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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,047

The distance of comet d from a Jupiter's moon is determined by the equation d = 47.9t² + 0.03t - 908.7 whereas t is how many hours passed ever since that comet appeared during June 28 2001. Determine the date when that comet reaches the moon again.

I tried to draw the illlustration as follows:

According to the illustration, t = 0 June 28 2001. After reaching the turning point located in the symmetrical axis, the comet appears once more from that moon. That means, the value of t when the comet comes again to the moon is twice the symmetrical axos, but I got negative number which is -0.03/47.9. Or was my method for solving this question learn?

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

Hi, I cannot say how you calculated that (although it is half of a rational constant in the solution), but the parabola is not quite symmetric.

The correct answer is about 8.7, which leads me to believe t is in days, not hours

*Last edited by Relentless (2016-02-05 20:33:08)*

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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,047

It's just my poorly made illustration.

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 53

The equation provided is a parabolic equation, but it is not the equation of the comet trajectory!! Since to derive the equation of the comet path, you have to introduce first a plane coordinate system (choosing a certain point as the origin) in which both variavbles are distances, or if you like to use time, you can use a parametric representation of the trajectory with time taken as the parameter. The equation which is provided has nothing to do with the trajectory (if it is the equation of the trajectory in a coordiate system other then the cartesian, then it must has different form). It just relates time to the distance of the comet from the center of jupiter's moon, and since the comet is at equal distances from jupiter's moon at symmetrical points (taking the axis of the parabola as the symmetrical line), each distance from the moon has two values, "equal in magnitudes and different in signs" depending on the side which the comet occupied. Therefore, since the comet in its nearest approach, when t=0, was at the distance d1=-908.7 then after passing the turning point it will approach jupiter's moon again at the distance d2=908.7 (since it is the symmerical point of d1) hence using the given equation we have:

Now since time taken to be zero when the comet first approaches jupiter's moon, so the time will be positive in the next approach hence t=6.1594.

Note: If t is in hours, then the next approach will be after

therefore the next nearest approach to jupiter's moon will be on "July 2011".

Q.E.F

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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,047

Grantingriver wrote:

The equation provided is a parabolic equation, but it is not the equation of the comet trajectory!! Since to derive the equation of the comet path, you have to introduce first a plane coordinate system (choosing a certain point as the origin) in which both variavbles are distances, or if you like to use time, you can use a parametric representation of the trajectory with time taken as the parameter. The equation which is provided has nothing to do with the trajectory (if it is the equation of the trajectory in a coordiate system other then the cartesian, then it must has different form). It just relates time to the distance of the comet from the center of jupiter's moon, and since the comet is at equal distances from jupiter's moon at symmetrical points (taking the axis of the parabola as the symmetrical line), each distance from the moon has two values, "equal in magnitudes and different in signs" depending on the side which the comet occupied.

But... Isn't the longest distance the turning point when the comet goes back to the moon?

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 53

When a celestial object (e.g. Comet, asteroid, planet...) orbet another object, the orbet could be one of the three types of conic sections (circle, ellipse or parabola it could also be a hyperbola in some hypothetical universes where the gravity repulsing instead of attracting) in the case of comets the orbets are parabola and the longest distance is "∞" (the comet does not return again). This fact is reflected in the equation that relates time with the distance from the center of jupiter's moon which you have provided, so the distance from the moon at the turnig point is not the absolute longest distance from the center of jupiter's moon, It is a local longest distance between the successive nearest approach of the comet.

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