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Q: Sun's current volume and Earth's orbit ( No Answer,   8 Comments )
Question  
Subject: Sun's current volume and Earth's orbit
Category: Science > Astronomy
Asked by: criley-ga
List Price: $10.00
Posted: 28 Oct 2002 21:32 PST
Expires: 27 Nov 2002 21:32 PST
Question ID: 91682
What if a giant trash compactor in the sky came and compressed the sun
to half its current volume? What would happen to the earth's orbit
then? Why?
Answer  
There is no answer at this time.

Comments  
Subject: Re: Sun's current volume and Earth's orbit
From: ishmaiel-ga on 28 Oct 2002 23:17 PST
 
Well, I can't say that I agreen with webadept-ga. The gravitational
force between two bodies of mass m and m' is

F = G*m*m' / r^2

where r is the distance between them. If the size of the bodies is
much smaller than r, as is in this case, then the force that m and m'
experience is not significantly related to the density, since you can
approximate the bodies as points of mass m and m', disregarding
internal structure. Thus, if the sun's volume were to decrease by 1/2
and its mass were to remain the same, then the gravitational
attraction would remain the same.

As an aside, if the bodies are sufficiently close to one another so
that r is on the order of the size of the bodies, you must perform in
integral. Suppose that m is a point mass and m' has some shape where
m'(x,y,z) is the density at x,y,z.

F = G*m*integral( m'(x,y,z)/r(x,y,z)^2 * dxdydz )

where r(x,y,z) is now the distance between (x,y,z) and m.

Finally, it should be pointed out that if the sun's volume were
crunched, its stability would be affected. I know too little about
astrophysics to comment, but my main concern would be that the sun
would automatically bounce back to its previous size with some adverse
effects like massive corona, solar winds and other radiation. The
current size of the sun is a result of a direct balance between force
of gravity, pulling the sun into itself, and radiation pressure,
caused by the fusion reaction that drives the core.

Specific gravity has very little to do with this. Specific gravity is
the density of an object relative to water.

Conclusion: force of gravity would remain the same - but put on your
sunglasses!
Subject: Re: Sun's current volume and Earth's orbit
From: rbnn-ga on 28 Oct 2002 23:33 PST
 
ishmaiel-ga is correct; the answer isn't.

Specific gravity is not relevant to the problem at hand, despite the word "gravity".
Subject: Re: Sun's current volume and Earth's orbit
From: rbnn-ga on 28 Oct 2002 23:52 PST
 
ishmaiel-ga: how much would the sun have to be compressed to form
black hole? I guess we aren't close to that in the problem at hand.

By the way, for whatever it's worth, one of the greatest short stories
ever written in my opinion involves solar activity: "Inconstant Moon"
by Larry Niven.
Subject: Re: Sun's current volume and Earth's orbit
From: javit-ga on 29 Oct 2002 07:44 PST
 
You would need to compress the Sun to approximately 1.8 miles [2.9 km]
in radius in order to make its escape velocity higher than the speed
of light hence become a black hole.
Please note that Sun will never be a black hole during its lifetime
due its relatively low mass. You will need some serious machinery to
compress it to 1.8 miles in radius :)
Subject: Re: Sun's current volume and Earth's orbit
From: archangel-ga on 31 Oct 2002 21:44 PST
 
If the sun was crushed to a smaller volume then it would greatly
increase its internal presure which would translate to an increase of
heat.  If the center was heated up it would increase the rate of
fusion reactions that occurs and would lead to more energy being
released.  If this theoratical trash compactered was left in place for
very long then when it was removed the energy that was stored up would
be rapidly released and we would get something akin to a nova.  Which
would problay blow us out of our current orbit.
Subject: Re: Sun's current volume and Earth's orbit
From: dannidin-ga on 01 Nov 2002 03:52 PST
 
ishmaiel-ga, you write

"As an aside, if the bodies are sufficiently close to one another so
that r is on the order of the size of the bodies, you must perform in
integral. Suppose that m is a point mass and m' has some shape where
m'(x,y,z) is the density at x,y,z.
 
F = G*m*integral( m'(x,y,z)/r(x,y,z)^2 * dxdydz ) 
 
where r(x,y,z) is now the distance between (x,y,z) and m. "

I believe this is false: For two spherically symmetric bodies that are
not in contact with each other, the gravitational pull between them is
PRECISELY
F = G*m*m' / r^2. This is not an approximation (at least within
classical mechanics). In other words it can be shown that this is the
value of the integral above, regardless of how close the two bodies
are, as long as they are not touching.
 
-dannidin
Subject: Re: Sun's current volume and Earth's orbit
From: physsed-ga on 04 Nov 2002 12:11 PST
 
Hi - 
This is quite an easy question - the short answer is NOTHING.
Since the mass of the Sun does not change when its volume is
compressed, the effect on the Earth's orbit will be nil. When dealing
with spherical objects (Earth and Sun are good enough approximations),
the graviational attraction of the two objects can be considered to be
between each object's center of mass. The Sun could be squashed into a
black hole and the Earth would still orbit as it does today!
Subject: Re: Sun's current volume and Earth's orbit
From: ishmaiel-ga on 07 Nov 2002 00:32 PST
 
dannidin - you are probably right, but I never assumed that the bodies
were spherical :) it sounds like one of those proofs I could do long
time ago.

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