Flight to Pluto

To calculate this, the speed and the acceleration must be known.

Ahhh! I am using the New Horizons spacecraft as my example in the book. It was launched on January 19, 2006 directly into an Earth-and-solar-escape trajectory. It had an Earth-relative velocity of about 16.26 km/s or 58,536 km/h (10.10 mi/s or 36,373 mi/h) after its last engine shut down.

Does this help?

Hi brbooker, you wrote: "It was launched on January 19, 2006 directly into an Earth-and-solar-escape trajectory. It had an Earth-relative velocity of about 16.26 km/s ..."

It is interesting to note that this was actually a 'super-Earth-and-solar-escape speed'. It only requires some 13.8 km/s relative to Earth for a minimal 'Earth-and-solar-escape trajectory.'

I know you said that you do not go for the math, but for reference, it comes from V²_esc = v²_e(sun) + v²_e(earth) ≈ 42.12² + 11.2², giving V_esc ≈ 43.6 km/s. Subtract Earth's orbital speed around the sun (29.8 km/s) and we get the 13.8 km/s. Here v_e(sun) is the escape speed from the Sun at Earth's orbital distance and v_e(earth) is Earth's escape speed from the surface.

The reason for the high speed departure was to shorten the mission time to Pluto (and beyond). Then the Jupiter flyby added another bunch of speed, so New Horizons is presently traveling at super-duper-solar escape speed (for where it is).

-J

Just curious, does the Jupiter fly-by increase the final velocity or does the increase occur only on the approach to Jupiter? I presume the deceleration as it leaves Jupiter's gravitational field bleeds off the velocity gained on the approach.

I think it was a real gravity assist where a fraction of the orbital energy of Jupiter is transferred to the spacecraft.

The end result is that the spacecraft gained more velocity (almost 4 km/s) upon exit than it lost after the encounter.

Hi corelite.

The Jupiter flyby ('behind the planet') increased the total orbital energy of the probe (relative to the Sun) - it robbed a tiny amount of Jupiter's orbital energy. This is the main point of 'gravity assist' flybys.

A flyby 'in front of the target planet' can also be used to decrease the total orbital energy of a probe. This is useful if you want to go into an orbit around the planet, because it reduces the amount of fuel needed for the inevitable 'braking burn' of the rockets.

-J

It all seems like lining up a golf putt>wondering if it could be modelled that way to derive the math solution?

UGH

Here's a gross approximation, with a lot of assumptions:

Say the distance to be covered is 3,000,000,000 mi, and it was going to go in a straight line from Earth to the point where Pluto should be when it gets there.

It would be off-target by 3 x 10⁹ x tan(1°) ≈ 5 x 10⁷, or 50 million miles.

Pluto is an average of 5,913,520,000 km (39.5AU) from the Sun. Earth is an average of 149,600,000 km (1.0AU) from the sun. This would indicate a distance of over 5.7 billion kilometers or 38.5 AU between the two. However, Pluto's orbit is highly elliptical and varies from 4.4 to 7.4 billion kilometers (30 to 49 AU) from the Sun.

Now, using the 5.7 billion kilometers we get distance of miss equals 5.7 x tan(1 deg) miss and we get .099 Billion kilometers, or 99 million kilometers of miss.

We have a "But" here.... Seeing that Pluto is moving and the time to close the distance is "36,373 mi/h) after its last engine shut down" it would take 2,721.799 hours to complete the trek.

not knowing the direction of orbit, or the speed of orbit, the space ship could have a direct hit with Pluto or it could miss by an additional 99 Million, making the miss 198 million Kilometers.

Laby

"not knowing the direction of orbit, or the speed of orbit..."

Assuming the guys who launched the probe had some kinda idea (unless they were all fekkin idiots) I think the expected position of Pluto when the probe was/is due to arrive can be taken as read.

The problem is how long after the craft goes off the planned trajectory is the error detected to be off course. I suspect that a single error of one degree could be corrected by fuel on board a deep space probe. I would be surprised to find that a one degree error was the lower threshold limit of a trajectory error detection.

From the OP - "if there is no trajectory corrections made along the way"

Totally agree that getting any kind of idea of the real situation is an enormous computational task - without any corrections, even 0.001° error could leave the probe heading out to visit Alpha Centauri or setting the controls for the heart of the sun.

I was just trying to give a feel for the minimum possible error, which was about half the distance from Earth to the Sun.

"or setting the controls for the heart of the sun."

Obviously a reference to Pink Floyd there. :-)

Hi AH, you wrote: "The long and short of it is your question would required a lot of computing time on a a specialized computer program with a very knowledgeable computer operator to get your answer."

Yup, the problem is normally computing time, because the desired precision requires integration over very small time increments, especially during flybys. This requires a supercomputer to do in reasonable time. I understood that NASA still uses a variant of the weak field, low velocity approximation to general relativity that Einstein originally proposed.

-J

Nice links Annon - I think that answered it - as best it can be.

Also makes the point on the 'orbital distance' given there are only really two practical zones you can intersect the rascal and both are inside 'average'.

I'm kinda curious what effect the Kuiper belt may have and will there be time to compensate, given the inclination of Pluto's' plane is an extra 3D variable.

Pity I can't vote - GA to you and JD.

Kyzine

I dont know either but in those hostile enviornments an inch is as good as a mile

Don't dig yourself a hole you can't get out of...

Just say "when they set off, they aimed at Pluto, but when they got there, Pluto had moved and was nowhere to be seen!"

By a country mile... and yes, steer clear of any statements that need to rest on scientific calculations... unless you're writing science fiction. Even then, it's probably the norm. these days to speculate on an extension of current theories, far out, though, they may be.

You might have your explorers actually arrive at Pluto and find a structure that looks like a face -- eerily resembling Neil deGrase Tyson.

My apologies to Mr. Tyson; Neil deGrasse Tyson

No, it will be Elvis.

I think that it is too difficult to calculate and it's not possible to assume a linear trajectory and neither a constant velocity. Although it's science fiction, it should be believable. Perhaps you could write about another civilization, I mean not writing about human beings of this time (maybe of the future), so you could invent some kind of technology that is still not invented. good luck with your book.

Of equal significance to the precise alignment of the spacecraft is its precise speed at any moment. If it is going too fast or too slow, then it may miss Pluto however correct the launch alignment is. Pluto is not the fastest object in the night sky, though it is still a moving target. If the spacecraft is off time, Pluto will have moved relative to the spacecraft's trajectory, and the spacecraft will miss it.

The deceleration needed to achieve orbit around Pluto, assuming that is the desire rather than impact with it as impact is rather pointless, is going to be substantial, and a lot of calculation in Celestial Mechanics will be needed to determine the options for spacecraft delivery at Pluto, which will impact upon the mass of the spacecraft, its size, its speed, its launch time, etc., and to select the best one.

Further, real-time operation at such distances precludes the use of course control taking place on Earth. All guidance systems will need to be on-board while the spacecraft flies, and remain fully functional without human intervention during the journey.

Gravity assist, i.e. using various planets' gravitational fields to change the velocity of the spacecraft, is a primary factor in keeping down its weight. The more the "slingshot effect" can be used, the less fuel it will need for acceleration, deceleration and course correction.

This small question has opened up a very large design topic!