Kerbolar probes

[Temstar]

Ahh yes, the challenges of exploring Kerbol at close range, let's see one's design for such high delta-V probes.

Here's my, with about 20,000m/s of delta-V in two stages. For the first launch of this probe I'm sending it to a direct Kerbol collision course and I managed to get into a direct dive with the nuclear stage and the ion stage still full. The next one of these will be launched at the next Eve transfer window to use Eve for a powered slingshot for a low as possible Kerbolar orbit before circularising. The sun shield at the front provides protection to the rest of the probe.

[Tw1]

That's a very straight trajectory! Why did you chose that, rather than just a low periapsis?

[Temstar]

There is a reason for that "straight down" trajectory, thanks to it I've made a discovery:

Here we see my probe as before, falling vertically at 431m/s straight down to the surface of Kerbol

Note what happens when I switch velocity to surface mode. The "surface" of Kerbol (it's corona?) is apparently moving relative to me underneath at a fantastic 197,430.2m/s. The prograde indicator shows westward motion as prograde, since I'm falling straight down that must mean the surface of Kerbol is rotation eastward (so all planets in the Kerbolar system are in prograde orbit) at 197,430.2m/s

Kerbol has an equatorial radius of 261,600,000m, this works out to be a circumference of 1,643,681,276.35m

If we work this out:

1643681276.35 / 197430.2 = 8325.38s

Thus, we've discovered that Kerbol rotates at 8325.36 seconds, or 2 hours, 18 minutes, 45.36 seconds per revolution. This is tremendous rate of rotation when you consider our sun has a sidereal rotation period of 25.05 days.

[automcdonough]

so do those panels act supercharged being so close, or do you run out of juice instantly with 3 ions?

[Temstar]

I'm still at roughly Kerbin orbit so the panels are not yet supercharged. But once you get very near to Kerbol they generate almost 10 times the charge they would normally generate around Kerbin.

[automcdonough]

dang. I'm gonna have to try a a lightweight solar sail and see how that goes

[Flixxbeatz]

Diving directly to the Sun. FOR SCIENCE!

[SunJumper]

There is a reason for that "straight down" trajectory, thanks to it I've made a discovery:

-Picsnip-

Here we see my probe as before, falling vertically at 431m/s straight down to the surface of Kerbol

-picsnip-

Note what happens when I switch velocity to surface mode. The "surface" of Kerbol (it's corona?) is apparently moving relative to me underneath at a fantastic 197,430.2m/s. The prograde indicator shows westward motion as prograde, since I'm falling straight down that must mean the surface of Kerbol is rotation eastward (so all planets in the Kerbolar system are in prograde orbit) at 197,430.2m/s

Kerbol has an equatorial radius of 261,600,000m, this works out to be a circumference of 1,643,681,276.35m

If we work this out:

1643681276.35 / 197430.2 = 8325.38s

Thus, we've discovered that Kerbol rotates at 8325.36 seconds, or 2 hours, 18 minutes, 45.36 seconds per revolution. This is tremendous rate of rotation when you consider our sun has a sidereal rotation period of 25.05 days.

Your physics is incorrect. Your surface velocity is 197km/s, but for a radius of 261e6 PLUS 13.312e9. I'm not sure how to calculate the correct period, though. I'll probably do it by guess and check.

Anyway, why does your impact probe have ion engines? Surely the nuclear stage is enough?

[Temstar]

Your physics is incorrect. Your surface velocity is 197km/s, but for a radius of 261e6 PLUS 13.312e9. I'm not sure how to calculate the correct period, though. I'll probably do it by guess and check.

Anyway, why does your impact probe have ion engines? Surely the nuclear stage is enough?

I don't think radius has anything to do with surface velocity. Can someone with a geostationary satellite confirm? If I understand correctly when you switch to surface mode on a geostationary satellite it should read somewhere close to 0m/s, hence geostationary.

If that's correct then my 197,430.2m/s surface velocity can be divided into only two components:

1. My orbital velocity of 430.8m/s going straight at Kerbol

2. Kerbol's sidereal rotational velocity, let's call it Xm/s

Since I'm going straight at Kerbol the equation is simply:

197432.2^2 = 430.8^2 + X^2

Since 197432.2^2 is so huge compared to 430.8^2, we can say X is pretty much 197,432.2m/s

As for the ion stage, the probe is a prototype for upcoming low kerbol orbit probe, I figure I'll just use it for impactor experiment rather than designing a new probe for that job. Using the prototype will allow me to get an idea on how much electricity I can get close to Kerbol.