# “The Sun is the most difficult object in the solar system to send a probe to.”

I’m trying to learn about orbital mechanics, and it seems that a popular bit of trivia in this field is the statement in the subject line. It comes as a surprise to many, because you’d think that because of the Sun’s strong gravitational pull, it would actually be an easy object to send a probe to, and the most difficult one would be Pluto or one of the outer planets where you have to go against the Sun’s gravity. (This is only about the energy required to send a probe on the desired trajectory, disregarding things such as solar wind and the obvious heat from the Sun that might pose additional obstacles.)

I don’t doubt the result, but I’m trying to come up with an intuitively understandable explanation. My attempt so far is this: Precisely because of the Sun’s strong gravity, everything that orbits the Sun (including the Earth) must be really fast relative to it (otherwise it would fall into it). And any probe we launch from Earth inherits this speed relative to the Sun. So to launch a probe from Earth and send it to the Sun, we must burn a lot of propellant to slow it down sufficiently from this speed that it started with. It’s a bit counter-intuitive at first glance but sounds reasonable when one thinks about it.

Is this explanation accurate to explain the effect, or is there something else at play?

If this Delta-V map of the solar system (starting at Earth) is to be believed, then yes, the sun is really, really difficult to get to, like a couple of orders of magnitude.

I’m trying to find the video where he said it, but what you said is almost word for word how Neil DeGrasse Tyson explains it in his podcast “Startalk”

That’s pretty much accurate. The concept that physicists use to talk about this is angular momentum, which is to say (roughly) that objects going around the sun tend to keep going around the sun. To actually go towards the sun, you need to get rid of this angular momentum, and to do this you have to slow down enough so that you’re not really going “around” the sun any more.

Yes, that’s an accurate statement. You have to go a bit into the mathematical weeds to show why it’s actually MORE energy to do that than to go to points outwards, but that’s the gist of it.

Although, of course, there’s more to it. If you’re patient enough, you can get to the Sun by going outward very far, where your orbital speed will be very low, and then killing your orbital speed. Not to mention all of the clever complicated things you can do by interacting with all of the other solar system objects. If you’re really patient, then about the same amount of energy it takes to reach the Moon can get you anywhere in the Universe, through the right sequence of gravitational slingshots.

Unmanned space probes frequently use gravitational assists from various planets, including the Earth, to get to various places in the Solar System. For example, the Juice mission, currently on the way to study Jupiter’s satellites, will slingshot off the Moon, then Earth, then Venus, then two more Earth flybys on the way to Jupiter. Once it gets to Jupiter, it’ll use slingshots off the various moons to change its orbit as needed.

Other missions, e.g. Galileo and the Parker Solar probe, have used multiple gravitational assists as well.

This seems counterintuitive to me. Can you expand on it?

Offhand, for one thing, it seems like it should take as much or more energy to go outward from the sun to very distant orbit as it would take to slow down nearer to the Sun.

What?

Seven responses in and not one joke about sending probes to Uranus?

I’m very disappointed, people!

Well, it is in FQ, and the factual discussion is still ongoing.

At this early stage, such a joke is IMHO more likely to draw a modnote than a chuckle.

As others have said, your intuitive explanation here is exactly right. An easy way to intuitively see the energy cost of sending a spacecraft into the sun or into a very close orbit around it is to compare it with sending a spacecraft to Mars.

Earth has an orbital velocity of just under 67,000 mph. Mars orbits at just under 54,000 mph, but it’s in a higher energy state because it’s “higher” above the sun in terms of the sun’s gravitational well. But one can see that the two orbits have similar kinetic and potential energies (per unit of mass) which is very very different from an orbit very close to the sun or falling into it, where all or most of the 67,000 mph has to be bled off. I suspect that the vast majority of the energy required to get a spacecraft to Mars is consumed by getting it off the surface of the earth and then accelerating it to escape velocity. After that it only needs enough propulsive thrust to send it into a higher orbit.

At any given distance from the Sun, there’s a speed you need to be traveling at to be in a circular orbit. This is basically what the Earth (and everything on it) is doing (our orbit isn’t quite a perfect circle, but close enough for this discussion). Call this speed the circular orbit speed. Circular orbit speed is very fast, very close to the Sun, and slower the further away you are.

You can also be at that distance and going a different speed. You’ll still be in orbit, but it’ll be a different orbit. If you’re going faster than circular orbit speed, then your orbit will be an ellipse that goes out further away from the Sun, and if you’re going slower, then your orbit will be an ellipse that goes in closer.

Also at any given distance from the Sun, there’s a speed you need to be traveling at to go away from the Sun forever and never return. You’ll continually slow down, but never quite stop, and eventually get as far away as you’d like. This is called escape speed. Escape speed also depends on distance from the Sun, and in fact, escape speed is equal to \sqrt{2} times circular orbit speed (about 1.41 times).

So, you’ve got the Earth, or something else that’s moving along with the Earth. To fall into the Sun, you need to lose all (or very nearly so) of your orbital speed, 1.0 times the circular orbital speed. But to get very far away, you only need to add on 0.41 times circular orbital speed. It’s less than half of the change in velocity needed to fall in.

And then, once you’re very, very far away, you get to a place where circular orbital speed is very, very low. And your speed is actually even less than that, because you’re not on a circular orbit, you’re on an elliptical orbit that extends inwards from that distance. So since your speed is very, very low, it now only takes a little bit of change in velocity to kill all of your speed, so you fall straight in.

Nor has anyone mentioned how to land a probe on the sun without it burning up.

Also, Uranus jokes are quite passé now that the new name of the planet has been announced.

A small change in velocity can be used to make the minor axis of your orbit smaller. Make it small enough and you will hit the sun off-center. Make the minor axis equal to zero and you fall straight in.

Land at night, of course.

This is not a perfect analogy, but if you were driving in a convertible with the top down at 60 mph, and you had to jump into another convertible. Would you rather jump into the one that’s moving down the road the same way you are at 50 mph, or the one that’s stopped? The biggest part of getting between two places in the solar system is matching velocities, and of course that’s going to be easier with anything moving the same way as you (all planets are on the same plane moving in the same direction) than something that’s essentially stopped.

I don’t dispute your figure of escape velocity at any given orbit being about 1.41 times the orbital velocity. But your conclusion seems wrong for the following reason.

Suppose you expend an enormous amount of energy so that a spacecraft orbiting the sun in the earth’s orbit is slowed from 67,000 mph to zero, so it falls straight into the sun. Then suppose you execute your plan to presumably expend less fuel by climbing to a far outer orbit and slowing to zero there, and falling into the sun from there.

The spacecraft falling into the sun from the distance of earth’s orbit will reach the sun with a considerably lesser velocity than the one falling from a very high distant orbit. IOW, with your distant-orbit plan, you’ve expended less energy, and yet somehow ended up with a spacecraft arriving at the sun with much more energy.

I think the flaw in this reasoning, unless I’m totally missing something, is that you’ve calculated the energy it takes to achieve escape velocity but not the large amount of energy it takes to reach higher, distant orbits. Those orbits represent proportionately higher potential energy and that energy has to come from somewhere.

With a rocket, most of the energy you spend ends up wasted in hot exhaust gas. The go-high-before-going-low plan just wastes less.

I don’t know that I have a lot to add; the o.p., despite his reservations, has the essentially correct understanding; everything in orbit around the Sun (and for that matter around any body) is moving so fast that it can’t fall into the massive body. The same is true for satellites and spacecraft in orbit of the Earth, which is why space launch vehicles turn sideways once they get above the thickest part of the atmosphere; they are developing enough angular momentum such that they can’t fall back to the surface but rather continue to fall across the horizon.

One concept that may help intuitively (and was alluded to by @wolfpup) is to consider the specific orbital energy (also often referred to as the vis-viva energy and associated equation which relates velocity, planar orbital elements, and the gravitational parameters). Using this will show that in order to cause a spacecraft to fall into the Sun, this quality has to be reduced to almost nothing, which is (seemingly) paradoxically challenging because the Sun is so massive. Without going through an entire derivation I’ll just note that studying the vis-viva equation gives a lot of insight as to why the specific relative angular momentum, h, is key to why it is so difficult to send a spacecraft into the Sun, every object in orbit, or caught in passing, will have some significant amount of h.

Prussing addresses this concept in the first chapter (Section 1.7 in the linked first edition); I’m sure Danby or any other text on orbital or celestial mechanics will also cover this in the introductory material. In doing practical orbital mechanics calculations, Lagrangian ‘energy methods’ are often used to find optimal solutions or to perform perturbation studies without explicitly calculating velocity or momentum at specific points except as the result of applying the optimized parameters to the equations of motion. (Lagrangian mechanics and their bigger brother, Hamiltonian mechanics are generally very useful in classical mechanics to reduce the complexity of problems that do not easily resolve in a Cartesian coordinate system, and as such are incredibly useful in celestial mechanics where nothing moves at right angles or along ‘straight’ lines.)

Stranger

What is being proposed is essentially increasing the eccentricity of the orbit to e=1, which means it just falls straight into the Sun. This is still reducing h to essentially zero but doing so by requiring very little impulse because the momentum of the spacecraft is at a minimum at aphelion (the farthest point in the orbit from the Sun).

The energy state required to achieve escape is where the characteristic energy, C3>=0, and there are no possible orbits with higher energy; any object with a C3>0 will be in a parabolic or hyperbolic trajectory.

Stranger

That still leaves the Sun as the most difficult object to send a probe too, since to get to any other orbit outside Earth will take less energy than required to get to the outer reaches of the solar system and then kill tangential velocity, and to get to any orbit inside Earth’s orbit, you can pull the same trick you describe, but change your tangential velocity a little less than required to reach the Sun (right?)