How much energy would it take to shift earth's orbit further out from the Sun by 100m?
Bonus question: And by how much would it reduce the avg temp?

Depend where on the flat earth you are speaking off

It's not just energy, it's reaction mass, what are you going to eject to do it.

An oblate spheroid..... orbiting in an ellipse....... and influenced by other celestial bodies.
You do the math.


But at a rough guess.... we are 149600000000m away

so shifting Earth 100m is a change of 1/1496000000th - totally insignificant.

Now if I dropped 1/1496000000th of my weight (100kg) that would be wayyyy less than 1g.
So if you think 1g makes me way or less hotter,come talk .

Short answer = dumb question. No change.

But Mark we're dealing with the inverse square law here, it's not just a 100m. But my Maths cct has gone to sleep, I'm not sure how much difference that makes.

Indeed and I failed to account for that


But the square root of 1/1496000000 is still ridiculously small so I think Flyman needs to chase another condundrum, like how much stuff to blow up a building with aeroplane wings and not be noticed

It would take the amount of energy required to change the gravitational forces of the entire solar system.
Or you could just shrink the sun. Good luck.

Just hack into the worlds data bases and shift it 100m. Much simpler.

It would be more effective to put a reflective foil sheet into orbit to shield the earth from sunlight. Paint a few logos on it, and you won't even need to spend any tax payers dollars to get it up there.

I don't know how regular and smooth you perceive the orbit of the earth to be, or where the point on the sun and earth you are measuring to, but as a concept of what I assume you are getting at :


The sun is about 150 million km away.

The energy received decreases relative to the square of the distance.

So moving 100m would decrease the heat from the sun by sqrt(100/150,000,000,000) or by about 0.0000025.

So, to somehow put that in some sort of context perhaps if you assume the average temp on the earth is 25 degrees and 100% of that temperature comes from the sun then the average temperature would decrease from 25 degrees to 24.9993545 degrees.

Given the average temperature year on year and century on century and epoch on epoch for the billion years before man burnt coal, varied by far more than 0.00065 degrees the answer to your questions are :

The energy expended to move it 100m would increase the temperature significantly more than the decrease from being further from the sun.


If you don't think this is the answer you want then answer this yourself first :

How much colder will I feel if I quickly run 100m up a flight of stairs ?

It's the classic HSC physics question. A variation on it has been on every HSC physics exam since the year dot. You've all forgotten it! So have I. But being a bit of a nerd back in 1970 I'd memorised every combination and permutation of the problem by studying up on past exam papers. (Not that it did me any good?) It's easy to work out. The quirky thing was that gravitational energy is referenced from infinity so that the integral of the 1/r^2 force is calculable. It boils down to a -1/r. And then you have to balance the centrifugal force mv^2 /r to the gravitational force. We'd travel a bit slower in a higher orbit and I seem to recall the kinetic energy of a higher orbit also boils down to a 1/r thing. All multiplied by big G, mass of the earth and mass of the sun. It'll be a lot of joules. No practical way of applying them.

But it's happening anyway. The decreasing mass of the sun and the tidal forces are slowly lofting us into a higher orbit.

Same with the earth and moon only stronger. The moon induces a tidal bulge on earth. And the earth, rotating once every 24 hrs, drags that tidal bulge ahead of the moon that only orbits once every 28 days. Like winding up a tennis ball in a stocking. 3.8 cm higher each year. (This bit wasn't in HSC exams it's on the internet)

curious.astro.cornell.edu/physics/37-our-solar-system/the-moon/the-moon-and-the-earth/111-is-the-moon-moving-away-from-the-earth-when-was-this-discovered-intermediate

Select to expand quote

decrepit said..
It's not just energy, it's reaction mass, what are you going to eject to do it.

His mother.

Select to expand quote

Mastbender said..
It would take the amount of energy required to change the gravitational forces of the entire solar system.
Or you could just shrink the sun. Good luck.

This. People think the earth orbits the sun. It doesn't.

It orbits the "gravitational center" of the entire solar system; the sun, all planets, all moons, all asteroids combined. If you have only two large bodies floating around in space they will (given the right conditions) orbit each other, the smaller of the two orbiting the other "more".

The sun orbits this center too. A star's wobble tells us about its planets.

Another fun fact: if you drop a feather and a bowling ball on the moon the bowling ball will fall faster than the feather (albeit perhaps imperceptibly so), for the same reasons as above. e.g. two neutron stars will fall toward each other faster than two bowling balls.

That's old Newtonian physics. Here's an Einsteinian problem that is going round between my ears at the moment.
A couple of weeks ago we saw the black hole photo.
The press release said that the M87 object has a mass of 6.5 billion suns and a diameter of 38 billion km.
The sun has a diameter of ~1.4 million km.
So the black hole is about 27 000 times the diameter, and its volume should be the cube of that bigger.
And that volume contains the mass of 6.5 billion suns.
My difficulty is that results in a density for the black hole that is only about 1/3000 of the solar density.
It makes no sense

Select to expand quote

Mr Milk said..
That's old Newtonian physics. Here's an Einsteinian problem that is going round between my ears at the moment.
A couple of weeks ago we saw the black hole photo.
The press release said that the M87 object has a mass of 6.5 billion suns and a diameter of 38 billion km.
The sun has a diameter of ~1.4 million km.
So the black hole is about 27 000 times the diameter, and its volume should be the cube of that bigger.
And that volume contains the mass of 6.5 billion suns.
My difficulty is that results in a density for the black hole that is only about 1/3000 of the solar density.
It makes no sense

Just sip the Koolade.'
The media reported it, so we know it must be right.

Select to expand quote

Mr Milk said..
That's old Newtonian physics. Here's an Einsteinian problem that is going round between my ears at the moment.
A couple of weeks ago we saw the black hole photo.
The press release said that the M87 object has a mass of 6.5 billion suns and a diameter of 38 billion km.
The sun has a diameter of ~1.4 million km.
So the black hole is about 27 000 times the diameter, and its volume should be the cube of that bigger.
And that volume contains the mass of 6.5 billion suns.
My difficulty is that results in a density for the black hole that is only about 1/3000 of the solar density.
It makes no sense

Could be referring to the diameter of the event horizon rather than the proper black hole in the middle of it.

www.space.com/black-holes-event-horizon-explained.html

That article has the black hole even larger, at 17.7 billion km and same mass. It might be sloppy writing, given that it does make a distinction between the event horizon and the black hole elsewhere, but it definitely gives that as the size of the black hole.
Personally, I was hoping for an explanation that might involve warping space-time so that the apparent length of a metre changes, or maybe even reverse inflation

Select to expand quote

Ian K said..
It's the classic HSC physics question. A variation on it has been on every HSC physics exam since the year dot. You've all forgotten it! So have I. But being a bit of a nerd back in 1970 I'd memorised every combination and permutation of the problem by studying up on past exam papers. (Not that it did me any good?) It's easy to work out. The quirky thing was that gravitational energy is referenced from infinity so that the integral of the 1/r^2 force is calculable. It boils down to a -1/r. And then you have to balance the centrifugal force mv^2 /r to the gravitational force. We'd travel a bit slower in a higher orbit and I seem to recall the kinetic energy of a higher orbit also boils down to a 1/r thing. All multiplied by big G, mass of the earth and mass of the sun. It'll be a lot of joules. No practical way of applying them.

But it's happening anyway. The decreasing mass of the sun and the tidal forces are slowly lofting us into a higher orbit.

Same with the earth and moon only stronger. The moon induces a tidal bulge on earth. And the earth, rotating once every 24 hrs, drags that tidal bulge ahead of the moon that only orbits once every 28 days. Like winding up a tennis ball in a stocking. 3.8 cm higher each year. (This bit wasn't in HSC exams it's on the internet)

curious.astro.cornell.edu/physics/37-our-solar-system/the-moon/the-moon-and-the-earth/111-is-the-moon-moving-away-from-the-earth-when-was-this-discovered-intermediate

Could be interesting calculation to find out what will be total energy required to release completely Earth from Sun orbit. Suppose the one day our scientific will predict incoming catastrophe like our Sun blowing up, our civilisation may decide to fly away with Earth instead of packing everything on ships and escaping. Having such energy to move the Earth, lack of Sun radiation to power our solar panels and plants could be lesser problem.

Select to expand quote

Mr Milk said..
That article has the black hole even larger, at 17.7 billion km and same mass. It might be sloppy writing, given that it does make a distinction between the event horizon and the black hole elsewhere, but it definitely gives that as the size of the black hole.
Personally, I was hoping for an explanation that might involve warping space-time so that the apparent length of a metre changes, or maybe even reverse inflation

Your dreams have been answered
www.abc.net.au/news/science/2019-04-30/black-hole-swirling-jets-in-space-as-star-ripped-apart/11050396

Black holes have no diameter.

Select to expand quote

Mr Milk said..
That's old Newtonian physics. Here's an Einsteinian problem that is going round between my ears at the moment.
A couple of weeks ago we saw the black hole photo.
The press release said that the M87 object has a mass of 6.5 billion suns and a diameter of 38 billion km.
The sun has a diameter of ~1.4 million km.
So the black hole is about 27 000 times the diameter, and its volume should be the cube of that bigger.
And that volume contains the mass of 6.5 billion suns.
My difficulty is that results in a density for the black hole that is only about 1/3000 of the solar density.
It makes no sense

The rest of the density will be all the spaceships and such like that have been sucked in there..... True story... I saw it on a movie once

Select to expand quote

evlPanda said..
Black holes have no diameter.

Hmmm, so a singularity has mass but no dimensions, is that correct?

Is that because, there's no way of measuring it?
Or space time is so distorted, dimensions are meaningless?
Or It's only theory, and it's where relativity and quantum mechanics collide, so basically we have no idea what's going on?

This bit from the www tells me we don't know what happens behind the event horizon so we might as well use its dimensions to calculate the density.

curious.astro.cornell.edu/86-the-universe/black-holes-and-quasars/general-questions/423-what-is-the-density-of-a-black-hole-advanced



"General relativity predicts that as an object collapses to form a black hole, it will eventually reach a point of infinite density. What that really means is that the theory of relativity breaks down at this point, and no one knows what happens at the center of a black hole - we would need a viable theory of quantum gravity in order to understand this."