The sun creates this heliosphere by sending a constant flow of particles and a magnetic field out into space at over 670,000 miles per hour, which is also known as solar wind. If the speed of the wind is that great, why doesn't it disrupt the planets in our solar system? Does it just prevent foreign particles from interstellar space entering the heliosphere?
[Physics] Why doesn’t the solar wind disrupt the planets
astrophysicssolar systemsolar-wind
Related Solutions
Background
In the absence of an electric field, a charged particle experiences a force that is perpendicular to the magnetic field and its velocity relative to that field, called the Lorentz force. This is given by: $$ \mathbf{F}_{s} = q_{s} \ \mathbf{v}_{s} \times \mathbf{B} \tag{1} $$ where $q_{s}$ is the charge of species $s$, $\mathbf{v}_{s}$ is the velocity of species $s$ with respect to the magnetic field, $\mathbf{B}$. We can approximate the plasma as a fluid, i.e., magnetohydrodynamics, which will help simplify some things.
In the fluid limit, we can show that Ohm's law, in a frame of reference moving relative to the fluid at velocity $\mathbf{U}$ (or equivalently, one can say we are at rest and the fluid moves at $\mathbf{U}$), is given by: $$ \mathbf{j} = \sigma \left( \mathbf{E} + \mathbf{U} \times \mathbf{B} \right) \tag{2} $$ where $\mathbf{j}$ is the current density and $\sigma$ is the electrical conductivity. Note that this is in the non-relativistic limit, where we can approximate by using gallilean transformations (e.g., $\mathbf{E}' = \mathbf{E} + \mathbf{U} \times \mathbf{B}$) instead of Lorentz transformations. We can then use Ampère's law combined with Ohm's law (e.g., Equation 2 above) to show: $$ \nabla \times \mathbf{B} = \mu_{o} \ \sigma \left( \mathbf{E} + \mathbf{U} \times \mathbf{B} \right) \tag{3} $$ where $\mu_{o}$ is the permeability of free space. We can take the curl of Equation 3 and combine with Faraday's law to show: $$ \partial_{t} \mathbf{B} = \nabla \times \left( \mathbf{U} \times \mathbf{B} \right) + \frac{1}{\mu_{o} \ \sigma} \nabla^{2} \mathbf{B} \tag{4} $$
In the limit as $\sigma \rightarrow \infty$, one can further show that: $$ \frac{d \Phi_{B}}{dt} = \int_{S} \ dA \ \left[ \partial_{t} \mathbf{B} - \nabla \times \left( \mathbf{U} \times \mathbf{B} \right) \right] \cdot \hat{\mathbf{n}} = 0 \tag{5} $$ where $\Phi_{B}$ is the magnetic flux and $dA$ is an arbitrary surface with unit normal vector, $\hat{\mathbf{n}}$. Equation 5 is known as the frozen-in condition because it states that the magnetic fields are tied to the fluid.
Answer
Why do both the positive and negative particles follow these field lines and flow into space?
The answer is two fold relating to the frozen-in condition being partially satisfied and the Lorentz force (Equation 1 above). Equation 1 shows that if a particle tries to move orthogonal to a magnetic field, the field will turn it back towards the field resulting in roughly circular motion (when considering only magnetic fields and only the perpendicular components of the velocity). There are intuitive reasons for why this should be so, as I previously stated in this answer.
If, however, a particle's velocity ($\mathbf{v}_{s}$) is parallel to the magnetic field, it will experience no force in the absence of electric and gravitational fields. Thus, it is very easy for particles to move along the magnetic field but difficult to move across it.
The frozen-in condition also shows that the field and particles are tied to each other, such that if one changes the other changes to compensate (assuming the changes are slow enough).
When the particles that make up cosmic rays get boosted up to the speeds at which we see them, they're ionized. The electrons and nuclei are separated from the start.
Both types of particle are going to be releasing energy every time they accelerate (synchrotron radiation) and when they interact with photons (inverse Compton scattering). When it comes to inverse Compton scattering, the electrons scatter much more than the alpha particles - there's a $\frac{1}{m}$ factor in both the energy exchange and the angle involved.
What this results in is basically the electrons going in all directions, gradually losing energy. Meanwhile, the alpha particles continue on with less deflection and with most of their energy intact. The electron is the ping-pong ball to the alpha particle's bowling ball. (4 bowling balls, really, but you get the idea.)
We still get electrons from space, but more of them are coming in slow, which puts them out of the "cosmic ray" category.
Best Answer
The Solar wind does indeed exert a force on the planets, however it turns out that the force is so small that it has no measurable effect.
The force can be calculated using the fact that force is equal to the rate of change of momentum. Suppose the total mass of all the Solar wind particles hitting the Earth per second is $M$, and the average velocity of the particles is $v$, then the force the solar wind exerts on the Earth is simply:
$$ F = Mv $$
Off-hand I don't know what the mass flux and velocity are, but the Wikipedia article on the solar wind reports the pressure, $P$, produced by the wind at the Sun-Earth distance to be 1 to 6 nano-Pascals. The total force on the Earth is this pressure multiplied by the cross sectional area $\pi r^2$. The radius of the Earth is about 6,371,000 metres, so we get:
$$ F = P \times \pi r^2 \approx 130 \,\text{to}\, 800 \,\text{kN} $$
To see why this is negligible, let's compare it with the gravitational force between the Sun and the Earth. This is given by Newton's law of gravity:
$$ F = \frac{GM_\text{Sun}M_\text{Earth}}{r^2} $$
and it works out to be:
$$ F \approx 3.54 \times 10^{22} \,\text{N} $$
so the force from the Solar wind is only about 0.000000000000001% ($10^{-15}\%$) of the gravitational force.