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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? |
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There is no answer at this time. |
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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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