Targeting the Sun

I don't think it matters whether it is in Jan or July, because if we are trying for minimum amount of energy, not shortest trip, all we need to do is boost it out of earth's gravity field, which is independent of the distance from the sun. After that it is a coast to the sun, we just need to aim it in the general direction, and so that it will avoid the inner planets during the trip.

I'm not sure. At aphelion the speed of Earth is it's slowest around the sun. At perihelion the speed of Earth is it's fastest about the sun. On the launchpad the rocket will be moving the same speed as the Earth about the sun since it is in the same orbit as the Earth about the Sun. The goal is to increase the eccentricity of the Rockets orbit.

The rocket will burn fuel in order to create this higher eccentricity orbit. The kinetic energy the rocket creates to change orbit will be a larger percentage of it's total kinetic energy at aphelion than at perihelion since the tangential velocity will be smaller at aphelion than at Perihelion for the launched rocket. So less fuel would be needed at aphelion than at Perihelion to create the necessary course that takes the rocket into the sun. Less fuel makes the rocket easier to launch.

So I think aphelion is the better launch choice. I also think there are things I'm missing and overlooking, so have at it.

Uhmmm..... What he said....
I guess.
Personally I was going to ask how the equation for work changes when you get into something like space. If I remember right there is no such thing as work in space.

Don't worry Ben, I'm not sure what I said either. I'm hoping it's in the ballpark but I wouldn't tell Regis it's my final answer.

As for Work, it's Force times distance W=Fx, and exits in space as well as anywhere else. Also, the Work done on an object is equal to the change in kinetic energy of an object. Work is in units of Joules, just like energy. Think of it as Energy added to the system or taken away from the system.

I guess I'm getting hung out on the fact that when you hit the object with it's initial force it will continue, theoretically, along the force vector for an infinite distance, barring real world things like planets and asteroids. That would mean that the object would have a near infinite amount of work done as the distance increases.
And yes I know the space isn't empty and that there is outside forces and even friction that will make it eventually stop, I'm over simplifying.

Keep going guys, we may still get the rocket off the ground on schedule! What schedule?

Think of it this way. F=ma says that force produces an acceleration. In order for there to be a continuous acceleration, a continuous force must be applied. However, if a temporary force is applied, the acceleration lasts only for the duration that the force is applied, after that the object will travel with constant velocity (objects in motion tend to stay in motion).

You are only doing work while the object is being accelerated, during the application of force. Once it is coasting the only work that is being performed is from the effects of gravitational field incident on the vessel. These forces are very weak until say it gets near the sun at which point it will be accelerated more strongly and the rate of work is increased. Since gravitational fields are present, albeit weak, everywhere in space, work will always be performed on the ship, though at a very small rate except in the event that it nears a large mass.

The bulk of the fuel is expended to break the escape velocity of Earth, which is 1120 m/s.

I am not sure what the actual requirements are for this shot, but it appears that it must be a "direct" shot from Earth to the Sun with minimal energy. Some assumptions must be made as to what direct means. I will assume that the trajectory shall not use any other planets for gravity assist. I will also assume that the trajectory does not need to be in a straight linear line from Earth to Sun; in other words, can be hyperbolic.

The difference between apogee and perigee for the Earth is small. The eccentricity for the Earth's orbit is 0.017, which is only a 3% difference between the two.

Kepler's Second Law states that planets sweep out equal areas in equal time. That is, when closer to the Sun our orbital velocity is slightly faster than when at apogee.

I would assume that the direct path is one where the craft circles the Earth and its final boost phase imparts enough energy to just exceed the escape velocity of Earth. That would mean that the spacecraft's trajectory is no longer elliptical, but has become hyperbolic.

One method is to set the spacecraft's trajectory directly opposite the Earth's orbital trajectory and at the exact opposite velocity (essentially cancel out all velocity relative to the Sun) and let the spacecraft fall into the Sun. If you take that tact you would want to wait until the Earth is at apogee and its orbital velocity is at its lowest speed (about 29.4 km/s). That would be the lower energy.

However, I don't think that would be the least energy required. A better tact would be to achieve Earth escape velocity with a hyperbolic trajectory that vectors it toward the Sun. That resultant vector would be a combination of a vector that is opposite to the Earth's orbital trajectory and a vector that is toward the Sun. Whether the Earth is at apogee or perigee (or someplace in-between) should not matter since the energy required is really that which is required to break the escape velocity of the Earth and the resulting trajectory places the spacecraft in a decaying orbit about the Sun. However, the relative velocity of the spacecraft to the Sun must be lower than the Earth's orbital velocity, so a trajectory that has a vector component that is negative to the Earth's orbital trajectory would do that.

I am not an expert on decaying orbits, but there a number of them on the Starship Enterprise. If memory serves, about every third episode had either the Enterprise or one of its shuttles stuck in a decaying orbit and they had less than one hour to fix it. Surely, they would know best how to get into one.

So far, not even Scotty had it quite right! He thinks decaying orbits - only useful if you can venture through part of the atmosphere of the Sun repeatedly. Our probe won't like that, hence Scotty is fired. Roger was closest with aphelion (i.e., July), but his orbital design is not spot-on.
But 'Hero', that "... exact opposite velocity (essentially cancel out all velocity relative to the Sun) and let the spacecraft fall into the Sun" of yours sounds promising! Can you or somebody else motivate?

PS. Orbits around the Sun of whatever shape does not quite meet the "direct" requirement. Yes, the requirement was vague (story of the development engineer's life), but 8/8 hindsight will show, it's irrelevant anyway.

Hmmm, Okay, a little reading leads me to a few more tid-bits of knowledge. First, the spacecraft's trajectory must be parabolic (e=1) for a minimum energy burn to escape the the Earth's gravity field.

The next question is how to vector our craft so that it does not become a satellite of the Sun. I am still a fan of launching at apogee and vectoring the craft so that it is retrograde to the Earth's orbit as it reaches Vesc.

I think apogee is better because the orbital velocity of the Earth is at its slowest. If we want a direct trajectory into the Sun we need to negate the spacecraft's orbital speed relative to the Sun, which is imparted by the Earth since we are lifting off on that planet.

So minimal Vburn should be when the Earth is at it's slowest orbital speed. I agree with Roger, that more fuel will be expended after the craft exceeds escape velocity because the orbital velocity that the Earth has is far greater than that of Vesc for the Earth. That being said, you still want to start out on the right foot and launch easterly at as close to the Equator to get the advantage of the Earths rotation. Then when you have reached Earth orbit you cab burn to reach Vesc with a parabolic trajectory that is predominately retrograde to that of the Earth's orbit. That burn will continue until almost all of the angular velocity component relative to the Sun is depleted.

My other thought is that the spacecraft trajectory should be launched both retrograde to the Earth's orbit and away from the Sun! My logic is that the higher the orbit, the lower the orbital velocity. So by kicking the craft into a slightly higher orbit about the Sun you can reduce the angular velocity even more and your Vburn should be less.

The trajectory would be parabolic referenced to the Sun with the craft first obtaining a higher orbit relative to the Sun in a retrograde direction relative to the Earth's orbit.

Am I getting closer?

I think you are close enough! I believe the plan for minimum energy is as follows: put the craft in the optimum elliptical 'escape transfer orbit' around Earth (in July, at Earth's aphelion, of course). At the apogee of that Earth orbit, boost it into a direction that will directly oppose Earth's orbit around the Sun. Shape this orbit until the angular momentum relative to the Sun approaches zero and leave it to fall into the Sun.

Any shooting it outwards (away from the Sun) will eat more energy... that is unless one brings in a massive planet, like Jupiter. That will change the character of the project completely (no longer a 'direct' to the Sun type of problem!)

The 'drop it directly into the Sun' scenario has its own set of problems – there is no rocket available that can drop even its empty casing into the Sun like that! It will require a velocity change of around 32 km/s, which is just not practically available. So, back to the drawing boards!

I will go with the July group. Since they are shooting a probe directly into the sun, the more directly overhead the sun is the less atmosphere you would have to go through. Although, I am not sure you could even shoot a probe directly into the sun. By the time it got there the sun would have moved.

You are right, the July group wins - well, sort of! Read reply to 'Hero' above, with the conclusion: 'back to the drawing boards'.

Nice puzzle! Thanks for sharing!

It is a nice puzzle. But are there Photographic or other scientific measures which would be available during the longer distance trip, ie more rotation of the sun, etc. After all, this is being done to investigate the sun - not escape velocities of earth, right?

Right! There must obviously be a scientific payload of some sorts. In practice, dropping the payload directly into the Sun is probably no good. An orbit like that of http://ulysses.jpl.nasa.gov/ is better for science.

I was thinking last night, would it not be better to do a lunar injection for a slingshot to gain additional velocity for our retrograde Earth orbit?

There is possibly a tiny bit to be gained by swinging it around the moon. Since, for economy, one have to burn almost all of your fuel while leaving Earth orbit, the speed past the moon would be so high that very little "gravity assist" is possible. The moon is too light for serious gravity assist. Mars is also rarely used due to its lower mass. It is far more profitable to think Venus, Earth or Jupiter!

I discovered a mistake in my statement (reply #5347) that you need 32 km/s delta speed to drop into the Sun. It's actually around 41 km/s! I used a orthogonal vector addition, but I now think it must be a straight linear addition of Earth's orbital velocity (~30km/s) and the escape velocity from Earth.

Are you sure? If you break from Earth's orbit in the retrograde direction you already are traveling 11 Km/s. Wouldn't you only need to add another 20 Km/s to achieve the desired velocity.

Not quite! Escape velocity near Earth is 11 km/s, but it decreases as you climb out of Earth's field so that by the time you break from Earth's influence, there is no relative speed left. Given that you are still at the same distance from the Sun, you are orbiting the Sun at the same speed as Earth (30 km/s). So you need that extra initial 30 km/s retrograde as well! Or do you see a flaw in my argument?

Nope. I think you are right. I was just confused on the 41 km/s figure. Your velocity never gets that high. You only need a final velocity of 30 km/s, but you still need to burn fuel for the escape and another burn to scrub off the Earth's orbital velocity.

True, the velocity never gets quite that high, but it gets close! If you could impart a 41 km/s energy boost instantaneously, that would be the least energy scenario. Practically, the probe/rocket must be accelerated over time, but NASA does it as quickly as possible, to minimize so-called gravity drag.
Anyway, this is just playing around - no rocket is likely to ever exist that can impart such a "delta-V". NASA uses various planets for gravity assists or 'swingby' to add energy and course changes.