Leaderboard

Wow, Earth has really gotten fat!

1. It is meaningless to say "spin at 28,437 km", as rotation needs to be measured as angular velocity. (deg/hr, rad/sec. etc.) I know that it is common to express this in terms of tangential speed( in this case, at the equator) But it is sloppy and can lead to misunderstanding. For example if the Earth had a tangential speed of 28,437 kph at the equator, then at 45 north latitude it would only be 20,105 kph As swansont has pointed out there is a speed where the centripetal force (the force required to keep an object moving in circle) and the gravitational force balance out. This would result in the object going into orbit around the Earth. Gravity is still in play. In fact it is gravity that would prevent someone standing on the equator from just shooting out into space at in a straight line instead of just hovering over the Earth. And by the way, your number is a bit low, the equatorial speed would need to be 28463 kph And because of what I alluded to earlier, only someone on the equator would even go into orbit. People elsewhere will feel lighter, and the ground would seem to tilt a bit under their feet ( And even this is an over simplification which assumes the Earth maintains its present shape. If the Earth was indeed spinning this fast, its very shape would change, making it much more of an oblate spheroid.

No ma = GMm/r^2 the mass m cancels, as it’s on both sides, but M remains, so a = GM/r^2 Gravitational acceleration doesn’t depend on the mass of the object, but does depend on the mass of the (usually celestial) object exerting the gravity

There is a speed such that the centripetal force is equal to the gravitational force. v = sqrt(GM/r). That’s the speed of a circular orbit at r. You would be weightless, but not lift up. (but the earth would fall apart before this could happen) Far from throwing a wrench into it - the above equation uses the equation to solve for v The mass of the earth would be slightly larger, increasing the pull on the moon by a small amount. Any other effect on the moon would be found in GR

Okay, let's look at the first "equation", F = ma= (G*M1*M2)/R2 It is not an equation, but rather a shortcut of two equations: F=ma ma=\(\frac {GM_1M_2}{R^2}\) The first equation is the Newton second law. It is fine. However, the second equation is meaningless, unless you have a reasonable interpretation for it. PS. You are only a bit older than me.

First of all I was not talking about beta decay from fusion because that was not the scenario you brought up, however I did incorrectly say beta- decay instead of beta+ decay. Addressing your OP you said: Alpha radiation consists of a He nucleus, so the neutrons and protons are 'fused' together. Let's assume we can somehow remove the neutrons from the nucleus. Your question then is could the 2 protons undergo fusion. The 2 protons are already fused together, so I'm not sure what your asking. In reality if the 2 neutrons were removed from the nucleus the resulting helium isotope would be extremely unstable and one of the protons would immediately beta+ decay to a neutron forming deuterium. No fusion would occur.

Your premise appears to be false. Renewable generation is now competitive with fossil fuel electricity production: https://www.weforum.org/agenda/2020/06/cost-renewable-energy-cheaper-coal/

But that process has nothing to do with the one you were asking us to consider. What you were proposing was conversion of ²He into D. That process, which is only followed in <0.01% of cases, is indeed β+ decay, but it is not fusion. The net conversion achieved by your proposal, starting from α-particles, is ⁴He -> D. This is a convoluted fission process, not fusion and, surprise, surprise requires a net input of energy to achieve it.

That graphic does not show what you described in the OP.

A lot of importance is given to Faraday's new electromotive motor and there is considerable hype that it might be amenable to practical applications where Mr Watt's engine proves cumbersome. However it's hard to see this ever going beyond a few niche markets, given the considerable expense and labor involved in the fabrication, not to mention the problems of creating a wide availability of electrical generation and transmission. Mr Faraday's device appears to be more a toy than a real solution in providing mechanical power in quotidian uses.

It takes energy to remove the neutrons and He-4 is more tightly bound than other light atoms. deuterium has a binding energy of ~2.2MeV. That tells you the nuclear binding between nucleans. The coulomb repulsion at typical nuclear separation has an energy of at least this amount - the p-p system isn't bound (there are other considerations; there isn’t a n-n bound state, either) p-p forming deuterium (and a positron and neutrino) only releases 0.43 MeV. You might note that that’s less than the binding energy, because of that excess of Coulomb repulsion. If you want to assess the feasibility you need to run the numbers.

I was not expecting that awnser. It depends on the reaction I am doing. Wow. I'am doing many reactions. There are many. -Alchols recting with acidified di chromate -Alchols reacting with hot sulfuric acid -oxidation and reaction reactions -acetic acid reactions etc The way I understand is that the product that that used to be a reactant and had the nucluphile has to have no formal charges but evreything else that was formed by leaving groups can have a formal charge. How far away I am from being correct