I have solved it using v^2=(GM/R)^1/2 and M=(4/3*3.14*R^3)1/R
If density is proportional to 1/r, the keplerian velocity at 1000 parsecs is 75km/s
And if the density is 1/r^2, then the speed is constant at all radiuses.
I Think this does make sense?
Silly me, haven't noticed the "nucleon" part in a "dark matter-nucleon" expresion...
Now there is only 4.65*10^-3 events per year, that is a bit low i think, as you will need 215 square metres of water to observe one event per year.
If the radius of a Xe atom is 10^-10m or 100pm and there is 2.5*10^25 atoms in a cubic metre, then the area* is 900000 square metres. this means that a particle will interact with 900000 Xe atoms when crossing a metre of a gass, that doesn`t make sense to me...
*area = all atoms in the cubic...
I tried to take a sqare metre box, and calculate the total area of the xenon atoms in the box, and i found out that the area is 165 000 square metres, which gives 100% impact probability. Thats not good...
The sun orbits the galaxy at 220 km/s at a distance of 8500 parsecs from the centre of the galaxy. If the density of matter in the galaxy drops off proportional to 1/r, what would be the Keplerian velocity of a star orbiting at a radius of 1000 parsecs? What would be the Keplerian velocity of...
Assume that the density of dark matter near the Earth is 0.3 GeV / cm3 and that the dark matter particle has a mass of 100 GeV and a velocity of 200 km/s. If the dark matter-nucleon cross section is 10-44cm2 calculate how many events you would expect to see every year in a metre cubed volume of...