Question

A spherical body of radius r consists of a fluid of constant density and is in equilibrium under its own gravity. If p(r) is the pressure at r(r < r), then the c

Answer 1

Explanation:

dp/dr=−ρg(r)

                g(r)=Gm(r)/r²;    m(r)=4/3πρr²

                 ⇒g(r)=4/3πρGr

                ⇒dp/dr=4/3πρ²Gr

                ⇒p(r)−p(R)=4/3πρ²G(R²−r²/2)

                ⇒p(r)−p(R)=2/3πρ²G(R²−r²)

                put p(R) = 0

                ⇒p(r)=2/3πρ²G(R²−r²)

               

p(3R/4)/p(2R3)=1−9/16/1−4/9=9/16×7/5=63/80

                p(3R/5)/p(2R/5)=1−9/25/1−4/25=16/21

                 p(R/2)p(R/3)=1−1/4/1−1/9=9/4×3/8=27/32