[Just thinking (OP)]

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?

[Bored chemist]

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).
There might be the odd factor of 2 or pi or something in there.
Orbiting is just falling, but missing.

[Just thinking (OP)]

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).

Orbiting is just falling, but missing.

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.

[Just thinking (OP)]

If the earth started out where Jupiter is it would build up much more velocity just throwing that out there. Comets that come very close to the sun or even do the sun dive are travelling at a huge speed much faster than orbiting planets.

[Bored chemist]

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.

[Just thinking (OP)]

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.

[Colin2B]

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

[Just thinking (OP)]

“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.

[Bored chemist]

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.

And if you remove the orbit...?

[Halc]

I'm not sure but I think it would take 6 months (half a year) to fall

Maybe 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.

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.

[Eternal Student]

LATE EDITING:   The elliptical orbit shown is incorrect, sorry.  I assumed a symmetry where there isn't one.   See post #28 for a correction.


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 = gravity
This 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.

[Just thinking (OP)]

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.

[Just thinking (OP)]

Alancalverd is an engineer, he might enjoy doing the calculation,

I will try Alan I'm not up to the task.

[Eternal Student]

Hi again.
  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 mph

    There'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.

[Eternal Student]

Hi again.

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.

[Just thinking (OP)]

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.

[Janus]

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.

[Just thinking (OP)]

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?

[Petrochemicals]

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?

[Halc]

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?

That figure assumes that, yes, just like it would take a feather the same 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.