Mass lumosity relationship stars and bars

The Sun and Stellar Structure

ble 2; note that the error bars are smaller than the plotted symbols. Relative to A91, 5 The mass-luminosity relation for the stars in Table 2. The correlation coefficients are 0~ for the i i~ high-mass stars (M ~ o~ I The empirical mass-luminosity relationship derived from data in Popper's review article'. show the mean error bars for the four groups of stars identified in the key. If you plot the masses for stars on the x-axis and their luminosities on the y-axis, you can calculate that the relationship between these two quantities is.

So what does this have to do with planets affecting solar variation? To understand this we need to consider why the Mass-Luminosity relation holds. From Penn State University: Given our theory for the structure of stars, you can understand where this relationship comes from. Stars on the Main Sequence must be using the energy generated via nuclear fusion in their cores to create hydrostatic equilibrium. The condition of hydrostatic equilibrium is that the pressure is balancing gravity.

If you increase the pressure inside a star, the temperature will also increase. So, the cores of massive stars have significantly higher temperatures than the cores of Sun-like stars. At higher temperatures, the nuclear fusion reactions generate energy much faster, so the hotter the core, the more luminous the star.

Why should there not be more massive and hence brighter and less massive hence fainter stars?

The Mass-Luminosity Relationship

An 80 solar mass star is not that much bigger than the Sun, but its luminosity is times greater. This means that even a slight difference in the mass among stars produces a large difference in their luminosities.

For example, an O-type star can be only 20 times more massive than the Sun, but have a luminosity about 10, times as much as the Sun. Putting together the principle of hydrostatic equilibrium and the sensitivity of nuclear reaction rates to temperature, you can easily explain why. Massive stars have greater gravitational compression in their cores because of the larger weight of the overlying layers than that found in low-mass stars. The massive stars need greater thermal and radiation pressure pushing outward to balance the greater gravitational compression.

The greater thermal pressure is provided by the higher temperatures in the massive star's core than those found in low-mass stars. Massive stars need higher core temperatures to be stable! The nuclear reaction rate is very sensitive to temperature so that even a slight increase in temperature makes the nuclear reactions occur at a MUCH higher rate. This means that a star's luminosity increases a lot if the temperature is higher.

Mass–luminosity relation

This also means that a slight increase in the mass of the star produces a large increase in the star's luminosity. Mass Cutoff Explained The principle of hydrostatic equilibrium and nuclear fusion theory also explain why stars have a certain range of masses. The stars have masses between 0. The potential energy of a spherical mass distribution is This and the volume are substituted in to give.

Here is the average mass of gas particles within the star. Substituting this equation into the initial luminosity equation, along with yields. A slightly more exact result can be achieved by considering that the above equation gives the average temperature based on the average pressure, but what is actually needed is the surface temperature. Because stars are much hotter in the center than near the surface, one also needs to estimate the relationship between the surface temperature and internal temperature.

Mass Luminosity relation

The center is so much hotter since energy takes a long time to escape, otherwise, thermodynamic equilibrium would be achieved quickly and the temperature would be nearly uniform throughout.

A random walk analysis can be used to estimate the "delay factor", i.

Mass–luminosity relation - Wikipedia

In reality, the mean free pathof photons within the Sun depends on the density and temperature, but here it will be approximated as a constant. After N interactions, resulting in N vector displacements in random directions, the distance traveled is: The square of the net displacement is: Averaging over many random direction changes, the terms involving dot products cancel since the direction is random.

Hence, for largeThus, to escape from the Sun, on average steps are required.