# what Mass would be required to create measurable resistance in this circuit?

sorry I'm not a physicist or student in this is not homework I was just thinking about this.

say you have a spherical Mass of a certain density I'm thinking say planet-sized and then tangent to the surface of that sphere you have a circuit running say 100 kilometers into space and it's a superconducting wire and also runs a hundred kilometers back to create a complete circuit because I'm guessing a complete circuit is required for this.

and then say you run a current through that circuit, pick a number, of volts or amps I don't know how to pick because I'm not an expert, but there's zero resistance because it's super conducting so what would the mass of that planet sphere have to be to make the electrons flow be measurably impaired (buy measurably let's say we're measuring the thing on a normal voltmeter ampmeter or whatever instrument would be required I don't actually know that, and you can see all the flow is impaired somehow by 1/10 of a unit of whatever our measure is), I'm guessing that impairment is resistance created by the gravity of that sphere.

I guess the reason I'm asking this is to have some sort of intuition or picture in my mind that I can go okay that's how eletricity and gravity and Mass all interact and I think knowing what the size or density of that sphere would be in relating that to planets that are known or other celestial objects then would be a good thing to know.

then I can think oh that's how much mass would be required to affect these electrons traveling in that circuit.

## 1 Answer

You can get an idea of this by comparing electrical and gravitational potential energy.

The energy required to move an electron against a potential V is given by:

$$E = eV \tag{1}$$

where $$e$$ is the electron charge. So to move an electron against a potential of one volt means it takes an energy of $$1.6 \times 10^{-19} \text{J}$$.

Now compare this with the gravitational potential energy. The energy required to move an electron upwards a height $$h$$ in a constant gravitational field $$g$$ is given by:

$$E = m_e g h \tag{2}$$

where $$m_e$$ is the electron mass. Suppose we set this equal to the electrical energy we got from equation (1) we get:

$$eV = m_e g h$$

You are asking about the gravitational field so we'll rearrange this to:

$$g = \frac{eV}{m_e h} \tag{3}$$

You suggest your circuit is 100km long, so we'll set $$h = 10^5 \text{m}$$, and we'll set $$V$$ equal to one volt. Then we substitute these values into equation (3) and it gives us:

$$g = 1.76 \times 10^6 \text{ms}^{-2}$$

which is about 200 000 times the gravitational field of the Earth. So to generate this field (keeping the size of the Earth the same) we'd need to increase its mass by a factor of 200 000.