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According to http://en.wikipedia.org/wiki/Escape_velocity at the Earth/Moon distance from the sun, the escape velocity is 42.1 km/s.

According to http://en.wikipedia.org/wiki/Earth the average orbital speed of Earth is 29.78 km/s.

The difference is 42.1 km/s - 29.78 km/s = 12.32 km/s.

The escape velocity from Earth surface is about 11.2 km/s. The velocity in lowest Earth orbit is about 8 km/s. Take a delta-v of 9.5 km/s to become realistic. Hence an additional 11.2 km/s - 9.5 km/s = 1.7 km/s is needed to escape the gravitational field of the Earth.

The sum of the two deltas is 1.7 km/s + 12.32 km/s = 14.02 km/s.

To dive into the sun, from the orbital speed of 29.78 km/s of Earth the probe needs to be slowed down to zero. The gravity from lower Earth orbit has to be overcome by additional 1.7 km/s. Together 29.78 km/s + 1.7 km/s = 31.48 km/s.

Are there fundamental errors (besides minor aphelion-perihelion discrepancies)? If so, which ones?

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    $\begingroup$ For a first-order approximation and launching on the ecliptic plane it seems about right to me. For more precise formulae see here. $\endgroup$ – TildalWave Feb 6 '14 at 13:02
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One thing you're missing seems to be Oberth's effect. To go from LEO to solar system escape velocity, you have to counter Earth's escape velocity, but after that, you get an additional multiplier by doing the burn at a higher initial velocity (at LEO).

Your method here also has a problem:

Hence an additional 11.2 km/s - 9.5 km/s = 1.7 km/s is needed to escape the gravitational field of the Earth.

To get to LEO, the 9.5 or 10 km/s is the delta v you need delivered by the engines. But that doesn't mean you're that much closer to escape from Earth's gravity. That's because air drag and gravity drag are both "wasted" impulse. Ultimately, they just go to friction. So if you're in LEO:

  • It took 9.5 km/s to get there (we'll say)
  • It will take an additional 11.2-7.9 = 3.3 km/s to escape the gravity well

Now, going from LEO to hyperbolic orbits is a bit more difficult. I'll use the energy balance, because I find it easiest to understand. The specific orbital energy is:

$$ \epsilon={v^2\over2}-{GM\over{r}} $$

After it's escaped Earth's sphere of influence, the energy balance will be simply:

$$ \epsilon = \frac{ v_{\infty}^2 }{2 } $$

For either escape or dropping into the sun, we have a final velocity in mind. This is at 1 AU from the sun, after we get out of Earth's sphere of influence. Earth is moving at 29.78 km/s. so we need:

  • To get to the sun, we need 0 km/s net velocity, so moving 29.78 km/s relative to Earth
  • To get out of the solar system, we need 42.1 km/s in the direction of Earth's motion, so 42.1-29.78 = 12.32 km/s

Now we need to use those above energy equations to get those velocities after getting out of Earth's sphere of influence. Now let's imagine we're half way through the burn, and are at LEO altitude with escape velocity. So we've spent exactly 9.5+3.3 = 12.8 km/s so far. We need to figure out how much more we need in this same burn to shoot for our destination.

$$ v_{\infty} = 29.78 \text{ km/s or } 12.32 \text{ km/s} = \sqrt{ 2 \left( \frac{v^2}{2} - \frac{GM}{r} \right) } $$

Solve this for both cases in terms of $v$ now. For completeness, I use $r=6,354.82 \text{ km}$. Everything else is known. Now the results are:

  • for getting to the sun v = 31.8 km/s
  • for getting out of the solar system v = 16.65 km/s

These are the numbers for the total velocity you need at LEO altitude. In the story I'm telling, you're at 11.2 km/s at the end of the previous leg, so subtract that number in order to calculate the final burn. Again, the trip is broken up into 3 segments according to my organization, but the last 2 are really the same burn. Let me focus on the escape from our solar system. The three legs are:

  • A launch that requires a 9.5 km/s burn
  • The first part of the burn at LEO to get to escape velocity is 3.3 km/s
  • Continuing that same burn, an additional 16.65-11.2=5.45 km/s to get the solar system's escape velocity after you're out of Earth's sphere of influence

The total of all these comes out to 18.25 km/s. If your propellant exhaust velocity is 4 km/s, then your ultimate mass fraction on the launchpad will be about 96-to-1. So a million pound rocket could get 10,436 pounds out of the solar system with this method (I'm not saying it's a good method for this purpose).

I hope this clears up the "from lower Earth orbit" part. It's not as simple as adding things up, because you're trying to get the velocity to escape the sun's gravity well, while you're still in Earth's gravity well. To do that, you have to include the Oberth effect due to your location within Earth's potential well. I hope I've demonstrated that correctly.

EDIT: here is a different set of numbers that starts with the radius of the Earth, instead of the "11.2" and "7.9" numbers, which I only used because they were in prior discussion.

  • Base Earth radius 6378.1 km
  • LEO altitude 300 km
  • LEO radius 6678.1 km
  • V at LEO 7.725529305 km/s
  • Escape V from LEO 10.92554832 km/s
  • Burn from LEO to escape 3.200019015 km/s
  • V needed at LEO to get 29.78 km/s 31.81648047 km/s
  • Extra burn needed past escape V 20.89093216 km/s
  • Oberth ratio 1.425498861 unitless
  • V needed at LEO to get 12.32 km/s 16.64999789 km/s
  • Extra burn needed past escape V 5.724449572 km/s
  • Oberth ratio 2.152171985 unitless
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  • $\begingroup$ There are some details I didn't understand: The 9.5 km/s contain the 8 km/s for a circular orbit on zero height plus the kinetic energy equivalent of the potential energy of the height of the orbit relative to the surface. How gets the air drag into these 9.5 km/s? Shouldn't on the gravitational potential of a lower earth orbit the escape velocity be below 11.2 km/s? How is the height of the low earth orbit considered in "r=6,354.82 km"? Where do the 11.2 km/s come from in 16.65-11.2=2.26 km/s? $\endgroup$ – Gerald Feb 6 '14 at 15:15
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    $\begingroup$ @Gerald I need to be more precise when I'm talking about the physical sequence of events. What happens: (1) rocket blasts into orbit (2) while in this orbit, fires engines again, enough to get to destination. That means that the burn is finished long before it gets out of the gravity well. You do it that way because of the Oberth effect. If you waited until you were away from Earth, then fired the 3rd Delta V, it would take more by the Oberth ratio I gave. "extra burn needed past escape V" is the relative velocity after out of the gravity well. But burn is done before leaving gravity well $\endgroup$ – AlanSE Feb 6 '14 at 18:29
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    $\begingroup$ I think, I've undestood what you meant. Thinking of the burn outside the gravity well has been my essential flaw. The burn should take place at the lowest gravitational potential to get maximum energy by the same velocity gain. Thanks again! $\endgroup$ – Gerald Feb 6 '14 at 19:28
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    $\begingroup$ Note 1: You don't need to completely cancel your velocity to dive into the Sun. You only need to reduce your periapsis to the radius of the Sun. That takes 26.9 km/s at the orbit of the Earth. Not 29.8 km/s. $\endgroup$ – Mark Adler Feb 7 '14 at 4:24
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    $\begingroup$ Note 2: The question was from "lower Earth orbit", which we can only presume means low-Earth orbit (LEO). In that case, you would not include the $\Delta V$ to get from Earth's surface to Earth orbit. Then the $\Delta V$ to get from LEO to barely solar escape is your $3.3\,\mathrm{km/s} + 5.45\,\mathrm{km/s} = 8.75\,\mathrm{km/s}$. $\endgroup$ – Mark Adler Feb 7 '14 at 4:27
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A different way to get to the sun is 9.5 km/s for getting to LEO, 3.5 km/s (less if you slingshot around moon) to escape to solar orbit, then 8.8 km/s for solar escape minus Jupiter/Saturn slingshot then before you escape burn retrograde till perisol is below solar surface. Total should be less than 21 km/s.

Pluto's orbit is on average 10km/s so I'd guess 6-8 km/s.

Sources: Wikipedia, other answers, and I play lots of Kerbal with RSS.

Note: launch into a prograde orbit near equator, otherwise it takes 10 km/s to get to orbit.

Note 2: use a fuel that does not evaporate over time.

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  • $\begingroup$ Note that it is more efficient to combine the Earth escape manoeuvre with the solar escape manoeuvre into a hyperbolic escape from LEO. $\endgroup$ – Hohmannfan Aug 14 '16 at 11:34

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