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brainTeaser[1]

 
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Norway tr1sth3t
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PostPosted: Mon Mar 22, 2010 6:53 pm   Post subject: brainTeaser[1] Reply with quote


The elongation (we'll say α) of a planet is the angle formed by the planet, Earth and the Sun. It is known that the distance from the Sun to Venus is 0.723AU and from the Earth to the Sun is 1AU. At a certain time the elongation of Venus is found to be 39.4°. Find the possible distances from the Earth to Venus at that time in Astronomical Units (AU).

Hint: Use the law of sines.

sin(A)/a = sin(B)/b = sin(C)/c



ANSWER TO BRAIN TEASER 0:
Quote:
r = 10m, Δr = ±1.2m
h = 15m, Δh = ±2.4m


If we logically think about these changes (Δr and Δh) in r and h they are simple the limits on the instrument.

ΔV ≈ dV = (dπr²)
= πr²dh + hd(πr²)
= πr²dh + h2πrdr
= πr²dh + 2πrhdr


|ΔV| ≈ |dV| = |πr²dh + 2πrhdr|
≤ |πr²dh| + |2πrhdr|
= |π(10)²(±2.4)| + |2π(10)(15)(±1.2m)|
= 240π + 360π = 600πm³1,884.66m³


Now we need to compute the relative error.
V = πr²h = (10)²(15)π = 1500π ≈ 4,712.4m³
|ΔV ÷ V| ≈ |dV ÷ V| = 600πm³ ÷ 1500πm³


Here the units cancel along with the symbol π (pi) leaving:
600 ÷ 1500 = 0.40 or a 40% relative error.

(a) The maximum possible absolute error in the calculated volume is 1,884.66m³
(b) The maximum possible relative (percentage) error in the calculated volume is 40%.


So this instrument sucks and a new one needs to be bought.

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Last edited by tr1sth3t on Sun Apr 04, 2010 3:43 am; edited 5 times in total
Canada irule
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PostPosted: Mon Mar 22, 2010 10:34 pm   Post subject: Reply with quote


Yay trig. I'm actually okay at this. I got a minimum distance of 0.02013 au and a max of 1.15325 au. And I'm probably completely wrong.
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Norway tr1sth3t
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PostPosted: Mon Mar 22, 2010 10:58 pm   Post subject: Reply with quote


irule wrote:
Yay trig. I'm actually okay at this. I got a minimum distance of 0.02013 au and a max of 1.15325 au. And I'm probably completely wrong.


I will say that the correct answer fall in between your max and min.
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PostPosted: Mon Mar 22, 2010 10:59 pm   Post subject: Reply with quote


Isn't the possible distance any value between those two?
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Norway tr1sth3t
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PostPosted: Mon Mar 22, 2010 11:52 pm   Post subject: Reply with quote


irule wrote:
Isn't the possible distance any value between those two?


Yes, but which value is it? Specifically the one where the angle formed by the Earth is 39.4°.
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Last edited by tr1sth3t on Sun Apr 04, 2010 3:37 am; edited 2 times in total
Canada irule
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PostPosted: Tue Mar 23, 2010 12:07 am   Post subject: Reply with quote


I thought the 39.4 degree angle was formed by Venus?

Nvm... the key lies in the fact that Earth is farther away from the Sun than Venus is... so it is between 1.1173728 and 26.849632? I don't see how there is enough information to get a single value.
Norway tr1sth3t
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PostPosted: Sun Apr 04, 2010 4:04 am   Post subject: Reply with quote


ANSWER:

V1 and V2 are the positions of Venus, E is the earth and S is the Sun. Since the distance from the Sun to Venus are the same consider SV1 = SV2 = 0.723AU. Consequently, the corresponding opposite angles are equal. So, the remaining angles are measured.

Now, using law of sines we have
[ V2E ÷ sin(xº) ] = [ 1 ÷ sin(140.6 - x)º ] = [ 0.723 ÷ sin(39.4º) ]
⇒ V2E = sin(xº)[ 0.723 ÷ sin(39.4º) ]

also, sin(140.6 - x)º = [ sin(39.4º) ÷ 0.723 ]
⇒ xº = 140.6º - sinˉ¹[ sin(39.4º) ÷ 0.723 ] = 79.208º
∴ V2E = 1.11891 astronomical units.

Again from the triangle V1SV2 , using the sine law we get
[ 0.723 ÷ sin(140.6 - x)º ] = [ V1V2 ÷ sin(101 - 2x)º ]
⇒ V1V2 = -0.708079 AU
∴ V1E = 1.11891-0.708079 = 0.42 AU


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PostPosted: Sun May 22, 2011 3:47 pm   Post subject: Reply with quote


hmm i dont know the anwser
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