{kg}}{6.387 \times 10^6 \ \textrm m}\\ &=-6.24 \times 10^{10} \ \textrm J\\ \therefore 5.0 \times 10^6-1.25 \times 10^{10} &= 500v_f^2 -6.24 \times 10^{10}\\ 500v_f^2 &= 4.99 \times 10^{10} \ \textrm J\\ v_f &= 9989 \ \textrm{m/s}\\ &= \boxed{9.99 \ \textrm{km/s}} \end{align} {/eq}


The amount of energy the object is going to release in this collision would the kinetic energy of the object at the impact, thus;

{eq}\begin{align} E &= \frac{1}{2} mv_f^2\\ &= 500 \ \textrm{kg} \times (9989 \ \textrm{m/s})^2\\ &= \boxed{4.99 \times 10^{10} \ \textrm J} \end{align} {/eq}


Learn more about this topic:

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Conservation of Mechanical Energy

from AP Physics 1: Exam Prep

Chapter 8 / Lesson 8
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