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It takes the Earth about 365.25 days to orbit the Sun on its current elliptical path. That means it moves about 30 km/s, 2×π×(149,600,000 km)/(1 year).

So, given that, how high could Earth's velocity go before its orbit changes?

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    $\begingroup$ Logan's answer is actually referring to perihelion or the closest point of Earth orbit around Sun using Kepler 2nd law. The orbit is actually elliptical according to Kepler 1st law so velocity Max out when Earth is closest to Sun. $\endgroup$ – user6760 Oct 15 '17 at 6:30
  • $\begingroup$ Yes, the 30 km/s is an overall average along the elliptical path. At some points of the orbit, it's faster and slower at others. I was hoping to make a storyline in which Earth's solar year became faster (shorter) without a change in orbit as I needed other environmental factors to remain unchanged. I see that's not a possibility without ignoring or wand-waving the laws of physics. $\endgroup$ – Leezard Oct 15 '17 at 17:33
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    $\begingroup$ @Leezard: you could sorta do it with really big engines and active course correction, but at that point you’ve just handwaved engines powerful enough to move the earth, so... $\endgroup$ – Joe Bloggs Oct 16 '17 at 8:05
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    $\begingroup$ @JoeBloggs It's a disaster scenario. Of course that's a fun idea, no doubt I can have someone say, "Big engines, what could go wrong?" :) $\endgroup$ – Leezard Oct 16 '17 at 18:30
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About 30km/s; that it to say, the velocity cannot change at all if you wish to maintain the same orbit.

Every circular orbit is associated with exactly one orbital velocity. Every general elliptical orbit is associated with exactly one velocity profile--one specific apoapsis velocity, one specific periapsis velocity, and one specific curve in between.

If you speed the Earth up at all, its orbit will be different.

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    $\begingroup$ For clarity you might point out the Earth is travelling at 30 km/s currently and this isn't some different or higher orbital velocity. Sorry if I'm being pedantic and trying to point the obvious. $\endgroup$ – a4android Oct 15 '17 at 4:12
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    $\begingroup$ If you add a large enough rocket engine to aim outward (from the sun), it could go much faster in the same orbital track. ;) $\endgroup$ – Yakk Oct 15 '17 at 5:41
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    $\begingroup$ @Yakk Adding radial acceleration outward in that way would force the Earth into a spiral orbit away from the Sun while the rocket was active and then a final orbital path (some conic path) after the rocket stopped. To offset the outward radial acceleration you would need to slow down the Earth (not speed it up). To speed up the Earth would require you add force towards the Sun. $\endgroup$ – StephenG Oct 15 '17 at 13:57
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    $\begingroup$ @StephenG The rocket engine aims outward, the Earth (the rocket) is pushed inward. $\endgroup$ – Yakk Oct 15 '17 at 15:24
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    $\begingroup$ @Agent_L The trouble is, it isn't "precisely". The Earth's orbit is slightly elliptical, so the Earth's orbital speed does in fact vary a bit above and a bit below 30 km/s. At any given point, it must have exactly one velocity, and changing that velocity will change the orbit; but over the whole orbit, it doesn't have just one velocity. $\endgroup$ – Logan R. Kearsley Oct 15 '17 at 22:23
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If the Sun were more massive than it is now, then the Earth would have to move faster to maintain the orbit that it currently has. The formula for the rotational velocity is $v^2$ = (G • M) / R, where v the velocity, G is the gravitational constant, M is the mass of the sun, and R is the radius of the orbit. As an example, if the Sun had four times the mass that it currently has, then the Earth would have to be moving twice as fast as it is now to be in the orbit it currently occupies. Of course if the Sun had four times the mass that it currently has then other properties about it would be very different as well.

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    $\begingroup$ Exactly my idea - a black hole is dumped into the sun, possibly first passing by Earth and conveniently speeding up its orbital velocity, so it fits the new sun mass...well, a bit too much of a coincidence. $\endgroup$ – Klaws Oct 16 '17 at 15:04
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You can't pick the velocity and distance from the sun independently: either one fixes the other. A planet is in orbit because the gravitational attraction of the star accelerates it, causing its path to be a curve rather than a straight line. If you make the planet move faster, then it would require more acceleration to keep it in its orbit (consider whirling a ball on a string around your head – make it spin faster and you feel more pull on the string), but the only way to get more acceleration is to be closer.

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Depending on the orbit, say if it was more elliptically "extreme" than now, something like this:
elliptical orbit and minor body at periapsis:  it appears like a big cartoon eye with pupil far to one side and staring at the smaller dot at corner of the eye; a tangential arrow on the smaller dot indicates direction of its velocity

Then the speed would be much greater when Earth reaches the point closest to the sun since it will have accelerated on its way towards the sun. The orbit would be the "same" but the speed throughout the full orbit would differ dramatically.

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  • $\begingroup$ So an orbit more like a comet's than a planet's. $\endgroup$ – David Richerby Oct 16 '17 at 16:40
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As mentioned by Itsme2003, v2 = (G • M) / R, where v is Orbital Velocity, G is the Gravitational Constant, M is the mass of the central body, in this case our Sun, and R is the Radius of Orbit.

Since the mass of the sun isn't changing in this scenario, G and M are both constants, so we can conflate the two into a single value, C:

v2 = C / R

This can be rearranged by multiplying both sides by R to give us:

v2 • R = C

What this shows us is that the Orbital Velocity and the Radius of Orbit are inversely proportional. As one increases, the other decreases, and vice versa.

Therefore, we can say with certainty that if you were to hack into the Earth's properties and change its "Orbital Velocity" value, the radius of the Earth's orbit will also change, taking us closer or farther from the Sun, depending on whether you increase or decrease the value.

This has been explained by others in other answers, but I wanted to show it through the inverse proportion relationship of the mathematical formula.

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