106
Because the earth goes very fast around the sun.
If you want to get to the sun, you need to slow down almost completely so that your speed relative to the sun becomes almost zero.
If you don't slow down (almost) completely, your probe will miss the sun when you 'drop' it, so it will eventually come back and you'll end up in an elliptical orbit.
Kind of like ...
54
No, because Mars can't have eclipses. Strictly speaking, Mars has only transits. The difference is that Mars's moons are smaller than the Sun as viewed from Mars, thus they don't block out the entire sun. Eclipses are defined as only occurring if the entire sun is blocked, or at least the vast majority. Phobos blocks out only about 60% of the sun at most. ...
44
Wouldn't i inevitably spiral to sun surface even if i was faster than 0km/s ?
No. On reasonable timescales, an orbit will have a fixed distance of closest approach, called "periapsis." (These timescales shorten if you're close enough to what you're orbiting that an atmosphere can drag you down).
You don't really need to "drop in straight line" (which ...
35
We can look at what happened when this actually occurred.
The geomagnetic storm of March 1989 was caused by a Coronal Mass Ejection.
Here are just a few of the many effects on satellites.
One satellite lost 3 miles in altitude (not 30 km! don't believe that
legend).
Another began uncontrolled tumbling.
GOES 7 lost communications and imagery for a time.
...
33
Changing orbits requires delta-v. To reach the Sun, you need to subtract delta-v such that your velocity relative to the Sun is near zero, which allows you to "fall straight down" into the Sun - your required delta-v is nearly equal to your orbital speed. To escape the solar system, you need to add sufficient delta-v in order to reach escape ...
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Looks pretty darned quiet to me right now:
You can find that here, along with other measures of space weather.
By the way, cosmic rays and solar activity are two entirely different things. Cosmic rays originate from outside of our solar system. The flux of cosmic rays is relatively constant.
As for your question about stress, indeed, I just experienced ...
31
Addressing
is the sun's mass and other quantities known well enough for this to be absolutely accurate?
Well, the key to this is the vis-viva equation in your question. It's not actually important for us to know the mass of the Sun precisely, so long as we know the product $GM$ (another answer makes mention of this).
And that product is, of course, the ...
30
I spent a couple of years working in the Astrophysics and Space Physics Section of JPL. Working with the Space Physics folks taught me a lot about the solar wind and other space weather phenomena. Later on, working with Hank Garrett of JPL's Space Environments group taught me more, especially concerning effects on spacecraft.
I'll start out with a ...
29
No, not yet. The Parker Solar Probe became the closest ever artificial object to the sun on October 29th, 2018, surpassing Helios 2 which held the record since 1975 [1].
No other human-made object has been closer to the Sun. The probe will repeatedly touch the outer corona until mission end in 2025, with the closest approach being 3.83 million miles [2]. It ...
28
You can find the image on Flickr.
On August 31, 2012 a long filament of solar material that had been hovering in the sun's atmosphere, the corona, erupted out into space at 4:36 p.m. EDT. The coronal mass ejection, or CME, traveled at over 900 miles per second. The CME did not travel directly toward Earth, but did connect with Earth's magnetic ...
28
As noted in another question, the ISS faces some pretty hot temps. Remember, the Sun heats radiantly. When you're sitting in that much radiant heat, without an atmosphere to dissipate it, you're going to get hot really quick (emphasis mine)
Without thermal controls, the temperature of the orbiting Space Station's Sun-facing side would soar to 250 degrees F (...
24
I suspect you mean objects that haven't left the solar system (and what is the boundary?) and that aren't orbiting another planet.
Every object en route to another planet that has left the Earth's sphere of influence and has not yet entered the target planet's sphere of influence is orbiting the Sun. Whether the Pioneer 10, Pioneer 11, Voyager 1, and ...
24
The Vis-viva equation is
$$ v = \sqrt{ GM \left(\frac{2}{r} - \frac{1}{a} \right) }, $$
The $GM$ product for the Sun is 1.327E+20 m^3/s^2.
If 1 AU is 150E+09 meters, then when you are in a circular ($r=a$) Heliocentric orbit at Earth escape/capture point your velocity is 29.7 km/s.
If you then change to an ellipse with aphelion still at 1 AU and ...
24
It doesn't really work that way. We can use the Sun to change direction, but we need rocket thrust to increase speed with the msneuver.
To begin with, the closest stars (apart from the Sun) are not close. If we were somehow to reach escape velocity from the Solar System (which this method won't do, see below), we would still be moving at only a small ...
23
While the solar corona is very hot, it also has very low density: Wikipedia gives a ballpark figure of about 1015 particles per cubic meter, which, at 1 million Kelvins, translates to a pressure of about 0.01 Pa. That's a pretty good vacuum, comparable to that in low Earth orbit.
The low pressure means that the coronal plasma doesn't hold much heat that it ...
23
The sun is not "burning" in the sense you are used to: there is no chemical reaction going on. Instead, there is a very high pressure in the core of a star (like our sun) due to the high mass that starts/sustains a nuclear fusion process. In our sun, hydrogen is fused to helium and the energy that's released in this process is what makes the sun "glow" and ...
23
The "gravitational" (slingshot) maneuvers space probes are performing are actually not so much about gravity. The gravity is method to "tie" temporarily these two bodies, but you could (purely hypothetically of course) use something else, some superstrong tether or so ... "Slingshot maneuver" is in fact much better name in this regard.
What actually happens ...
22
The closer an object to the source of light, the larger the shadow it casts.
That's true if we're talking about a point source or at least a compact source of light and "shadow" refers to the "umbra" or area of complete shadowing. But it no longer makes sense in this case where seen from Earth the obscurer (spacecraft) is tiny compared ...
20
Escaping the solar system requires adding orbital velocity to the spacecraft. Similarly, getting closer in the solar system requires removing orbital velocity. It turns out Earth is more out of the Sun's gravity well than it's in it.
In other words, the simple answer is that Mercury is "farther away" in terms of the change of velocity that's ...
18
In addition to the definition given by David H, I would add Ulysses, which is specifically observing the sun from out of the plane of the ecliptic and might mean what the questioner is asking for.
Ulysses made two passes around the sun, one over each pole, and its mission is officially over after the second pass. In order to get out of the plane of the ...
18
You need below 2866 m/s of orbital velocity at 1 AU to crash into the Sun.
You technically don't need to slow down exactly to 0 m/s relative to the Sun in order to crash into it. Let's calculate the approximate velocity required to graze the "surface" of the Sun. This is an excellent answer on how to calculate apoapsis and periapsis of an orbit.
So first, ...
16
Solar Probe, planned to launch in 2018, will get to within 8.5 solar radii of the Sun's surface. For comparison, Mercury gets no closer than 65 solar radii. Solar Probe will use a thick carbon-carbon, carbon foam shield when in close proximity.
The record so far is held by Helios-2 in 1976, which got just inside Mercury's orbit at 0.29 AU, about 61 solar ...
16
And note that if you want to hit the sun the cheaper (but slow!) way to do it is to head out. 12.32km/sec will take you to infinity, at infinity a burn of 0m/sec will kill your orbital velocity and you'll come straight in. Of course this will take infinite time, but even going only as far as Jupiter's orbit means you use less energy to drop your periapsis ...
15
Spacecraft Overview: "preliminary designs include an 8-foot-diameter, 4.5-inch-thick, carbon-carbon carbon foam solar shield atop the spacecraft body..., radiators for the solar array cooling system, ... actively cooled solar arrays". Low albedo for the heat shield isn't mentioned explicitely.
A highly elliptical orbit leads to relatively short periods of ...
14
If you're already in a solar orbit, then yes. You can use a sail at an angle and send the reflections prograde. The result is to reduce your orbital energy and you spiral in.
I recall it was a standard physics problem to find the angle that maximized the energy transfer (it's not 45 degrees).
Time of journey depends on the mass of your item and the ...
13
Getting hard numbers about how accurate measures we can get from current systems, adapted to the Sun instead of far away stars is difficult, bordering to impossible. But we can get data about the relative difficulty of the solar system planets.
First off, we can do some cheating for Mercury and Venus, as they occasionally go in front of the Sun. Given your ...
answered Dec 13 '15 at 9:34
SE - stop firing the good guys
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Thermodynamics says there's no such thing as a completely internal cooling system. The best you can do is pump heat to a cooler location. That's very difficult to come by in the photosphere. Besides trying to keep cool enough for some alloy to remain solid, keeping electronics or a power source cool enough to function would be significantly more difficult....
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I'm posting these images as a supplement @Hobbes's accepted answer and @TildalWave's comments (which includes links to these images). I started reading some of those links. The gallery is a good starting place but there are different tabs to check out.
The values 171 and 304 represent the central wavelengths used, in Angstrom units. Our visible spectrum is ...
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Insulation can function in both ways, keeping heat on whichever side is desired. In space limiting the amount of thermal input from the sun is very valuable since that heat is easily acquired but hard to get rid of.
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Based on the calculations presented by @uhoh I generated a plot showing the necessary delta-V for
a fly-by mission, i.e. entering into a Hohmann transfer with a far point intersecting the orbit of a planet
to get into a circular orbit at the same radius as a planet
Note that this does not include any methods to save fuel (aero-braking, swing-by) and ...
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