find the Maximum energy released in the completely inelastic collision of this object with Earth

Question

On Oct 31 2015, massive asteroid TB145 nicknamed "Spooky" passed near the Earth vicinity. The measured diameter of the asteroid: 400meters , and its speed relative to the Sun: 33.0 km/s. 

(NOTE you are not given the parameters of collision, so they become important part of the "worst case –best case scenario.") Treat the asteroid as spherical object with the density of between 3g/cm³ to 4g/cm³. Treat an Earth's orbit around the Sun as a perfect circle of radius =150 x 10⁶ km. Take one year to be 365.24days.

Find the worst case scenario for the collision of such object with the Earth and find the Maximum energy released in the completely inelastic collision of this object with Earth. State your answers to the nearest mega-ton of TNT. ( 1 megaton of TNT= 4.184 x10¹⁵J). 

Answer 1

Carlee,

This question has two Laws in it: the Conservation of Momentum and the Conservation of Energy.

Let's deal first with the collision from a Conservation of Momentum perspective.

You have to find the mass of the asteroid, you can do this using the relationship m = ρV, or density (ρ) ⋅ Volume. to find the volume, use V_sphere = 4/3 π r³ Watch your units!

One problem here is that the density of the asteroid is given as 3 - 4 g/cm³ which can be assumed to be 3.5 g/cm³.

The issue is one of what we call "significant digits" which are digits that have meaning. Your instructor gives you the time of one year to 5 significant figures, but then saddles you with one, maybe two for the density of the asteroid. The best you can hope for in this answer with the given data is only, maybe, two, but I digress....

So now you have enough information to find the momentum of the asteroid, m₁v₁ .

Next, you need to calculate the momentum of your home planet of Earth. Its mass is ~6 x 10²⁴ kg (actually 5.97, but the earlier significant digit issue.... ya know).

Its velocity can be found from the radius and time data. v = (2π radius)/time. Watch your units!

This will be m₂v₂ .

Solve this as an inelastic collision where two masses become one.

m₁v₁ + m₂v₂ = (m₁+ m₂) v_both

Be sure to make one of the velocities negative for this "worst case scenario".

Finally, calculate the energy lost using the Conservation of Energy:

½ m₁v₁² + ½ m₂v₂² = Energy Before the collision.

½ (m₁+ m₂) v_both² = Energy After the collision.

The difference between these two will be the energy released.

Carefully convert these Joules to "megatons"

Watch your units.