New Horizons flyby of Pluto [updated for Ultima Thule]

  • Thread starter rootone
  • Start date
Very odd. Closest approach was on Wed 3rd Jan but their NH twitter feed is still pumping out the boring "snowman" image from 1st Jan.

What is going on here? Has the data link gone down or are NASA playing hide-and-seek with the publicly owned images again?
From: http://www.planetary.org/blogs/emily-lakdawalla/2019/mu69-baby-comet-contact-binary.html

No further images will come to Earth from New Horizons until January 10 because the spacecraft is behind the Sun, making radio communication error-prone.
 
187
13
Ah thanks. Since we are at perihelion now that means this MU69 object is almost exactly along the Earth's line of apsides. Coincidence, or did they chose that path out of the inner SS for a specific reason? It will be interesting to see more detail over the next few weeks.

At least I can stop going back every day hoping there an update. Shame they did not explain this on the NH page where they post the new images.

It's also a shame they only give up crappy jpegs instead of the option of a non lossy format which would be more use for playing around with image processing.

Thanks for the info.
 
Last edited:

OmCheeto

Gold Member
2,023
2,366
Well, I found a gravity bug in my method, yielding somewhat different numbers.

2019.01.05.Ultima.Thule.update.png


At the far end of Thule, the required centripetal force increases to 0.00025 m/s2 and gravity decreases to 0.0007m/s2
My gravity number is 4 times larger than yours at the Thule end.
Even discounting the gravity from Ultima, my value is still 3 times higher than yours.
g = 1.5e15 kg * 6.674e-11 / (7000 m)^2 = 0.002043 m/s^2

0.002043 / 0.0007 = 2.9


What am I doing wrong?
 

Attachments

32,690
8,556
Ah thanks. Since we are at perihelion now that means this MU69 object is almost exactly along the Earth's line of apsides. Coincidence, or did they chose that path out of the inner SS for a specific reason? It will be interesting to see more detail over the next few weeks.
Coincidence. The main trajectory was based on the Pluto approach, afterwards they searched for something that can be reached with minimal fuel consumption.

The trajectory at Pluto was chosen to get a gravity assist at Jupiter, to get a good view of the day side with a close fly-by, a good view of Charon, a pass through Pluto's shadow (to measure the atmosphere), to have Pluto between probe and Earth (for radio measurements) and to have Charon behind Pluto to have some light on its night side. That a trajectory exists with all these features is amazing already.
 
187
13
Well, I found a gravity bug in my method, yielding somewhat different numbers.

My gravity number is 4 times larger than yours at the Thule end.
Even discounting the gravity from Ultima, my value is still 3 times higher than yours.
g = 1.5e15 kg * 6.674e-11 / (7000 m)^2 = 0.002043 m/s^2

0.002043 / 0.0007 = 2.9


What am I doing wrong?
You are using billiard ball physics but your "centre" of mass seems to be at the point they touch.

If you want to simply to two point masses they are separated by 7+9.5km
If you want the effect of one mass at the outside to the other it's its radius plus the diam of the other.

BTW inverse square law can only be applied "far field" , ie. where the diameters are much smaller than the separation between them.
 
32,690
8,556
Inverse square law is exact everywhere outside of spherically symmetric objects (within Newtonian mechanics). Didn’t pay much attention to that discussion part so far but if there is still something unclear I can have a look.
 
187
13
"Inverse square law is exact everywhere outside of spherically symmetric objects"
It is good enough for a first approximation but it is certainly not exact.

Consider a test mass one radius from a spherical body. Work out the contributions form two points diametrically opposed on the surface, once is distant by r , the other by 3r. Compare that to the same masses placed a the centre of mass.

1/r^2+1/(3r)^2 = 2/(2r)^2 ??

A similar inequality and of the same sign will apply for all pairs of points symmetrically placed either side of the plane through the c.o.m. and perpendicular to the line joining the test mass and the centre. Summing all such pairs to represent the whole shows the inequality holds for the whole body.

The inaccuracy reduces as the separation becomes much larger than the diameter of the sphere.
 
Last edited:
32,690
8,556
It is exact. You can't limit the calculation to two points, you have to integrate over the whole shell. Do it and see what you get.
Alternatively: It is a direct consequence of Gauss' law.

This is a well-known result from classical mechanics and not the topic of this thread. If you have more questions about it please start a separate thread.
 

OmCheeto

Gold Member
2,023
2,366
And, we're back in business.

2019.01.08.DSN.NHPC.downloading.png


Not sure how long it will take to download the next image. I guess it depends on how big the file is.
 

Attachments

  • Like
Reactions: mfb

Borg

Science Advisor
Gold Member
1,820
2,032
Could you post the link to that page, Om?
 

OmCheeto

Gold Member
2,023
2,366
32,690
8,556
Canberra (43) is receiving data from New Horizons right now. 2 kb/s downlink at 8.44 GHz, 500 b/s uplink at 7.18 GHz.

Edit: Canberra tweeted it, too
 
Last edited:

Want to reply to this thread?

"New Horizons flyby of Pluto [updated for Ultima Thule]" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top