6 ms·

> It still takes a delta-V of ~30m/s to hit the Sun. That number looks like it should be much much bigger to me.

dabluecaboose 2y ago
Indeed, it's off by orders of magnitude (so probably a simple typo). The Earth orbits the Sun at between 29 and 30 km/s. It would take roughly that much dV to successfully create an orbit that intersects the sun.
tialaramex 2y ago
The straight forward way is typically given as 30km/s so yeah, it's the wrong unit. But, you'd never actually do that, the calculations involved assume you have decided to specifically go towards the sun, maybe to get there quickly, but going to the sun is just a means to an end, so you'd actually pull shenanigans to get the same result indirectly for "only" a few km/s of delta V with carefully chosen launch windows. Which is like how it'd be easier to build a bridge from Greenland to Iceland than from Greenland to Spain. "We can't do that" versus "We can't do that either".
- ninkendo 2y ago
  Even without gravitational slingshots, the simple solution would be to just take advantage of the oberth effect and do many, many successive burns, retrograde at aphelion and prograde at perihelion, until your perihelion is low enough to intersect the sun. It wouldn’t be crazily less delta-V but it would be appreciably lower. Another related fun fact from GP’s is that it’s way less delta-V to hit the sun from Pluto than it is from the Earth… 4.63 km/s according to wolfram: https://www.wolframalpha.com/input?i=hohmann+exit+delta+v%2C+Sun&assumption=%7B%22FS%22%7D+-%3E+%7B%7B%22HohmannExitDeltaV%22%2C+%22deltav2%22%7D%2C+%7B%22HohmannExitDeltaV%22%2C+%22r2%22%7D%2C+%7B%22HohmannExitDeltaV%22%2C+%22r1%22%7D%7D&assumption=%7B%22F%22%2C+%22HohmannExitDeltaV%22%2C+%22r2%22%7D+-%3E%2240+au%22&assumption=%7B%22F%22%2C+%22HohmannExitDeltaV%22%2C+%22r1%22%7D+-%3E%22695000000m%22 https://www.wolframalpha.com/input?i=hohmann+exit+delta+v%2C...

Indeed, it's off by orders of magnitude (so probably a simple typo). The Earth orbits the Sun at between 29 and 30 km/s. It would take roughly that much dV to successfully create an orbit that intersects the sun.

tialaramex 2y ago

The straight forward way is typically given as 30km/s so yeah, it's the wrong unit. But, you'd never actually do that, the calculations involved assume you have decided to specifically go towards the sun, maybe to get there quickly, but going to the sun is just a means to an end, so you'd actually pull shenanigans to get the same result indirectly for "only" a few km/s of delta V with carefully chosen launch windows. Which is like how it'd be easier to build a bridge from Greenland to Iceland than from Greenland to Spain. "We can't do that" versus "We can't do that either".

ninkendo 2y ago
Even without gravitational slingshots, the simple solution would be to just take advantage of the oberth effect and do many, many successive burns, retrograde at aphelion and prograde at perihelion, until your perihelion is low enough to intersect the sun. It wouldn’t be crazily less delta-V but it would be appreciably lower. Another related fun fact from GP’s is that it’s way less delta-V to hit the sun from Pluto than it is from the Earth… 4.63 km/s according to wolfram: https://www.wolframalpha.com/input?i=hohmann+exit+delta+v%2C+Sun&assumption=%7B%22FS%22%7D+-%3E+%7B%7B%22HohmannExitDeltaV%22%2C+%22deltav2%22%7D%2C+%7B%22HohmannExitDeltaV%22%2C+%22r2%22%7D%2C+%7B%22HohmannExitDeltaV%22%2C+%22r1%22%7D%7D&assumption=%7B%22F%22%2C+%22HohmannExitDeltaV%22%2C+%22r2%22%7D+-%3E%2240+au%22&assumption=%7B%22F%22%2C+%22HohmannExitDeltaV%22%2C+%22r1%22%7D+-%3E%22695000000m%22 https://www.wolframalpha.com/input?i=hohmann+exit+delta+v%2C...