Another challenge, as **Stranger **pointed out in post #25, is that your aim at apsis better be very, very, very good.
If you don’t quite zero the tangential velocity, net of all the perturbations you’ll experience on the way back “down” to the Sun, you’ll miss it. And end up in a very long period narrow elliptical orbit. So a few millennia later when you next hit apsis you can try again if you still have fuel and batteries and …
You sure? Vis-viva equation is:
v[sup]2[/sup] = u(2/r - 1/a)
And orbital speed in a circular orbit is:
v[sup]2[/sup] = u/r
Where:
u = GM
a = semi-major axis
If r is the starting radius and R is the periapsis, then a = (r+R)/2
We’re trying to minimize:
sqrt(u(2/r - 1/a)) - sqrt(u/r) + sqrt(u(2/R - 1/a))
Change units so u=1 and r=1
sqrt(2/1 - 1/a) - sqrt(1/1) + sqrt(2/R - 1/a)
Remove the constant term and substitute a:
sqrt(2 - 2/(1+R)) + sqrt(2/R - 2/(1+R))
Remove the factor of sqrt(2) and simplify:
sqrt(R/(1+R)) + sqrt(1/(R(1+R)))
As R goes to infinity, the first term goes to 1 and the second to 0. I’m too lazy to prove it via calculus, but it seems sorta obvious that it’s also the minimum (and Wolfram Alpha agrees).
I believe one mission was already done.
I don’t have a link to the mission but Gary Larson documented it. There was an alleged theft of sunglasses involved.
To finish the proof–we’re still trying to minimize:
sqrt(R/(1+R)) + sqrt(1/(R(1+R)))
for positive R. Substitute Q=1/R:
sqrt(1/(Q(1+1/Q))) + sqrt(Q/(1+1/Q))
Simplify:
sqrt(1/(Q+1)) + sqrt(Q[sup]2[/sup]/(Q+1)) =
(sqrt(1) + sqrt(Q[sup]2[/sup])) / sqrt(Q+1) =
(1+Q) / sqrt(Q+1) =
sqrt(Q+1)
Take the derivative to get:
0.5 / sqrt(Q+1)
Ignoring imaginary numbers, the function is defined for Q > -1, and always positive. Therefore the original function is monotonically increasing, and Q=0 must be a minimum for non-negative Q. So R is (waves hands) infinity (yadda yadda limits and stuff).
Let’s hope I didn’t make some crucial error earlier…
Or I could have just simplified it properly in the first place instead of doing a weird substitution (bleh):
sqrt(R/(1+R)) + sqrt(1/(R(1+R))) =
(sqrt® + sqrt(1/R)) / sqrt(1+R) =
(sqrt(R[sup]2[/sup]) + sqrt(1)) / (sqrt(1+R) * sqrt®) =
(R+1) / (sqrt(1+R) * sqrt®) =
sqrt(1+R) / sqrt® =
sqrt((1+R) / R) =
sqrt(1 + 1/R)
Obviously this reaches a minimum as R goes to infinity.
Can someone explain, as if to a child, why something launched in the Sun’s direction from Earth with a force powerful enough to be free of Earth’s gravity would not just get sucked into the Sun by it’s infinitely more powerful gravitational forces? Is it something along the lines of the momentum or force needed to “break through” the Sun’s gravity so the object doesn’t get “stopped” by gravity and forced into orbit around the Sun instead of just “hitting” our mediocre star?
That was explained earlier, but to restate: Anything you launch directly at the Sun from the Earth is also going sideways at 66,000 mph (the Earth’s orbital velocity) and it keeps doing that after leaving the Earth - much like if you pitch something sideways out of your car window, it leaves the car going as fast forward as the car was when you threw it. There’s a small but crucial difference: Objects you throw out of your car will be slowed by friction with the air and ground. Objects you throw into space will not.
Just leaving the Earth at its escape velocity doesn’t put you on a straight-line path to the Sun. It puts you into another orbit around the Sun more or less identical to the Earth’s.
Thank you Your Great Darsh Face
I have been trying to grok why it would take so much energy to fire something at the sun and what you said finally clicked. I read the other posts and I was still not getting it but now finally this duncecap(me) gets it. thanks
It is amazing to me how resistant the mind is to grasping certain things. Everyone knows what gravity is. Everyone knows(!*!$?) that the gravity of the sun, because its mass is so great that it will suck anything into it that gets close. It is beyond counter-intuitive that you can’t simply fire at the sun and hit it.
Perhaps another analogy might make this stuff more clore to more people:
Sure Kathmandu is only X miles from Mount Everest, but given the terrain it might as well be on the far side of the world in terms of what it takes to get there.
So if I am now grasping this correctly one simply needs to picture the gravity wells which form space in which things orbit around something as troughs and that in order to get from point A to point B in a straight line one must traverse not only the distance between the troughs but also the troughs themselves and how much energy is required to do so is really determined by the scale of the trough to be traversed.
Once one has picture of the terrain this stuff is simple to grasp. But if you just tell someone that you need to decelerate by 29km/s from escape velocity to hit the sun it makes absolutely no sense at all and runs counter to everything one knows from lay physics. For you guys/gals who already have a clear picture in your head this is just duh simple. But I imagine most people have a two dimensional grasp of space, not three dimensional. Think of it this way: as earthlings we always have the ground underneath us, the three dimensionality of the world is in essence, if you will, superimposed on a two dimensional structure (we always have a clear distinction between up(top) and down(bottom), and even though we have left vs right and front vs back each of these 3 orientations is defined in terms of two). However in space there is no top or bottom, they are meaningless. My picture of troughs still doesn’t capture it because there is nothing in earthly experience to draw a comparison, there is literally nothing “on” the earth that is surrounded in all three dimensions by the same something that can be seen by the naked eye ie. privy to direct human experience without the aid of measuring devices(sure if we are talking at the atomic or sub atomic levels sure, but humans don’t experience atoms qua atoms.)
So just to sum this up again and show how dumb I am: at least within a solar system where there is a star around which bodies orbit, where each of those bodies potentially has other bodies orbiting them, there is no such thing as a straight line, except for things which have no mass( like light/photons). So for all things which have mass a straight line in space is an orbit and to traverse from one orbit to another orbit means changing ones speed to match the the orbit of the place you want to go. I am sure that there is much I am still not grokking, but if I am even 1% right, anyplace other than on a planet (ie. space) is truly alien, as in utterly incomprehensibly other/different from anything in the normal panoply of human expeience as earthlings. I am still left wondering if everything is space really is that way or all of this is a function of Newton having defined everything in terms of motion(just reflecting on how we would have no ability to find anything in space not in motion because every measuring device with which we study universe is based on detecting change/motion-or is that what a black hole is -a region of space without measurable movement…hmmm sorry more questions just keep popping up.)
Almost everything you’ve just said is utter nonsense. And irrelevant to throwing things from the Earth into the Sun.
The whole issue with throwing things into the Sun being unintuitive is that most folks sitting still in a chair don’t realize they’re actually going 66,000mph in the wrong direction to throw things into the Sun. They think they’re sitting still because their chair isn’t moving in their room.
If they were sitting still in space in their isolated chair, then throwing something into the Sun would be (almost) as easy as letting go of it and waiting awhile.
Understanding how motions look different from different points of view is all that’s needed. But that’s hard for many people; and very, very hard, bordering on impossible for some people.
This has almost zero to do with Einstein’s “relativity”, and truly zero to do with two versus three dimensional space.
Of course, if you were sitting still in a chair in space, the Sun’s gravity would pull you in at almost the same speed as whatever object you just chucked into the sun.
Here’s another way to put it. Almost all objects that could fall into the Sun have already fallen into the Sun billions of years ago. Almost all objects that could escape the Sun’s gravity and be ejected into interstellar space have already done so billions of years ago. The only objects left in the solar system are objects that are in stable orbits around the Sun.
Now you have to understand what an orbit is. The problem is that most humans spend most of their lives on or near the surface of planet Earth. If you throw an object near the surface of Earth, what happens? It starts to slow down due to friction from the air and eventually smacks into the Earth. Which leads people to believe that the natural state of things is for things to fall to the surface of the Earth.
But remember, it’s easy to throw things into the Earth if you’re already on Earth. In fact, it takes no energy at all. And likewise, it would be really easy to fire something into the Sun if you were already on the Sun.
Or, if you were just at some height above the Sun–like on a really really tall tower–it would be really easy to fire something into the Sun, you’d just have to drop it. The problem is that there is no such really tall tower on the Sun. Instead, we are on the Earth. And the Earth is in orbit around the Sun, otherwise it would already have crashed into the Sun, or flew off into interstellar space.
That means any objects we try to throw into the Sun are in orbit around the Sun. In order to make the object fall straight into the Sun you’d have to stop it from orbiting the Sun. As has been said, the Earth is orbiting around the Sun at 30 km/s, so to stop orbiting you’d have to throw it at 30 km/s. That’s a lot faster than a speeding bullet, which is usually less than 1 km/s.
Would it be easier to fire an object at a black hole, or the same?
It would depend on whether you were at rest with respect to the black hole, or in orbit around it. If you wanted to throw something into the black hole in the center of the Milky Way, that would be quite difficult- the Sun is orbiting it at 230 km/sec. If you’re talking about throwing something into a stellar-mass black hole closer to us, it would depend on your velocity relative to said black hole. If the Sun became a black hole but nothing about Earth’s orbit around it changed, throwing something into the new black hole would be the same as throwing something into the Sun.
It would be even harder – With the Sun you can miss by the radius of the Sun (probably even farther due to drag from the stellar atmosphere) and still get a “hit”. If the black hole was the same mass as the Sun, it would be a much smaller target.
If the black hole was somehow made the same size as the Sun, it would be a lot more massive, and Earth would have a lot more orbital velocity that the projectile would need to bleed off.
For a given mass, it’d be harder to throw something into a black hole than into a star. The energy requirements would be about the same (slightly higher for the BH), but your aim would need to be a lot better. If you calculate the exact rocket boost you need to hit the center of the Sun, but you’re a bit off, you might still hit the edge of the Sun, and friction will take it from there. With a black hole, though, if you’re off by even a few kilometers, you’ll just swoop wight past.
I just knew that black holes were portals to the plane of negative energy!
OK, now I’m glad that I didn’t bother to edit my post when I saw that right after submitting. Well played.
I think the question has been answered, so I’ll just tell the OP that the best way I know of to visualize this is to play Kerbal Space Program. 
You can play in sandbox mode and design and fly a ship. You’ll see it’s more difficult to fly to the inner planets than the outer ones, and it take quite a bit of delta V to get something to fly into the sun. It’s a pretty fun way to play around and learn about orbital mechanics that isn’t all mathy (well, ok…I admit it…I did have pages and pages of calculations and vectors when the game first came out and before mods like MechJeb made it so easy to do stuff ;)).
If you’re off by a few km from a black hole, you’re still likely to hit the accretion disc and go in. I don’t know how large accretion discs for black holes would typically be but you have a fair bit more leeway than just the event horizon.
I didn’t account for the accretion disk because they’re not part of the black hole. In principle, you can have a black hole without one, or an ordinary star with one. The only reason we associate accretion disks with black holes is that they’re really hard to detect without a disk.