Question 411: Properties of Matter

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 correct option(s) is(are)

Accepted Answer

The acceleration due to gravity at a radial distance r (r < R) from the centre of a sphere of constant mass density $$ho$$ is given by
$$g = {{G\left( {{4 \over 3}\pi {r^3}} ight)ho } \over {{r^2}}} = {{4\pi ho G} \over 3}r$$.
Consider a spherical shell of radius r and thickness dr.
Let P and P + dP be the fluid pressures inside and outside the shell. Consider a small element of area A and mass $$dm = ho Adr$$ on the shell. The forces on this element are gravitational force (dm g) direted inwards, the force due to pressure of the fluid outside the shell $$((P + dP)A)$$ directed inwards, and the force due to pressure of the fluid inside the shell (PA) directed outwards. In equilibrium, net force on the element is zero, i.e.,
$$(ho Adr){4 \over 3}\pi ho Gr + (P + dP)A = PA$$, or
$$dP =  - {4 \over 3}\pi G{ho ^2}r\,dr$$.
Note that the pressure decreases with increase in r. The pressure is zero at the surface of the sphere i.e., P = 0 at r = R. Integrate to get
$$\int_0^{P(r)} {dP = P(r) =  - {4 \over 3}\pi G{ho ^2}\int_R^r {rdr} } $$
$$ = {2 \over 3}\pi G{ho ^2}({R^2} - {r^2})$$.
Now, $$P(r = 0) = {{2\pi } \over 3}{ho ^2}G{R^2} 
e 0$$
$${{P(r = 3R/4)} \over {P(r = 2R/3)}} = {{{R^2} - {{(3R/4)}^2}} \over {{R^2} - {{(2R/3)}^2}}} = {7 \over {16}} 	imes {9 \over 5} = {{63} \over {80}}$$
$${{P(r = 3R/5)} \over {P(r = 2R/5)}} = {{{R^2} - {{(3R/5)}^2}} \over {{R^2} - {{(2R/5)}^2}}} = {{16} \over {25}} 	imes {{25} \over {21}} = {{16} \over {21}}$$
$${{P(r = R/2)} \over {P(r = R/3)}} = {{{R^2} - {{(R/2)}^2}} \over {{R^2} - {{(R/3)}^2}}} = {3 \over 4} 	imes {9 \over 8} = {{27} \over {32}} 
e {{20} \over {27}}$$

Answer

$$P(r = 0) = 0$$

Answer

$${{P(r = 3R/4)} \over {P(r = 2R/3)}} = {{63} \over {80}}$$

Answer

$${{P(r = 3R/5)} \over {P(r = 2R/5)}} = {{16} \over {21}}$$

Answer

$${{P(r = R/2)} \over {P(r = R/3)}} = {{20} \over {27}}$$

Properties of Matter

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 correct option(s) is(are)

Choose the correct answer: