# IPhO 2013 Solution Attempt

1. Jul 23, 2013

### theultimate6

1. The problem statement, all variables and given/known data

http://ipho2013.dk/ipho2013-theoretical-problem-1.pdf

3. The attempt at a solution

Are those correct?

1.1: Velocity in the x direction : $$v_x=ω_x r=(π*(θ_2-θ_1)*r/180)/Δt r$$
Velocity in the y direction : $$v_y = ω_y r=(π*θ'_1-θ'_2)*r/180)/Δt$$

Then use Pythagorean theorem to get the average $$v = 31128.6 m/s$$

1.2a: $$F_friction=kρ_atm Av^2=4kρ_atm πR_M^2 v^2=m_M dv/dt =(0.10m_M v_M)/dt$$

$$dt=(0.10m_M v_M)/(4kρ_atm πR_m^2 v_M^2 )=(0.10〖 m〗_M)/(4kρ_atm πR_M^2 v_M^2 )$$

1.2b: $$E_k=(mv^2)/2=(30*(2.91*〖10〗^4 )^2)/2=1.2615*〖10〗^10 j$$

$$E_melt= mL=30*2.6*〖10〗^5=7.8*〖10〗^6 j$$

$$E_kin/E_melt =(1.2615*10^10 )/(7.8*〖10〗^6 )≈1617.307692$$

1.3a: $$L=t^α M^β L^(-3β) L^2γ t^(-2γ) T^(-γ) M^δ L^δ t^(-3δ) T^(-δ)$$
$$L:1=-3β+2γ+δ$$
$$t:0=α-2γ-3δ$$
$$M:0=β+δ$$
$$T: 0=-γ-δ$$

Solving the system of equations

$$α= 1/2$$
$$β=-1/2$$
$$γ=-1/2$$
$$δ=1/2$$

thus :

$$x=√((tk_sm)/(ρ_sm c_sm ))$$

$$x=√((tk_sm)/(ρ_sm c_sm ))=√((5*2)/(1.2*〖10〗^3*3.3*〖10〗^3 ))=0.00158910 m$$

$$x/R_M =0.00158910/0.13=0.012223846$$

$$1.5: t_Encke=2π√((a_max^3)/(GM_sun ))=2π√((6.16*〖10〗^11 )^3/(6.67*〖10〗^(-23)*1.99*10^30))=2.636*〖10〗^8 s=8.353 years$$

2. Jul 23, 2013

### verty

1.1 doesn't look right. I think you need to find the distance the comet traveled.

3. Jul 23, 2013

### collinsmark

This particular post only concentrates on problem 1.1
(Maybe I can comment on the other problems/parts later.)

Your approach doesn't work that way for this problem. Your answer is in the ballpark, but it's off by a few km/s.

I see what you approach is: you're treating it as a rotating system, then calculating the tangential velocity. But that doesn't quite work for this problem since the meteor did not travel in a circular trajectory. The 'r' in your approach is constantly changing, the the direction of the meteor is not tangential to r. I'm not sure I follow you regarding the θ' variables.

I suggest a different approach. Break up the horizontal distance part of the problem into two separate triangles (each triangle uses a different azimuth data point). Use the following for each
• The three angles of a triangle always add to 180o.
• You won't hear me say this often, but use the "law of sines." With that you can get the distances of all three sides of each triangle.
The altitude part is a little simpler, since it always involves right triangles. But it does rely, in part, on results you obtained from the horizontal distances in your triangles.

Then use the Pythagorean theorem to combine the horizontal and vertical distances.

4. Jul 23, 2013

### theultimate6

An Update on 1.2 :

$$F_friction=kρ_atm Av^2=4kρ_atm πR_M^2 v^2=m_M dv/dt$$

Solve for $$dt$$

$$dt=m_M/(4kρ_atm πR_M^2 v^2) dv$$

integrate both sides from $$0.90v_M to v_M$$

$$∫ m_M/(4kρ_atm πR_M^2 v^2 ) dv = t$$

$$t= (0.00884194m_M)/(kρ_atm R_M^2 v_M )≈0.21925 seconds$$

5. Jul 23, 2013

### theultimate6

The three angles of the triangle are 50,75 and 55

so :

$$Sin (75) / 195 = Sin(50) / d_horizontal$$

d_horizontal = 154.65 m

6. Jul 23, 2013

### collinsmark

Good. You've found that the third angle in Fig. 1.1(c) is 55 degrees. You'll need that later.

Yes, but that's not the horizontal distance traveled between frames 155 and 161.

To calculate that distance, you'll need to create two more triangles. Both of the new triangles share the 55o angle, and they both share the 195 km distance side. However, the other sides and other angles are different. Use the given azimuth data in Fig. 1.1(b) to calculate the top angle of each of the new triangles.

Last edited: Jul 23, 2013
7. Jul 23, 2013

### theultimate6

But if I use the azimuth data the 55° angle changes

8. Jul 23, 2013

### collinsmark

No, the 55o angle stays the same. The top angle (originally the 50o in the original, big triangle) comes directly from the azimuth data, which of course changes. Consequently, the 75o angle will change.

But the 55o angle stays the same, as does the 195 km side of the triangle.

That's because the trajectory of the meteor with respect to the landing site is fixed. Also the distance and angle between the camera and the landing site are fixed.

[Edit: the distance between the camera and meteor changes with time, the azimuth angle of the meteor with respect to the camera changes with time (meaning the top angle and the lower right angle change with time), and of course the distance of the meteor with respect to the final landing site changes with time. Your initial goal is to measure the change in distance from the meteor to the final landing site between frames 155 and 161. But you'll also need the camera to meteor distances at these points too for later when you calculate the altitude (vertical) changes.]

Last edited: Jul 23, 2013
9. Jul 23, 2013

### theultimate6

The top angles are :35 and 41

and the bottom angles are : 90 and 84

10. Jul 23, 2013

### collinsmark

Sorry, I'm not following you here. :uhh:

The left side of the triangle is distance from the camera to the final landing site (it's 195 km long, and bounded by points C and M). The right side of the triangle is the distance from the camera to the meteor. The bottom side of the triangle is the distance from the meteor to its eventual landing site.

Remember, the meteor follows the path of the purple line (actually its more of magenta color, but I'll just say purple) in figure 1.1(c).

Last edited: Jul 23, 2013
11. Jul 23, 2013

### collinsmark

Okay, I sort of see what you are doing. You are keeping the bottom right angle of 75o constant. and keeping the right side constant.

That will work too. If you do it that way, the top angles are 35 and 41 degrees as you mentioned.

But first you'll have to calculate the distance of the right side of the triangle. It's not 195 km. You'll have to calculate what that is.

Also recalculate the bottom left angles (they're not 90 and 84).
-----

Alternately, you could use the method in my previous post. That way you get to use the 195 km value, which is easier to type into a calculator.

12. Jul 24, 2013

### voko

Alternatively, it could be solved in the following way. Assume that the camera is at (0, 0, 0). The velocity of the meteorite is $\vec{v}$. Assuming it is falling in a straight line, its trajectory is described by $\vec{x} = \vec{v} t + \vec{p}$. We know the following: at time $t = t_1$, the meteorite was at some unknown distance $A$ from the camera in the direction $\vec{a}$ ($|\vec{a}| = 1$), and at time $t = t_2$ it was at some distance $B$ in the direction $\vec{b}$ ($|\vec{b}| = 1$), and at some unknown time $t = T$ it was at the known location $\vec{c}$ (on the ground). Thus: $$A\vec{a} = \vec{v} t_1 + \vec{p}\\ B\vec{b} = \vec{v} t_2 + \vec{p} \\ \vec{c} = \vec{v} T + \vec{p}$$ These are 3 vector equations, i.e., 9 scalar equations, with 9 unknowns: 3 components of $\vec{v}$, 3 components of $\vec{p}$, and $a$, $b$, and $T$, thus this is all we need to solve the problem. Since $\vec{p} = \vec{c} - \vec{v}T$, $$A\vec{a} = \vec{v} t_1 + \vec{c} - \vec{v}T \\ B\vec{b} = \vec{v} t_2 + \vec{c} - \vec{v}T$$ Which yields $$\vec{v} = (A\vec{a} - B\vec{b}) / (t_1 - t_2)$$ Substituting all this back into the first equation: $$A\vec{a} = (A\vec{a} - B\vec{b}) / (t_1 - t_2) (t1 - T) + \vec{c}$$ This is three scalar equations for three unknows $A$, $B$ and $T$: $$a_1 A = (a_1 A - b_1 B) /(t_1 - t_2) (t_1 - T) + c_1 \\ a_2 A = (a_2 A - b_2 B) /(t_1 - t_2) (t_1 - T) + c_2 \\ a_3 A = (a_3 A - b_3 B) /(t_1 - t_2) (t_1 - T) + c_3$$ Because the meteorite hits the ground, $c_3 = 0$, which simplifies things a bit.

When this system is solved, the velocity may be determined.