Since the sphere is uniform, it must have a constant density, =3M/4R³

 

Consider a shell of width dr at radial distance of r from the centre of the sphere. This shell experiences a pressure due to gravitational force(a type of load from the mass above it i.e mass for rr'R)

 

Now, every shell at a radial distance of r experiencing gravitational force only due to the mass present for 0r'r. By Third law of motion, it is applying an equal and opposite compressing force on the inner portion

 

So, by Newton's Universal Law of Gravitation

dF= - (4G(4/3)r³)dr

 

Since pressure is defined as

P(r)=dF(r)/dA

 

Pressure due to a shell of width dr at radial distance of r:

P_i(r)= - (4G(4/3)r³)dr/8rdr= - (2/3)²Gr²

 

So, net pressure experienced by a shell of width dr at radial distance of r

P(r)=_{[R ]}^{[r ]} P'(r)dr=(2/3)²G(R²-r²)

 

Since  =3M/4R³

 

P(r)=(3GM²)(1 - r²/R²)/(8R⁴)

 

 

 

 

Substitute the following in the above equation:

r=0

R=R_E

M=M_E

 

So, P_centre=(3GM_E²)/(8R_E⁴)

 

One can also get the numerical value of above expression, which comes nearly 1.8*10¹¹ Pa

Do not be confused by questioning "what is the variable r?"

 

Variable is simply as radial variable, which is used in different context to refer to the radial distance. It is not something which is fixed to denote a particular thing.

 

 

 

PS: Earlier{ at the begining of 11th} i was also confused with use of such notation. But latter got used to it. Such notations are used in old books.