I need to know this for a test. Please help. The question states "If a spacecraft was parked on Venus and needed to make a flight to Jupiter, how far would it need to travel? (Assume both planets are aligned with the sun and are on the same side of the sun.)"
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4$\begingroup$ The question doesn't make any sense, since both planets are moving. If you just want to know the distance from the sun to Venus and from the Sun to Jupiter, those values are easily found by searching on the internet. If you want to know the actual distance travelled by a space craft on a hypothetical mission from Venus to Jupiter, the question is not complete, since there are several possible trajectories depending on how much fuel you have, and the type of rocket you are using. Also taking off on Venus is essentially impossible. $\endgroup$– James KApr 23, 2019 at 22:01
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$\begingroup$ OK, I just recently found the answer to the question, but thanks anyways. :) $\endgroup$– HelpApr 23, 2019 at 22:07
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2$\begingroup$ @Help go ahead and post an answer to your question. It is always okay to answer your own question in Stack Exchange, as long as the answer is reasonable. If there are no other better answers posted in a few days, you can even accept your own answer. Welcome to Stack Exchange! $\endgroup$– uhohApr 23, 2019 at 22:26
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2 Answers
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To save energy, let's consider a Hohmann transfer orbit, an ellipse tangent to opposite sides of Venus's and Jupiter's orbits. The spacecraft would follow the outbound half of this orbit, departing ahead of Venus and arriving as Jupiter overtakes it.
Depending on where in Jupiter's orbit the spacecraft meets it, this orbit has semi-major axis a = (rV + rJ) / 2 = 2.96±0.03 au and period of a3/2 = 5.10±0.08 years. The outbound half has a path length of 7.78±0.07 au and takes 2.55±0.04 years. As Jupiter moves 72° to 84° around the Sun in that time, the launch window is when Venus's heliocentric ecliptic longitude is 96° to 108° behind Jupiter's.
To convert au to km, multiply by 1.496 × 108 km / au.
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$\begingroup$ Thanks for providing those numbers. Of course, the original question doesn't specify the Hohmann method, nor does it take into account where either planet are relative to aphelion or perihelion. The Hohmann method provides for the smallest energy expenditure, which I assume produces the lowest delta-v. If you increase velocity, you can reduce the total distance traveled significantly, but this requires a greater energy expenditure. Again, the original question provides insufficient data as I read it. $\endgroup$ Apr 24, 2019 at 20:32
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1$\begingroup$ Yes, I added a constraint to an underconstrained problem. $\endgroup$– Mike GApr 24, 2019 at 20:51
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If you want the distance between the planets, then you can easily look up the orbital radius of both planets, subtract the orbital distance of Venus from that of Jupiter, and there you have the distance between Venus and Jupiter.... but this is problematic.
First of all, both planets orbit in an elliptical path. Their distances from the sun vary. For example, Venus's Perihelion, the closest it gets to the sun, is about 107,447,000 km. It's Aphelion (furthest from the sun) is 108,939,000. Jupiter's Perihelion is 740,520,000 km and its Aphelion is 816,650,000. If Venus is at Aphelion and Jupiter is at Perihelion when Jupiter is in opposition to Venus (meaning directly opposite from the sun), then the distance is about 631,581,000 km. If Venus is at Perihelion while Jupiter is at Aphelion while at opposition, then the distance is 709,173,000 km. If they're not both at opposition, the distance can be as far as 925,559,000 km if they're both at Aphelion and on opposite sides of the sun.
But these are straight line distances. If you're flying between the two, as you leave one and head to the other, you won't travel in a straight line. You'll follow a curved path to get from one to the other. The distance traveled will then be highly variable depending on the velocity. The higher the Delta-V, the "straighter" the path between them and, thus, the shorter the distance. But a smaller Delta-V, the longer the trip, and the wider the radius of the arc, and, thus, the longer the distance.
In the end, there's not enough information given to provide an accurate answer.