## All eyes on #Rosetta: About micro-gravity, solar radiation pressure and solar winds

Good Morning space geeks, I’m almost ready for my trip to the German Aerospace Center in Cologne to watch Philae’s commissioning in their Microgravity User Support Center (MUSC). But since I have a few hours to kill I thought I share a few thoughts with you about the microgravity in 67P’ vicinity and how the solar wind and solar radiation pressure affects Rosetta’s orbit around 67P. Warning: May contain mathematics :-)

So, what are the forces acting on Rosetta while lurking around 67P? The primary forces are:

• Gravity of 67P
• Solar wind pressure

There’s also the gravity of the sun, but since 67P and Rosetta are very close to each other and share the same orbit around the sun, we neglect this for now.

Before starting, I ask you to remember a few basic pinciples.

Newton’s Law of Universal Gravitation. This law say that the force (in Newton) between two masses is the product of their mass multiplied with the gravitational constant, divided by the square of their distance:

$F=\frac{G*m_1*m_2}{d^2}$

So, what’s the force acting between Rosetta and 67P if Rosetta is on a circular, 1000 meter orbit? From Wikipedia I learned that 67P’s mass is about $3.14*10^{12}\,kg$; Rosetta’s mass is about two metric tons, so $2000\,kg$. The gravitational constant is, as (as we think) everywhere in the observable universe, $G\,\approx\,6.67\,*10^{-11}\,m^3kg^{-1}s^{-2}$.

$F=\frac{6.67*10^{-11}\,m^3\,kg^{-1}\,s^{-2}*3.14*10^{12}\,kg*2000\,kg}{(1000\,m)^2}\approx 0.4191\,N$

So, only about 0.4 Newton; that is not a lot. As the ever helpful Wolfran|Alpha search engine suggests, this is the peak force your fingers exert on your keyboard while typing.  When I first did this calculation I really wondered how you could possibly keep Rosetta in a stable orbit around 67P with such little forces at work when Rosetta is constantly being bombarded with solar radiation and when the solar wind keeps blowing.

But as always with physics, assumptions are nice, but if you can do the math, just do it. So: Is the combined forces of the solar wind and solar radiation pressure big enough to affect Rosetta and would Rosetta need to do a lot of correction burns to keep it’s orbit stable?

Let’s head over to the radiation first. To figure out the force we need the solar radiation pressure and multiply this with the effective cross-section of Rosetta. I assume a cross-section of 1.5 m^2 – that might be wrong, but the principle of calculation stands.

First, a definition which was deliberately marked with “DO NOT CITE”, but I found this definition most easy to understand:

“Radiation pressure is the pressure exerted upon any surface exposed to electromagnetic radiation. If absorbed, the pressure is the power flux density divided by the speed of light. If the radiation is totally reflected, the radiation pressure is doubled.”

So, what is the power flux (measured in Watt per meters squared) at Rosettas current position? Rosetta is, as of today, about 4.25 AU away from the sun. The power flux on earth is around  $1370 Wm^{-2}$ so we need to use the inverse-square law to calculate the power flux at 4.25 AU.

The Inverse power law states:

$S=S_0*(r_0/r)^2$

Hence:

$1370\,Wm^{-2}*(1\,AU/4.25\,AU)^2=75.8\,Wm^{-2}$

However, Rosetta is not absorbing all the power 100%; I have no figures, so I assume it absorbs about 50% ($d=0.5$) of all the radiation. We will now calculate the effective radiation pressure from the power flux at Rosetta’s current position.

$P=S/(c/(1+d))$
$P=3.793*10^{-7}\,kg/m^2$

Now multiply by Rosetta’s effective cross-section to convert the pressure to force:
$F=P * 1.5\,m^2 = 5.69*10^{-7}\,N$

So… compared to those 0.4191 Newton, that’s not a lot! But what about the solar wind? The formular for calculatng the solar wind pressure at a distance of 1 AU is:

$P=1.6727*10^{-6}\,N/m^2*n*V2$

where pressure P is in nPa (nano Pascals), n is the density in particles/cm3 and V is the speed in km/s of the solar wind. Wolfran|Alpha told me today the parameter for n and V2:

$V2 = 438.7\,km/s$
$n = 1.4\,Protons/cm^3$

We also need to apply the inverse power law again to draw conclusions for the distance at 4.25 AU:

$P=P_0*(r_0/r)^2$
$P=1.6726*10^{-6}\,Nm^{-2}*(1\,AU/4.25\,AU)^2$
$P=9.26*10^{-8}\,Nm^{-2}$
$P*1.5\,m^2=1.39*10^{-7}\,N$

Not a lot either! Now add these forces and compare to the gravitational forces between 67P and Rosetta:

$F_{wind} + F_{radiation} =$
$1.39*10^{-7}\,N + 5.69*10^{-7}\, N = 7.079*10^{-7}\, N$

$7.079*10^{-7}\,N << 0.4191\,N$

So we showed that solar wind and radiation is so little that it won’t affect Rosetta’s orbit much.