How much stuff will neutralise the sun's pull on the earth

1. Jul 1, 2014

bland

I tried to make the title as descriptive as I could without being overly long.

Anyway, I've been trying to work this out but I'm having some difficulty. We all know the strength of the electromagnetic force is very much greater than the gravitational force. I'm not sure what it is as it's a number that seems to vary between 10^35 and 10^40 times depending on the source. And the mass of the earth is 6 x 10^24kg.

For the sake of the argument, let's call the gravitational force 10^36 times weaker than the EM force and the mass of the earth as 10^25kg

I was trying to work out given the above numbers, how many kg's of protons would need to be evenly spread through the sun and also evenly spread through the earth, such that the electromagnetic repulsion would be enough to overcome the pull of the gravitational attraction between the two objects (sun and earth) So that the earth would float away.

At first I thought it would be simple to work out but after thinking about it for a while I realise that I don't really know how to do this.

I hope this makes sense.

2. Jul 1, 2014

jbriggs444

One place to start would be Coulomb's law. That gives you force in terms of charges and distance.

You'd need to figure out the distance between Earth and Sun.

You'd need to get a conversion factor between Coulombs and kilograms of protons. There are several ways to get that conversion factor. One way is to look up how many elementary charges per Coulomb, and how many elementary charges per mole. Then, a bit of background knowledge in chemistry is that 1 mole of protons (aka 1 mole of hydrogen nuclei) is 1 gram.

Then you'd need to repeat the exercise for the force of gravity between Earth and Sun. You could use the period and radius of the Earths orbit to arrive at the acceleration of the Earth in its orbit around the sun and factor in the mass of the earth to get the force of gravity on the Earth by the sun.

3. Jul 1, 2014

dauto

The radius of the orbit cancels out of the equations. No need to waste time figuring that factor.

4. Jul 1, 2014

jbriggs444

Good point, thanks. Though it depends on which known values and you Googled up to work from when computing the gravitational force and which equations you used with them.

If you calculated the gravitational force using the approach that I mentioned then would do not have a factor of 1/radius2 in the gravitational force to cancel against the factor of 1/radius2 in the electrostatic force.

5. Jul 1, 2014

Matterwave

Newton's law of universal gravitation says:

$$F_g=\frac{Gm_1m_2}{d^2}$$

Coulomb's law says:

$$F_e=\frac{kq_1q_2}{d^2}$$

Setting the two forces equal, you will get very simply:

$$k q_1 q_2=Gm_1m_2$$

If you assume $q_1=q_2=q$ you can solve that equation very easily to find q. From q, you need to use jbrigg's suggestion to find the mass of the protons involved.

6. Jul 1, 2014

DrZoidberg

Imagine the earth and the sun both consisted completely of protons and had therefore a rediculous amount of charge. In that case the electric repulsion should be 10^36 times stronger than the attraction. That means the electric repulsion has to be reduced by a factor of 10^36 to make both forces equal.

That could for example be achieved by reducing the charge of both bodies by a factor of 10^18.
The sun's mass is 2 * 10^30 kg. Divided by 10^18 is 2 * 10^12 kg
Earth's mass is 6 * 10^24 kg. Divided by 10^18 = 6 * 10^6 kg

That means you'd have to add 2 * 10^12 kg of protons to the sun and 6 * 10^6 kg of protons to the earth.

7. Jul 7, 2014

elegysix

Interesting approach here. Has anyone worked it out the other way and gotten similar numbers?