A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its

Q: Why is the pressure at the surface of the fluid sphere $P(R) = 0$?

At the outer boundary of the sphere ($r = R$), there is no fluid layer above it to exert any downward force. Since pressure is defined as force per unit area due to the weight of the fluid above, the pressure at the free surface must be zero.

Q: Why does only the enclosed mass $M(r)$ contribute to the gravitational pull at radius $r$?

This is a direct consequence of Newton's Shell Theorem (or Gauss's Law for Gravitation). The gravitational forces exerted by all the outer shells (at radii greater than $r$) on a particle at radius $r$ perfectly cancel each other out, leaving a net force of zero from the outer mass.

Q: Why is there a negative sign in the hydrostatic equation $dP = -\rho g(r) dr$?

The negative sign indicates that pressure decreases as the distance $r$ from the center increases. This makes physical sense because as you move outward, there is less fluid above you, so the weight of the fluid column pressing down decreases.

Q: What would happen to the pressure profile if the density of the fluid was not constant?

If density $\rho(r)$ varied with radius (which is true for real stars where density is higher at the center), the enclosed mass $M(r)$ would be found by integrating the density profile. The resulting pressure profile would no longer be a simple quadratic function but would drop more steeply near the core.

Q: Is the pressure at the center of the sphere really maximum? Why?

Yes, the pressure at the center ($r = 0$) is maximum because the entire weight of the fluid sphere is pressing down toward the center. Even though the gravitational field $g(0)$ is zero at the center, the cumulative weight of all the outer layers creates the highest compression there.

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 options is/are

Options:

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

(B) $\frac{P\left(r = \frac{3R}{4}\right)}{P\left(r = \frac{2R}{3}\right)} = \frac{63}{80}$

(C) $\frac{P\left(r = \frac{3R}{5}\right)}{P\left(r = \frac{2R}{5}\right)} = \frac{16}{21}$

(D) $\frac{P\left(r = \frac{2R}{3}\right)}{P\left(r = \frac{R}{3}\right)} = \frac{20}{27}$

Sigma Percentile · Accepted Answer

### Introduction to Self-Gravitating Fluids

Imagine a massive celestial body—like a newly forming star or a giant gas planet—composed entirely of fluid. Unlike everyday fluids on Earth, which are held down by the Earth's gravity, this cosmic fluid body is held together by its **own self-gravity**.

Every single particle of fluid exerts a gravitational pull on every other particle. This mutual attraction pulls the fluid inward, threatening to collapse the entire sphere into a point.

So, what prevents this catastrophic collapse? It is **hydrostatic pressure**. As the fluid is compressed inward, the pressure deep inside rises, creating an outward force that perfectly balances the inward pull of gravity. This state of perfect balance is known as **hydrostatic equilibrium**.

In this article, we will explore the physics of a self-gravitating fluid sphere of constant density $ho$ and radius $R$, derive its internal pressure profile $P(r)$, and use it to solve a classic JEE Advanced problem.

---

### The Physics of Hydrostatic Equilibrium

To find how pressure varies inside the sphere, let us isolate a thin spherical shell of radius $r$ and thickness $dr$ concentric with the fluid sphere.

Let the pressure at the inner surface of this shell (at radius $r$) be $P$, and the pressure at the outer surface (at radius $r + dr$) be $P + dP$.

The outward force exerted on the inner surface of the shell due to pressure is:

$$F_{	ext{out}} = P \cdot (4\pi r^2)$$

The inward force exerted on the outer surface of the shell due to pressure is:

$$F_{	ext{in, pressure}} = (P + dP) \cdot 4\pi (r + dr)^2 \approx (P + dP) \cdot 4\pi r^2$$

Thus, the net outward force due to the pressure difference is:

$$F_{	ext{net, pressure}} = -4\pi r^2 dP$$

This outward pressure force must balance the inward gravitational force acting on the mass of the shell $dm$. If $g(r)$ is the acceleration due to gravity at radius $r$, the inward gravitational force is:

$$F_{	ext{gravity}} = dm \cdot g(r) = (4\pi r^2 ho \, dr) \cdot g(r)$$

For equilibrium, these forces must be equal in magnitude:

$$-4\pi r^2 dP = 4\pi r^2 ho \, g(r) \, dr$$

Dividing both sides by $4\pi r^2$, we obtain the fundamental equation of hydrostatic equilibrium:

$$\frac{dP}{dr} = -ho g(r)$$

This simple differential equation is the cornerstone of stellar astrophysics!

---

### Finding the Gravitational Field $g(r)$

According to Newton's shell theorem (or Gauss's law for gravitation), the gravitational field $g(r)$ at any point inside a uniform sphere depends only on the mass enclosed within that radius $r$. The mass outside exerts no net gravitational force.

The mass $M(r)$ enclosed within a sphere of radius $r$ is:

$$M(r) = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3}\pi r^3 ho$$

Using Newton's law of gravitation, the gravitational acceleration at radius $r$ is:

$$g(r) = \frac{G M(r)}{r^2} = \frac{G \left(\frac{4}{3}\pi r^3 hoight)}{r^2} = \frac{4}{3}\pi G ho r$$

Notice that $g(r)$ increases linearly with $r$ from the center to the surface. This is a beautiful result: at the very center ($r=0$), gravity is zero, and it reaches its maximum value at the surface ($r=R$).

---

### Deriving the Pressure Profile $P(r)$

Now, let us substitute our expression for $g(r)$ back into the hydrostatic equation:

$$\frac{dP}{dr} = -ho \left(\frac{4}{3}\pi G ho right) = -\frac{4}{3}\pi G ho^2 r$$

To find the pressure $P(r)$ at any radius $r$, we integrate this equation from $r$ to the outer boundary $R$:

$$\int_{P(r)}^{P(R)} dP = -\frac{4}{3}\pi G ho^2 \int_{r}^{R} r \, dr$$

At the outer surface of the fluid sphere ($r = R$), there is no fluid above to press down, so the boundary condition is:

$$P(R) = 0$$

Applying this boundary condition and integrating:

$$0 - P(r) = -\frac{4}{3}\pi G ho^2 \left[ \frac{r^2}{2} ight]_{r}^{R}$$

$$P(r) = \frac{2}{3}\pi G ho^2 (R^2 - r^2)$$

To make our calculations elegant, let us define a positive constant $c$ containing all the constant terms:

$$c = \frac{2}{3}\pi G ho^2$$

Thus, the pressure profile inside the self-gravitating fluid sphere is:

$$P(r) = c(R^2 - r^2)$$

This quadratic profile shows that the pressure is **maximum at the center** ($P(0) = c R^2$) and drops parabolically to **zero at the surface** ($P(R) = 0$).

---

### Verifying the Options

Let us now evaluate each option using our derived pressure profile $P(r) = c(R^2 - r^2)$.

#### Option (a): $P(r=0) = 0$
Substituting $r = 0$ into our profile:
$$P(0) = c(R^2 - 0^2) = c R^2 
eq 0$$
Thus, the pressure at the center is maximum, not zero. **Option (a) is incorrect.**

#### Option (b): $\frac{P\left(r = \frac{3R}{4}ight)}{P\left(r = \frac{2R}{3}ight)} = \frac{63}{80}$
Let us calculate the pressure at both radii:
$$P\left(\frac{3R}{4}ight) = c\left(R^2 - \frac{9R^2}{16}ight) = \frac{7}{16}c R^2$$
$$P\left(\frac{2R}{3}ight) = c\left(R^2 - \frac{4R^2}{9}ight) = \frac{5}{9}c R^2$$
Taking the ratio:
$$\frac{P(3R/4)}{P(2R/3)} = \frac{\frac{7}{16}c R^2}{\frac{5}{9}c R^2} = \frac{7}{16} 	imes \frac{9}{5} = \frac{63}{80}$$
**Option (b) is correct.**

#### Option (c): $\frac{P\left(r = \frac{3R}{5}ight)}{P\left(r = \frac{2R}{5}ight)} = \frac{16}{21}$
Let us calculate the pressure at both radii:
$$P\left(\frac{3R}{5}ight) = c\left(R^2 - \frac{9R^2}{25}ight) = \frac{16}{25}c R^2$$
$$P\left(\frac{2R}{5}ight) = c\left(R^2 - \frac{4R^2}{25}ight) = \frac{21}{25}c R^2$$
Taking the ratio:
$$\frac{P(3R/5)}{P(2R/5)} = \frac{\frac{16}{25}c R^2}{\frac{21}{25}c R^2} = \frac{16}{21}$$
**Option (c) is correct.**

#### Option (d): $\frac{P\left(r = \frac{2R}{3}ight)}{P\left(r = \frac{R}{3}ight)} = \frac{20}{27}$
Let us calculate the pressure at both radii:
$$P\left(\frac{2R}{3}ight) = \frac{5}{9}c R^2$$
$$P\left(\frac{R}{3}ight) = c\left(R^2 - \frac{R^2}{9}ight) = \frac{8}{9}c R^2$$
Taking the ratio:
$$\frac{P(2R/3)}{P(R/3)} = \frac{\frac{5}{9}c R^2}{\frac{8}{9}c R^2} = \frac{5}{8} = \frac{15}{24} 
eq \frac{20}{27}$$
**Option (d) is incorrect.**

---

### Conclusion

By combining the principles of **Newtonian gravity** and **fluid mechanics**, we successfully derived the quadratic pressure distribution inside a self-gravitating fluid sphere. The correct options are **(b)** and **(c)**.

Animated Solution for Physics - Gravitation: 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 options is/are

Select Answer:

Visualized Solution

Visualizing the Self-Gravitating Fluid Sphere

The Condition for Hydrostatic Equilibrium

Finding the Enclosed Mass and Gravity $g (r)$

Setting up the Differential Equation

Integrating to Find the Pressure Profile $P (r)$

Simplifying the Pressure Equation

Verifying Options (a) and (b)

Verifying Options (c) and (d)

The Sigma Insight: Gravitational Potential and Potential Energy

Introduction to Self-Gravitating Fluids

The Physics of Hydrostatic Equilibrium

Finding the Gravitational Field $g (r)$

Deriving the Pressure Profile $P (r)$

Verifying the Options

# Option (a): $P (r = 0) = 0$

# Option (b): $\frac{P ( r = \frac{3 R}{4} )}{P ( r = \frac{2 R}{3} )} = \frac{63}{80}$

# Option (c): $\frac{P ( r = \frac{3 R}{5} )}{P ( r = \frac{2 R}{5} )} = \frac{16}{21}$

# Option (d): $\frac{P ( r = \frac{2 R}{3} )}{P ( r = \frac{R}{3} )} = \frac{20}{27}$

Conclusion

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