Integrating dP/dr over the Sun

[CaptainEvil]

Homework Statement

Consider the Sun to be a sphere of hydrogen gas of uniform density equal to its
average density (1440 kg/m3).
a) Integrate the equation of hydrostatic equilibrium for dP/dr from the Sun’s radius to the
core to estimate the central pressure in Pa and in atmospheres. Assume the pressure at the
surface is zero. [Note: your estimate will fall far short of the true value that can be
derived using a more realistic density profile, about 250 billion atmospheres, but that’s ok
for our purposes]

Homework Equations

dP/dr = -Gmp/r^2 (p = density)

The Attempt at a Solution

I replaced m with 4/3pir^3p which yields
dP/dr = -4piGrp^2/3
and integrated from R to 0 yielding 2piGp^2R^2/3
and filling in the values and Radius of the sun, I get 1.4e6 atm of pressure.
Does this make sense? (Is it consistent with what the question is asking)

 

[CaptainEvil]

bump - need help please

 

[thomate1]

I would say that your attempt at a solution is a good start to estimating the central pressure of the Sun. However, there are a few things to consider when integrating dP/dr over the Sun.

First, the equation of hydrostatic equilibrium is usually written as dP/dr = -Gm(r)p(r)/r^2, where m(r) is the mass enclosed within a radius r and p(r) is the density at that radius. Since the Sun is a sphere of uniform density, your equation for dP/dr is correct, but it would be more accurate to write it as dP/dr = -4piGr^2p.

Second, when integrating over the entire Sun, you should use the total mass of the Sun, not just the mass enclosed within a certain radius. This means that you should use the mass of the entire Sun (M = (4/3)piR^3p) in your integration, not just (4/3)piR^3p.

Finally, your estimate of 1.4e6 atm is significantly lower than the true value of 250 billion atmospheres. This is because your calculation assumes a uniform density throughout the Sun, which is not the case in reality. The density of the Sun increases significantly towards the core, leading to much higher central pressures.

In conclusion, your attempt at a solution is a good start, but to get a more accurate estimate of the central pressure of the Sun, you would need to consider the varying density profile and use the total mass of the Sun in your integration.