3
$\begingroup$

I know that Earth is always falling towards the Sun, but due to its sideways momentum it just misses always. But how did it all start? To go into detail, when the gas cloud with zero angular momentum collided, our solar system was made with planets and sun having their own angular momentum which makes them rotate. But how did the revolution start in the beginning? Were they placed in such a position and distance that they started attracting each other but how ? .They were having only rotation . But a revolution started which forced them to follow a elleptical path so how this happened in the beginning? Did some external force act on them and revolution started? How was this revolution started?

i have assumed dust cloud was having zero angular momentum but it can have any value that does not affect the question i think ?

| improve this question | |
$\endgroup$
  • 7
    $\begingroup$ Why do you assume that the cloud that collapsed had no angular momentum? $\endgroup$ – HDE 226868♦ Jul 23 '18 at 13:44
  • $\begingroup$ @JanDoggen Question is clearly different as i am not asking about rotation but about revolution as why sideways momentum started ? how these tangential force started to apply on planets its a complete different question i think $\endgroup$ – Amey Jul 23 '18 at 14:00
  • $\begingroup$ The gas cloud will have initial angular momentum from the rotation of the Galaxy. This is conserved as the cloud collapses to form the star and planets $\endgroup$ – astrosnapper Jul 23 '18 at 14:06
  • $\begingroup$ It's not a duplicate of that question, @JanDoggen. The referenced question asks about a planet's rotation, with the questioner understanding why planets orbit in the same direction. This question is asking about orbits. $\endgroup$ – David Hammen Jul 23 '18 at 14:42
  • $\begingroup$ The sun is so much biggger than the earth because so much matter did fall into it. In fact all matter with too little momentum. - i have assumed dust cloud was having zero angular momentum in general nothing has zero angular momentum. $\endgroup$ – TaW Jul 23 '18 at 16:44
10
$\begingroup$

A particle's angular momentum around a point is the particle's mass times its distance from the point times its velocity around the point. So a small velocity at a large distance can still mean a lot of angular momentum.

The bit of a Giant Molecular Cloud which collapses to form one solar system is typically on the order of a light-year across or somewhat less. It normally possesses some angular momentum -- just from random motions if nothing else.

When it collapses to form a solar system, its size drops from maybe a light-year down to maybe 10-100 AUs. A light year is about 60,000 AUs (astronomical units), so it is collapsing to about 1/1000 the size. To conserve angular momentum, the circular velocity of the matter must increase by a factor of a thousand. And the fraction of the matter which will form the star collapses by another factor of a thousand or so.

Not only does this give you plenty of speed, but for the star to form it actually needs to shed angular momentum.

See the Wikipedia article on angular momentum for a lot of good details.

| improve this answer | |
$\endgroup$
  • $\begingroup$ The tidal angular momentum from the galaxy's gravity well ensures there's always some to start with. $\endgroup$ – Joshua Jul 23 '18 at 15:33
  • $\begingroup$ ... which would mean that we could just reframe the question "why does the galaxy's gravity well have an angular momentum"... ;) @Joshua $\endgroup$ – AnoE Jul 23 '18 at 15:36
  • $\begingroup$ @AnoE: Because the cloud is orbiting it rather than having fallen to the center. The required orbital velocity for the inside and the outside of the cloud aren't the same. $\endgroup$ – Joshua Jul 23 '18 at 15:38
2
$\begingroup$

Imagine the initial gas and dust cloud before it collapsed. It was enriched in heavy elements by supernovaes from other stars. As we could simplify a supernova as an "explosion", the shock waves of the novae alone will provide some momentum.

Additionally the sun did not form alone but is surrounded by other stars, which will also exert some force on the initial dust cloud.

The other fact to bear in mind is that even if the cloud collapsed towards it's center, the individual particles and gas molecules can and will have their own path through this cloud. The will collide with each other, be pushed around by radiation and gravity of other bodies.

If we set aside these forces and trace the path of one of this particles towards the center there are two things that could happen. It "hits" the center or it doesn't. If it doesn't hit the center it will continue it's path on a elliptical path around the center. As the gas cloud already has an initial momentum by the forced described above, all particles that do not follow this momentum are more likely to suffer a head-on collision, loose some or all of their momentum and therefore fall towards the center. The other particles will be more likely to follow their path. This will lead to a situation where all particles, that do not follow the overall momentum of the cloud, will fall towards the center and unify with our Proto-Sun. Therefore only the particles that followed the initial momentum of the cloud had the cance to form planets and asteroids.

| improve this answer | |
$\endgroup$
1
$\begingroup$

Ontological/statistical answer:

Assume a great many clouds of matter, each of which collapses. They can start out in any configuration (speed, rotation, momentum; being more homogenous or more disturbed etc.) whatsoever.

There are comparatively few special overall configurations: the ones where there is no noticable net rotational momentum, and no appreciable local clumps with their own momentum. These few configurations will lead to not forming a solar system with planets; if all turns out well (highly unlikely), everything will just smack together in the middle and create a very lonely star.

All other cases will lead to some form of momentum, which will start to spin faster as matter gets drawn into a smaller volume (compare dancers who tuck their arms in).

So: lots and lots of statistical probability for a random start configuration to end up with some momentum; very very small probability for that being not the case.

| improve this answer | |
$\endgroup$
  • $\begingroup$ "There is exactly one special overall configuration: the one where there is no net rotational momentum" Why? Surely there are a very large number of states - the peak of the probability distribution - where the net angular momentum is zero. I understand the point that any non-zero net AM of the initial cloud will tend to be magnified as the cloud collapses into a star + planets, but something has to impart that initial AM. A cloud of particles with randomly distributed velocity vectors, collapsing under gravity, will not end up spinning. $\endgroup$ – Tom Graham Jul 23 '18 at 16:09
  • $\begingroup$ Clumps of matter with their own momentum will end up as planets - but all orbiting in different directions and planes. Which statement is wrong? $\endgroup$ – Tom Graham Jul 23 '18 at 16:09
  • $\begingroup$ Just as a note, I've converted Tom's answer to a comment, as it was meant to be. I had to do some editing and rolling back and such to get it to work correctly and obey the character limit - apologies if anything is confusing. $\endgroup$ – HDE 226868♦ Jul 23 '18 at 17:42
  • 1
    $\begingroup$ @TomGraham the largest number of states perhaps but that doesn't mean much here: Consider flipping a million indistinguishable fair coins. While the most common single outcome is exactly 500,000 heads and 500,000 tails, this is still only accounts for 0.08% of outcomes. There's way more than a million things being collapsed into a star system, and there's more than two states for each. $\endgroup$ – Dan Uznanski Jul 23 '18 at 17:56
  • $\begingroup$ @TomGraham, all of this depends on how we define the dimensions of the "state" (it does not need to have the individual atoms listed up etc.); but that is rather a technicality in regards to my answer. I've changed it a bit, maybe it's better now. $\endgroup$ – AnoE Jul 24 '18 at 11:15

Not the answer you're looking for? Browse other questions tagged or ask your own question.