How much charge would the Earth and Sun need to knock the Earth out of orbit

In summary, the conversation discusses the calculation of how much charge would be needed on the Earth and Sun to knock the Earth out of orbit. The equations used are F = gMm and F = kqq, and the attempt at a solution involves finding the number of protons needed to generate the necessary force. However, it is pointed out that the calculation does not take into account the kinetic energy of the Earth, and an alternative strategy is suggested involving dimensional analysis. It is also mentioned that the Earth would not crash into the Sun, but rather be expelled from the solar system, due to the opposite charges of the two bodies.
  • #1
g.lemaitre
267
2

Homework Statement



How much charge would the Earth and Sun need to knock the Earth out of orbit. Then, assuming the distance from one proton to another and the length of one electron to another is 197 pm, find out how large of a block you would need of protium located on Earth and how large of a block of pure electrons located on the Sun you would need to accomplish the aforementioned result. Ignore the fact that a block of protium cannot exist since all of the protons would be repulsed by each other. Same for electrons. I made up this question.


Homework Equations



F = gMm
F = kqq

The Attempt at a Solution



For the force of gravity the Earth's orbit around the Sun creates I have

6.7*10^-11 * 1.99 * 10^30 * 5.97 * 10^24 = 7.96 * 10^44 Nm^2

So to find out how much charge you need:

7.96 * 10^44 = kqq = kq^2 = (8.99 * 10^9)q^2

= 2.98 * 10^17

To find out how many protons you would need you divide 2.98 * 10^17 by the charge of one proton, which is 1.61 * 10^-19 which equals 1.85 * 10^36 protons.

I think 197 pm is 1.97 * 10^-10

First I multiply them

1.97 * 10^-10 * 1.85 * 10^36 = 3.64 * 10^26

but then to get it in three dimensions I'm pretty sure you need to take the cube root.

(3.64 * 10^26)^1/3 = 7.14 * 10^8 m^3 = 7.14 * 10^5 m^3

On the one hand the volume of the Earth is 1.08 * 10^12 km^3. So the block would be something 6 * 10^-5 % of the Earth's volume which is small. But on the other hand the block would be bigger than the United States which is large.

Let me know if you see any mistakes in my calculations.
 
Physics news on Phys.org
  • #2
g.lemaitre said:

Homework Statement



How much charge would the Earth and Sun need to knock the Earth out of orbit. Then, assuming the distance from one proton to another and the length of one electron to another is 197 pm, find out how large of a block you would need of protium located on Earth and how large of a block of pure electrons located on the Sun you would need to accomplish the aforementioned result. Ignore the fact that a block of protium cannot exist since all of the protons would be repulsed by each other. Same for electrons. I made up this question.

Homework Equations



F = gMm
F = kqq
What about the r^2 term?

Are you trying to see how much charge on the Earth and sun is needed in order to 1. make the Earth fall into the sun (ie its orbit < radius of the sun) or 2. to send the Earth out of the solar system? It sounds like it is the former case.

What is your reasoning for equating the electrical and gravitational forces? How does that make the Earth crash into the sun?

If you are trying to figure out the electrical force needed for the Earth to crash into the sun, it is a bit complicated, because you have to calculate the force needed to make the Earth's orbit such that on its closest point to the sun it hits the sun. That distance (ie. effective distance from the solar centre to its "surface") is a bit difficult to determine so you will have to state your assumptions.

Suppose that you posed the second question: sending the Earth out of the solar system. If the repulsive Coulomb force is equal in magnitude to the attractive gravitational force, what will happen?

AM
 
  • #3
Andrew Mason said:
What about the r^2 term?

In the equation gmm/r^2 and kqq/r^2 the r^2 is the same for both of them, so I figured it does not factor in.

Are you trying to see how much charge on the Earth and sun is needed in order to 1. make the Earth fall into the sun (ie its orbit < radius of the sun) or 2. to send the Earth out of the solar system? It sounds like it is the former case.
Being repelled out of orbit or being attracted to the sun it would be the same number of charges, just the sign on the charge would be different, so either way it takes the same number of charges.

What is your reasoning for equating the electrical and gravitational forces?
I'm trying to find the force and both the gravitational force and the electric force are measured in Nm^2.

How does that make the Earth crash into the sun?
Correct me if I'm wrong but if the electrical force were stronger than the gravitational force that would knock it out of orbit.
 
  • #4
Your proposed formulae will work (although the quantity you are calculating is not a force) to calculate the charge needed to neutralise the gravitational attraction of the Sun-Earth system. However, this will not cause the Earth to crash into the Sun - quite the opposite. You also need to calculate the extra charge needed to counter the kinetic energy of the Earth.

Equal sized spheres touching each other tend to adopt a configuration called face-centered cubic. If you look that up, and work out the volume of the cube and the number of spheres inside it (careful - look up fencepost error too) that will give you a volume per proton/electron. Take it from there.

An alternative strategy would be dimensional analysis, which might give you a hint where your volume calculation has gone wrong. What units have you got? What should they be?
 
Last edited:
  • #5
Ibix said:
Your proposed formulae will work (although the quantity you are calculating is not a force) to calculate the charge needed to neutralise the gravitational attraction of the Sun-Earth system. However, this will not cause the Earth to crash into the Sun - quite the opposite.
Why not? By opposite do you mean it will expel the Earth from the solar system? If opposite electric charges attract, then why would a positively charged sun not attract a negatively charged Earth?

You also need to calculate the extra charge needed to counter the kinetic energy of the Earth.
See, I was wondering if the velocity of the Earth through space matters. I'm pretty sure it's roughly 21 km/s. I would think it would be harder to knock a fast object out of orbit than a slow object.

So are you saying I should add the gravitational energy + the kinetic energy to get the total energy?

Spheres touching each other tend to adopt a configuration called face-centered cubic. If you look that up, and work out the volume of the cube and the number of spheres inside it (careful - look up fencepost error too) that will give you a volume per proton/electron. Take it from there.
I'll do this later.


An alternative strategy would be dimensional analysis, which might give you a hint where your volume calculation has gone wrong. What units have you got? What should they be?
the units are meters cubed and I think that's what they should be.
 
  • #6
From your calculation equating the electrostatic force to the gravitational force, I assumed you were trying to calculate the repulsion necessary to exactly counter the gravitational attraction of the Sun. If you did that, it would be analogous to swinging a ball round your head on a string and then cutting the string. The ball (the Earth) would go flying off, not spiral into your head.

If you want to know what force is necessary to cause the Earth to spiral into the Sun, I think that you need to
  • calculate the angular momentum of the Earth as it moves around the Sun
  • calculate the centripetal force needed to maintain an orbit with that same angular momentum in an orbit just grazing the Sun
  • subtract the gravitational force at that distance
  • calculate the charge you need to provide the remaining force at that distance.

With regard to the volume, you do not have a number in cubic meters, although you are correct that that is what you want. Your figure [itex]3.64\times10^{26}[/itex] is the length (in meters) of all of the necessary protons laid out in a straight line. Taking the cube root of that gives you a number ([itex]7.14\times10^8[/itex]) with associated unit [itex]m^{1/3}[/itex], which is not what you want. The simplest way to get the number you are looking for is to calculate the volume occupied by one proton and multiply by the number of protons. That's a slightly simplified variant on what I suggested with the face-centered cubes. Since the situation you are describing isn't really physically possible, worrying about crystallography is probably redundant.
 

1. How much charge is needed to knock the Earth out of its orbit around the Sun?

The amount of charge needed to knock the Earth out of its orbit around the Sun is not a well-defined concept in physics. The Earth's orbit is determined by the gravitational force between the Earth and the Sun, not by electric charge.

2. Can the Earth be knocked out of its orbit by increasing the charge of the Sun?

No, the charge of the Sun does not affect the Earth's orbit. The Sun's charge does not play a significant role in the gravitational force between the Sun and the Earth.

3. What about the charge of the Earth? Can increasing its charge affect its orbit around the Sun?

Similarly to the previous question, the Earth's charge does not significantly affect its orbit around the Sun. The Earth's orbit is determined by the gravitational force between the Earth and the Sun, not by electric charge.

4. Is there a certain amount of charge that could cause the Earth to leave the Solar System entirely?

No, the Earth's orbit around the Sun is stable and it is not possible for electric charge to significantly affect this stability. Other factors, such as the gravitational influence of other planets, would have a much larger impact on the stability of the Earth's orbit.

5. Would a large solar storm or solar flare have any effect on the Earth's orbit?

No, a solar storm or solar flare would not have any significant effect on the Earth's orbit. These events are caused by disturbances in the Sun's magnetic field and do not affect the gravitational force between the Sun and the Earth.

Similar threads

Replies
43
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
Replies
17
Views
2K
Replies
3
Views
907
  • Other Physics Topics
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
909
Replies
29
Views
5K
Back
Top