Recent content by TheMan112

Actually, as I wrote in the edit, I'm taking the less luminous early Sun into consideration by including a dimming factor of 40%. Are you suggesting that a Kelvin-Helmholtz effect would have pushed the frost-/snowline further away from the Sun? Would you know how to introduce such a heatsource...

The Frost Line is the distance from the Sun where the radiation from the Sun becomes too dim to make ice sublimate into vapour. See e.g. http://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29" . It is important because it explains why only terrestrial planets exist inside this boundary. I'm...

Homework Statement I'm trying to model the potential field in and around a symmetrically charged disc where the charge density drops exponentially from the center. Homework Equations This can be done by solving the double integral: \int ^{2 \pi} _{0} \int ^{\infty} _{0} \frac{r e^{-r/b}...

Yeah, it apprears the detectors on WFPC2 can resolve down to 0.04 arcsec, given that lamda = 500 nm we get an angular resolution for the telescope itself at roughly 0.05 arcsec. The other camera appears to have much worse resolutions. Thanks!

Thanks for your thoughts guys, I talked to my professor who agreed on the whole fuzzy criterion thing.

I'm trying to calculate the diffraction limit/angular resolution for the Hubble Space Telescope. I know this can be found using the formula: \theta = 1.22 \frac{\lambda}{D} Where \lambda is the wavelength of the light being observed and D is the diameter of the objective lens (2.5 m on...

Some later thought... 500 nm coincides with: 1. The irradience top of sunlight (and thus any G-type star). See: http://en.wikipedia.org/wiki/File:EffectiveTemperature_300dpi_e.png 2. The sensitivity top of the human eye. 3. The approximate middle wavelength of the visible spectrum...

I'm trying to calculate the diffraction limit/angular resolution for the Hubble Space Telescope. I know this can be found using the formula: \theta = 1.22 \frac{\lambda}{D} Where \lambda is the wavelength of the light being observed and D is the diameter of the objective lens (2.5 m on...

In analogue with my last post I come to the expression for the antiquarks: \bar q (x) = \frac{9 \pi}{8 G^2 M E} \frac{1}{x} \left(\frac{1}{3} \frac{d \sigma^{\nu N}}{dx} - \frac{d \sigma^{\bar \nu N}}{dx} \right) The valence quarks can be calculated by subtracting q(x) and \bar q (x) as...

Am I supposed to get q_V (x) from q(x) but keep \bar q(x)? I tried to extract q(x) from the system of equations, but I got a rather messy expression: q(x) = \frac{9 \pi}{8 G^2 M E} \frac{1}{x} \left(\frac{1}{3} \frac{d \sigma^{\bar \nu N}}{dx} - \frac{d \sigma^{\nu N}}{dx}\right) Not least...

I've been banging my head against the wall for days now trying to figure out how one determines the number of quarks inside the nucleon. I understand it comes from the fact that the Gross-Llewellyn-Smith sum rule is equal to three: \int ^1 _0 F_3 ^N (x) = \int ^1 _0 (u_V (x) + d_V (x)) =...

Using electron-neutron scattering I'm trying to find out how the three quarks (udd) behave inside the neutron. S.Kopeky (Phys. Rev. 1995) found that for small Q2 the equation for the neutrons rms-radius goes towards: -6 \hbar \frac{dG_E ^n (Q^2)}{dQ^2} \right|_{Q^2=0} = -0.113 \pm 0.005...

Seems like that did it, thanks!

Imagine we mounted the muzzle of a gun to a hole in the side of the watertank, fireing into the water horizontally. Haha, I know that, and from the following differential equation... m\ddot{x}-k\dot{x}=0 ...I get: \frac{dx}{dt} = e^{\frac{m}{k} t} Which in turn gives: s =...

Yes, my apologies. I started this thread before I noticed one should ask such questions i the HW-section. I'm going to post my reply to Andy there, you may remove this thread.