# Magnetism, Orbits, Friction-Is there a relationship?

1. Jan 5, 2004

### timejim

I hope this is the right Thread for this discussion. I was wondering by what actions do Planetary Bodies, such as the Earth, continue in their orbits, millenia after millenia and so on, without decaying and being "sucked" into the Sun, in a rather short period of Time. Something doesn't add up to me. If the Sun has "Magnetic" force radiating outwards from it in all directions, why, as the Earth orbits the Sun, does not its Rotation slow down, and its orbit Decay, as it moves thru this "field of magnetic attraction?". Wouldn't the surrounding "fields", as it interacts with the Earths magnetic field, create a sort of "Magnetic" friction? Would this not cause the Earths' roattion to slow? If not, why not? Likewise, would the Earths orbit be affected by the slowing of the Earths rotation from the friction induced by the magnetic field? If you pass 2 magnets close to each other they react in a physical way, i.e., moving, twisting, etc. Wouldn't this be the case with regards to the Earth traveling thru these radiated magnetic fields from the Sun? Just asking, it does interest me, though.

2. Jan 5, 2004

### Integral

Staff Emeritus
When the solar system fromed it had a significant amount of spinning motion, this motion was transfered to the planets as they formed. Thus the orbital motion of the planets is due to initial angular Momentum of the solar system.

Gravity is a weak but far reaching force, at the distances involved in the solar system magnetic forces are very small, thus have no signifigant effect on planetary motion.

If the earths rotation about the sun does slow it will not cause it to fall into the sun, just the opposite, a lower orbital velocity will mean that it must move to a orbit FURTHER from the sun.

Janus?

3. Jan 5, 2004

### Jimmy

$$V_o = \sqrt{\frac{G M}{R}}$$

If you arrange that to solve for R:

$$R = \frac{G M}{V^2}$$

It would seem that if V decreases, R would increase. But the first formula is for determining the velocity one would need to maintain a circular orbit at a particular distance.

You would have to increase your velocity to reach a higher orbit and then decelerate to the new orbital velocity depending on the height of the new orbit. * see http://users.commkey.net/Braeunig/space/orbmech.htm [Broken]

If the earth were to stop dead in its tracks, it would fall into the sun.

Last edited by a moderator: May 1, 2017
4. Jan 5, 2004

### Staff: Mentor

The forces involved in the interaction of the magnetic fields are not big enough to be relevant to the orbits of the planets. They lose strength quickly with distance.

edit: can't think and type at the same time.

Last edited: Jan 6, 2004
5. Jan 5, 2004

### waynet

This thread kind of makes me wonder of something else that I have thought about. What about the expansion of the universe? If the stretching of space is uniform, would it have an effect on smaller scales like the solar system? Say for instance, if the universe had doubled in size the last five billion years, wouldn't it have a similar effect on expanding the solar system? Or do you think local gravity would completely overcome this force?

6. Jan 6, 2004

### timejim

Good thought. I wonder about this, too. If everything is expanding away, what about solar systems or even Galaxies. Eventually you would be alone in the Universe. Doesn't make sense to me, formulas or no formulas.

7. Jan 6, 2004

### Janus

Staff Emeritus
Actually, you would have to accelerate to match match the orbital velocity of the higher orbit.

Assuming you are moving from an orbit of radius R1 to an orbit of radius R2.

You increase your velocity at some point of orbit 1. This will put you in a new orbit. This new orbit will be an eliptical one, with its perhelion (closest distance to the Sun) being at orbit 1 and its aphelion(furthest distance to the Sun) being at orbit 2.

Essentially, any change in velocity you make at an particular point of an orbit does not effect the orbit at that point but changes the orbital distance of the point 180Â° away.

At perhelion, it will will be traveling much faster (by the amount of the added velocity) than the orbital velocity of orbit 1.

As it climbs out to aphelion it will lose velocity such that by the time it reaches orbit 2, it will be traveling much slower than orbital velocity of orbit 2.

At which point you would accelerate to add velocity to match orbit 2. (By adding velocity at aphelion you increase the distance of perheion. Increase it enough, and you raise the perhelion to the same distance as aphelion and you get a circular orbit. )

One formula for determining the velocity at aphelion of an eliptical orbit is

$$V_{a}= \sqrt{\frac{2GM}{(R_{p}+R_{a})}\frac{R_{p}}{R_{a}}}$$

$$R_{a}$$ is the distance of the aphelion
and
$$R_{p}$$ is the distance of the perhelion

To get the velocity at perhelion, just transpose the values for perhelion distance and aphelion distance in the formula.

The formula for determining how much velocity you need to add to orbit 1 in order to enter the eliptical orbit needed to reach orbit 2 is

$$\Delta V = V_{R_{2}}\left[ \sqrt{\frac{2R_{1}}{R_{1}+R_{2}}}-1 \right]$$

$$V_{R_{1}}$$ is the orbital velocity of orbit 1.

Last edited: Jan 6, 2004
8. Jan 6, 2004

### alexbib

How do you get to these equations?

edit:

Oh wait, you have m*ra*va=m*rp*vp because angular momentum is conserved, and 1/2mva^2+-GMm/ra=1/2mvp^2+-GMm/rp and you just solve these two equations for what you want. Is this right?

Last edited: Jan 6, 2004
9. Jan 6, 2004

### Jimmy

Thanks for that correction Janus. You made it very easy to understand. I guess that's why you're a mentor, eh? I'll definitely study those formulas.

Last edited: Jan 6, 2004
10. Jan 6, 2004

### krab

Given that space junk in insufficiently high orbit eventaully comes crashing to earth, your comments surprise me.

11. Jan 6, 2004

### alexbib

yeah, he's mistaken. At distance r, the earth always accelerate towards the sun at the same rate. If it is moving sufficiently fast(tengentially), sqrt(GM/r), the acceleration will only make it stay in orbit. However, if it is slower, it will fall into the sun. (centrifugal force not strong enough to cancel the sun's attraction, if you look at it along the earth to sun |distance| axis)

Last edited: Jan 6, 2004
12. Jan 6, 2004

### Jimmy

Well, everyone makes mistakes from time to time.

Actually, whether the earth would fall into the sun depends on how much the velocity decreased. It could just fall into an orbit with a higher eccentricity if the change in V was small enough.

13. Jan 6, 2004

### Janus

Staff Emeritus
Correct. In fact, the Earth would have to lose about 99% of its present orbital velocity of 30km/sec in order to hit the Sun.

This means it would take a greater velocity change (and thus energy expended) to cause the Earth to fall into the Sun than it would take to throw the Earth completely free of the Solar System. (about 41% of its present orbital velocity.)

Last edited: Jan 6, 2004
14. Jan 6, 2004

### Staff: Mentor

Surprisingly enough, thats atmospheric drag! In space!

15. Jan 6, 2004

### alexbib

Cool, thx for precising that! Sounds logical, as long as the orbit has a perigee higher than the sun's radius, it should not fall into the sun. (It'd probably get a bit too hot for comfort though )

16. Jan 6, 2004

### Jimmy

Thanks Janus. I wasn't sure about the exact numbers. In fact, I was going to asked what the maximum change in velocity would need to be for the Earth to fall into the sun.

Gee, I'm not sure which fate would be worse.

17. Jan 9, 2004

### Vanja

Simple solution

These I like best. Simple solutions to questions.
Check out http://www.geocities.com/terella1/ regarding planets and magnetism. Fact is that magnetic fields do not rotate with the planet.
May sound silly. But ... Just check it out!

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