Earth splash into Sun

[Grok]

Right.  Think of the Earth and Moon as a closed mass-energy system.  Doesn't matter which object in the system you apply the energy to, as the effect will be the same amount of energy.

The incline example:  even without a push, the ball would still go down the incline.  The force of your push adds a finite energy amount to the ball system.

Drop a pencil.  Notice it falls to the floor.  The only reason something orbits instead of falls is its angular velocity keeps it going away from the Earth at the same speed it is falling.  But orbiting things ARE falling.

So again, all we have to do is reduce angular velocity by any amount, and the falling overtakes the going-away speed.

[Camel]

So again, all we have to do is reduce angular velocity by any amount, and the falling overtakes the going-away speed.

Would it not simply establish a new equilibrium, closer to the sun?

I was bored in my Calculus class today, and decided to write a simulator for this on my TI-83+. The numbers are completely wrong because I was just trying to get it to work correctly before inputting the actual data (which I didn't know at the time):

I've added whitespace and comments here for convenience

prgmSUN

Code Select Expand

:ClrDraw

:3->O //number of objects. each matrix row is one object; format is {X, Y, Vx, Vy, Mass}
:[
[0,0, 0, 0, 0,   100]
[1,0, 0, 0, 0.1, 1]
[1.01,0, 0, 0.6, 1/6]
]->[J]

//This will set up the window to display everything correctly; it's not essential
:1->Ymax
:-1->Ymin
:94/62->Xmax
:-94/62->Xmin
:AxesOff
:FnOff
//to be completely explicit, one might also use PlotsOff, et al.

:Lbl L //i had to many [if:then:..:end]s, and decided i would just use Goto. much easier to read

//begin the draw loop. will draw a line for each object's velocity (it "points" backwards so that the line will look continuous)
:For(I,1,O
:[J](I,1)->X //displacement
:[J](I,2)->Y
:[J](I,5)->M //mass
:[J](I,3)*M->S //velocity
:[J](I,4)*M->T
:X+S->X // move to new position
:Y+T->Y
:X->[J](I,1 //store new position
:Y->[J](I,2
:Pt-On(X,Y
:Line(X,Y,X-S,Y-T //in writing this, i realize i should have stored the new position in different variables
:End

:For(I,1,O
:For(J,1,0
:If I != J:prgmSUNIJ //SUNIJ sounded really cool
:End
:End

:Goto L


prgmSUNIJ

Code Select Expand


:[J](J,1)-[J](I,1)->X
:[J](J,2)-[J](I,2)->Y
:RtoPT(X,Y)->Θ //under 'ANGLE', 6
:RtoPr(X,Y)->D //under 'ANGLE', 5

:[J](I,5)/De-4->A

:[J](J,3)->X
:[J](J,4)->Y
:X-Acos(Θ)->X
:Y-Asin(Θ)->Y
:X->[J](J,3
:Y->[J](J,4


[Grok]

So again, all we have to do is reduce angular velocity by any amount, and the falling overtakes the going-away speed.

Would it not simply establish a new equilibrium, closer to the sun?

Hold an object out away from your body.  Drop it.

Did it go into low earth orbit at a reduced equilibrium?  Why not?  Why do you expect the Earth to go into a lower orbit around the Sun when you reduce the Earth's angular velocity?

[Camel]

Hold an object out away from your body.  Drop it.

Did it go into low earth orbit at a reduced equilibrium?  Why not?  Why do you expect the Earth to go into a lower orbit around the Sun when you reduce the Earth's angular velocity?

If it has an angular velocity of zero it will reach equilibrium at the center of the earth, which it cannot reach because the path is obstructed. The question was about a scenario with reduced velocity, no zero velocity.

Perhaps I should rephrase my question: What is the relationship between the angular velocity required to keep Earth orbiting around the sun and the distance between them? If it requires less velocity as it gets closer, wouldn't the Earth simply establish a new equilibrium with its new velocity?

[Banana fanna fo fanna]

Perhaps we should instead have the moon crash into the earth. Since it has much less mass than the earth, it would be easier to do so. I also believe that a moon impact would have sufficient power to kill everyone.