Post by: Just thinking on 25/09/2021 09:04:48
Post by: Bored chemist on 25/09/2021 11:26:38
There might be the odd factor of 2 or pi or something in there.
Orbiting is just falling, but missing.
Post by: Just thinking on 25/09/2021 11:36:39
I'm not sure but I think it would take 6 months (half a year) to fall and it would hit the sun at the current orbital velocity (about 67000 mph).I was thinking about 6 months too but not 100.000 Kph as the earth would be starting off from a standstill. But you may well be right it would be interesting to see the numbers crunched on this I wouldn't know where to start or even how for that matter.
Orbiting is just falling, but missing.
Post by: Just thinking on 25/09/2021 11:42:45
Post by: Bored chemist on 25/09/2021 12:32:38
earth would be starting off from a standstill.In a way, it always does.
Imagine looking at the Erath + Sun from a distance and from just above the orbital plane.
You would see the earth trace out a wide, thin ellipse.
It would move mainly left and right (from your point of view).
Each year, you would see it move to one side, slow down and stop, then start moving back again, only to slow down and sop once more.
Post by: Just thinking on 25/09/2021 12:49:32
Each year, you would see it move to one side, slow down and stop, then start moving back again, only to slow down and sop once more.But that is due to the viewers perspective the earth in reality is still moving only appears to stop as it comes towards or moves away.
Post by: Colin2B on 25/09/2021 12:54:45
The earth has great mass so I'm sure it will be slow getting started on its journey to the sun.🎶 “What’s mass got to do with it”🎶
Although it will be slower starting - that’s what happens with acceleration - both the moon and earth would accelerate towards the sun at the same rate if mass were the only consideration. Think Galileo.
There would be slight differences due to diameter.
You can work out the g due to the sun/earth using Newton’s laws
Post by: Just thinking on 25/09/2021 12:59:59
“What’s mass got to do with it”Yes you are spot on I forgot even a feather would move towards the sun at the same rate.
Post by: Bored chemist on 25/09/2021 13:50:46
And if you remove the orbit...?Each year, you would see it move to one side, slow down and stop, then start moving back again, only to slow down and sop once more.But that is due to the viewers perspective the earth in reality is still moving only appears to stop as it comes towards or moves away.
Post by: Halc on 25/09/2021 13:52:03
I'm not sure but I think it would take 6 months (half a year) to fallMaybe two months at best. 6 gets us all the way to the far side, and that's without any change in speed all the way.
Quote
and it would hit the sun at the current orbital velocity (about 67000 mph).615 km/sec in fact which is the escape velocity of sun minus a little for escape potential from Earth distance. This is about 2.2 million km/hr or 1,380,000 mph
If the earth started out where Jupiter is it would build up much more velocity just throwing that out there.Negligible difference. Maybe 616 km/sec
The earth has great mass so I'm sure it will be slow getting started on its journey to the sun.Usually the intuition is that big things fall faster than little things (feathers and such), but Galileo showed that the mass of the falling object has negligible effect on the time it takes to get to the much more massive thing attracting it.
Post by: Eternal Student on 25/09/2021 13:54:51
Hi.
Well, I'm tired but I made it about 91 days until the earth hit the sun (about one-quarter of a year). This is a simplification assuming the sun is just a point mass. Realistically, it has some radius and we want to know where you consider the earth to have actually hit the sun. It has a diffuse outer layer, so where do you decide the sun is solid enough (or hot enough) to say the earth has been destroyed? I'm also completely ignoring drag, the space around the sun should start becoming quite dense and offering quite a lot of drag. These are all fairly minor considerations because the heat from the sun is going to be the biggest problem.
The most interesting thing is how you could determine this. There are two reasonable methods:
1. Differential equations: We have,
from Newton acceleration = gravityThis d.e. is a bit tricky but can be solved.
2. A faster or slicker method is to realise the Kepler's thrid law will apply. Falling to the sun in a straight line is just an extremely degenerate elliptical orbit. Just consider that earth is allowed to have a small transverse velocity instead of being forced to have 0 transverse velocity, so it will just miss the sun and go in orbit like this:
ellipse.jpg (18.64 kB . 897x176 - viewed 3735 times)Anyway, the semi-major axis ≈ 1 A.U. ≈ the same as the semi-major axis in our current orbit. So the total Time period ≈ the same as our current orbit = 1 year.
The entire motion is extremely symmetric so we can split it up into quarters (it's symmetric about the centre in the x-axis and the y-axis), so the earth starts from the furthest point away from the sun (far left on our diagram) and comes closest to the sun after one-quarter of a time period (about 91 days). For the extremely degenerate orbit (the straight line into the sun rather than a narrow ellipse around it) the earth would have hit the sun then.
(Bored Chemist has written about viewing the orbit of the earth from a side profile..... it's actually very much like the idea presented above).
3. Other methods probably exist.
Anyway, other articles consider the journey toward the sun to be over in about 65 days but that seems to be because they reckon the tidal forces from suns gravity would rip the planet apart before we even reach the outer edges of the sun.
See: What Would Happen if the Earth Stopped In Its Orbit? https://www.wired.com/2014/12/empzeal-earthfall/
for a day-by-day diary of what could happen on earth as we approached the sun.
As for the velocity at the point of impact, I reckon it would be about ... well I can't be bothered to work it out. You can do this yourself Just Thinking:
The potential energy of the earth is
using Newton's gravity which will be more than good enough for us. Find G = universal grav. constant in Wikipedia. M = mass of the sun (also in Wiki).r = distance from the sun to the earth, this is going to vary as the earth moves closer to the sun.
Set r = average distance from sun to earth initially = 150 milion Km.
Then determine the pot. energy at somewhere we can call the edge of the sun ---> radius of the sun is about 696 thousand Km.
The general idea is that the difference in potential energy is converted to kinetic energy.
use kinetic energy of earth =
with m = mass of earth (find it in Wiki). and v = velocity of earth.I don't have a calculator and I usually get the things wrong anyway since I have fingers that don't do what they are told. Alancalverd is an engineer, he might enjoy doing the calculation, ask him/her if you can't do this yourself.
Best Wishes.
Late editing: Inserted diagram, fixed an error in a formula.
Post by: Just thinking on 25/09/2021 14:07:44
And if you remove the orbit...?Somehow I have missed your point I'm sure you are correct but it just isn't sinking in.
Post by: Just thinking on 25/09/2021 14:09:00
Alancalverd is an engineer, he might enjoy doing the calculation,I will try Alan I'm not up to the task.
Post by: Eternal Student on 25/09/2021 14:27:13
Well, if Halc's escape velocities are right (i.d.k. he probably found them in Wiki), then that would do the job for you.
Those escape velocities would have been determined using the potential energy and kinetic energy forumulae mentioned earlier.
What did he say?
575 km/sec in fact which is the escape velocity of sun minus escape velocity of solar system from Earth. This is about 2 million km/hr or 1,200,000 mphThere's a risk he was using a complicated solar system with all the masses of the planets, asteroids and whatever but let's just assume he was considering the sun as the only mass of interest. Never do a claculation yourself if someone else will do it for you.
Best Wishes.
Post by: Eternal Student on 25/09/2021 14:31:50
This is a hypothetical question and may have an interesting answer.It may have an interesting answer? You're not planning to try it are you? We can't encourage these kinds of experiments.
Best Wishes.
Post by: Just thinking on 25/09/2021 14:37:28
It may have an interesting answer? You're not planning to try it are you? We can't encourage these kinds of experiments.Don't tell anyone but I have located the earth's hand brake hold on I'm about to pull it.
Post by: Janus on 25/09/2021 21:34:42
If the earth stopped orbiting the sun. This is a hypothetical question and may have an interesting answer. Can anyone do the math and reveal the answer as to the arrival velocity and time period for the earth as it makes contact with the sun. The earth has great mass so I'm sure it will be slow getting started on its journey to the sun. I haven't calculated the speed or time for this event as I have no understanding of the equations involved. I hope someone can give the answer?The quick and dirty method of computing the fall time is to assume the Earth's orbit is an extremely eccentric ellipse with the Earth's present distance being at perihelion. This puts the Semi major axis at half the perihelion distance, and you can just calculate half the period of such an orbit. This works out to be about 64 days.
A more accurate answer which will give you the time to reach the Sun's surface can be found by using the equation at the bottom of this article:
https://en.wikipedia.org/wiki/Equations_for_a_falling_body
Just use the Sun's radius for x and the Earth's orbital radius for r.
Impact speed can be arrived at by taking the difference between the gravitational potential energy at the suns surface and that at Earth orbital distance, and solving for the velocity needed for the Earth to have a equal kinetic energy.
I get ~616.64 km/sec
The fall time from Jupiter distance is ~ 2 years.
Post by: Just thinking on 25/09/2021 21:44:45
This works out to be about 64 days.Thanks Janus. Have you taken into account that the earth is starting from a stand still it will take some time to get going?
Post by: Petrochemicals on 25/09/2021 21:51:17
Post by: Halc on 25/09/2021 22:00:09
That figure assumes that, yes, just like it would take a feather the same time to get going.This works out to be about 64 days.Have you taken into account that the earth is starting from a stand still it will take some time to get going?
If Earth is not starting from a stand-still, it can take pretty much any time you want, as measured by a clock on Earth. A day, one second, or 200 years, all depending on what velocity you give it. The 200 year one is difficult if the rest of the solar system isn't empty because in that much time, the pull from the other objects will have time to bend the path of Earth and make it miss the sun. So it's easier if you just consider a simple 2-body case.
Post by: Just thinking on 25/09/2021 22:05:35
How would 5he earth stop orbiting? Would it grind to a halt or suddenly start travelling toward the sun? Did the sun dissappear or was the earth ejected such as in an orphan planet?Good question but I only put this problem out there as a hypothetical situation. No mechanism for the earth stopping in its tracks.
Post by: Janus on 25/09/2021 22:08:21
Tidal forces wouldn't effect the answer significantly. The Roche limit( as measured in multiples of the primary's radius) is dependent on the relative densities of the bodies.
Anyway, other articles consider the journey toward the sun to be over in about 65 days but that seems to be because they reckon the tidal forces from suns gravity would rip the planet apart before we even reach the outer edges of the sun.
The Earth is some 4 times denser than than the Sun. If we treat the Earth as a fluid body, this puts the Roche limit at 1.55 Sun radii from the center of the Sun our just over 1/2 Sun radius above its surface.
Treated as a rigid body, the Roche limit ends up at ~0.8 the Sun's radius, or under the Sun's surface. The actual Roche limit for the Earth will likely fall somewhere between.
But even if we put it at 1/2 the Sun's radius away, the Earth is just ~350,000 km from the surface, and will be moving close to its final speed of over 616 km/sec. In other words, it is only around 10 min from surface impact.
Post by: Just thinking on 25/09/2021 22:08:56
So it's easier if you just consider a simple 2-body case.That was my intention just the earth and the sun with the earth starting from a standstill.
Post by: Eternal Student on 25/09/2021 22:27:23
The quick and dirty method of computing the fall time is to assume the Earth's orbit is an extremely eccentric ellipse with the Earth's present distance being at perihelion. This puts the Semi major axis at half the perihelion distance, and you can just calculate half the period of such an orbit. This works out to be about 64 days.Please check your use of the term perihelion. I think you want the earth at aphelion to start with. I'm also not sure that "perihelion distance" is a well defined term. See if the diagram in post #10 is what you were saying.
-------
The main thing is that Just thinking is getting some slightly different answers from different people, ranging from two months through to six months. Perhaps we should try and get some concensus on this.
I reckon that 91 days is the right answer assuming loads of things (the sun is a point mass etc.)
It's less than this is you consider other stuff - for example the possibility that the earth will be torn apart by tidal forces and volcanic activity before it actually hits the sun.
-------
How would 5he earth stop orbiting? Would it grind to a halt or suddenly start travelling toward the sun? Did the sun dissappear or was the earth ejected such as in an orphan planet?i.d.k. Our calculations were based on the velocity of earth being suddenly brought to 0 relative to the sun. How you do this is a problem for the engineers. Let's say a tractor beam was used by an alien spacecraft which brought the earth to a standstill (relative to the earth) and then we did the calculation to see what happens when they turn that beam off and allow the earth to fall into the sun.
Best Wishes.
Post by: Janus on 25/09/2021 22:28:54
Yes. It, like anything else, will start "falling" immediately, and at the same acceleration. However, that is not to say that the mass of the Earth is completely irrelevant. It can have an effect on impact time, just in the opposite manner that you seemed to imply. Increasing the mass of the Earth would decrease the time to impact. The reason for this is that not only is the Earth falling towards the Sun, but the Sun would fall towards the Earth due to Earth's gravity. Since the Sun is some 333,000 times more massive than the Earth, the amount it would "fall" towards the Earth is pretty insignificant, and we can safely ignore it.This works out to be about 64 days.Thanks Janus. Have you taken into account that the earth is starting from a stand still it will take some time to get going?
However, if the Earth had a mass that was a much larger fraction of the Sun's, you would have to account for it.
Post by: Eternal Student on 25/09/2021 22:32:16
Tidal forces wouldn't effect the answer significantly.You may very well be right but earth isn't a rigid body. It's a dangerously loose collection of techtonic plates. The possibility of earth breaking apart was an idea presented here:
https://www.wired.com/2014/12/empzeal-earthfall/
I certainly haven't done anything with the tidal forces, it was only mentoned because their answer was less than 91 days but they make it clear that earth never reaches the sun.
Best Wishes.
Post by: Janus on 25/09/2021 22:36:02
Hi again.Oops, I did mean aphelion. Just one of those mental lapses. It might have something to do with the fact that in orbital mechanics so many of the parameters and equations use perihelion as the reference, that I got used to using it.The quick and dirty method of computing the fall time is to assume the Earth's orbit is an extremely eccentric ellipse with the Earth's present distance being at perihelion. This puts the Semi major axis at half the perihelion distance, and you can just calculate half the period of such an orbit. This works out to be about 64 days.Please check your use of the term perihelion. I think you want the earth at aphelion to start with. I'm also not sure that "perihelion distance" is a well defined term. See if the diagram in post #10 is what you were saying.
Post by: Janus on 25/09/2021 22:40:35
Hi again.That's why I used the purely "fluid body" Roche limit to get the 10 min from impact answer. That is the furthest point at which the Sun could begin to pull the Earth apart.Tidal forces wouldn't effect the answer significantly.You may very well be right but earth isn't a rigid body. It's a dangerously loose collection of techtonic plates. The possibility of earth breaking apart was an idea presented here:
https://www.wired.com/2014/12/empzeal-earthfall/
I certainly haven't done anything with the tidal forces, it was only mentoned because their answer was less than 91 days but they make it clear that earth never reaches the sun.
Best Wishes.
Post by: Eternal Student on 28/09/2021 18:49:53
I've had a bit more time to consider the situation described in the OP.
The diagram in post #10 is incorrect. An elliptic orbit like that one is not a solution in the usual two body problem (where the sun can move about the barycentre).
Simplifying the problem to a one-body central force problem (where the sun is assumed to remain stationary) we should have the sun at a focus of the ellipse not at the centre of the ellipse.
Anyway.... re-working the time to impact, we need T/2 as our estimate and the semi-major axis a ≈ 0.5 A.U.
This gives Time to impact ≈ 64.7 days.
Which means that at least two contributors are agreeing on the impact time (Janus and Eternal Student), if that helps you (Just Thinking).
- - - - - - - -
For a brief aside: Consider a one body problem. A body is attracted to a fixed point in space with a force following an inverse square law on distance (e.g. Newtonian gravity). Assume the body is at rest initially but a distance r ≠ 0 from the point of attraction. The point of attraction is just a point in space, there's nothing solid here so the other body could pass through.
Diagram:
* *
body Point of attraction
<------------- r ------------->
At first glance it would seem that the body would be accelerated toward the point of attraction and pass through it, following a similar but mirror image motion on the other side until it eventually stops and starts to move back the other way.... etc. etc. an endless oscillation. (Take a moment to consider the problem yourself).
However, there is no way to determine such motion. Why? Because the force on the body would be undefined when it reaches the point of attraction and so Newton's F=ma can't be used. If we are going to use Newtonian Mechanics then we must avoid the position of the body ever being exactly at the point of attraction.
If you considered the motion as a limiting process of a proper elliptical orbit around the point of attraction it would seem that the body approaches infinite velocity as it closes on the point of attraction but cannot pass through it. Instead there is something more like an elastic collision, the body receives an infinite impulse and reverses direction of travel instantly. Fairly weird in my opinion.
Best Wishes.
Post by: Just thinking on 28/09/2021 19:06:53
This gives Time to impact ≈ 64.7 days.Thanks Eternal Student I'm in agreeance with this calculation and thank all that has taken the time with this as well as all the explanations given.
Which means that at least two contributors are agreeing on the impact time (Janus and Eternal Student), if that helps you (Just Thinking).
Post by: Halc on 28/09/2021 20:57:32
Instead there is something more like an elastic collision, the body receives an infinite impulse and reverses direction of travel instantly. Fairly weird in my opinion.Intuitive actually. Motion straight through would totally discard Keplerian mechanics. Regardless of the eccentricity of the orbit, the 'point of attraction' must remain at one focus of the elliptical orbit, and not in the center as you've depicted in post 10.
Of course, Newtonian mechanics breaks down if you get too close to the point of attraction since the mass ends up with speed faster than light. In reality, if the orbit gets too eccentric, it starts to precess. If it gets really close (as the photo-sphere), it never comes back.
Post by: Eternal Student on 29/09/2021 00:53:58
Intuitive actually.Well thanks for that. It didn't seem intuitive to me late one night. :-[
Best Wishes.