The mass density of a planet of radius R varies with the distance r from its centre as rho(r) = rho_

Q: Why do we use integration to find the mass?

Because the density is not constant. It varies with the distance from the center. We must sum up the mass of infinitesimally thin shells to find the total mass.

Q: Why does the field formula only use enclosed mass?

According to Gauss's law for gravitation (or Newton's shell theorem), a spherically symmetric mass distribution exerts no net gravitational force on an object located inside it. Only the mass closer to the center than the object contributes to the net inward pull.

Q: Why do we set the derivative to zero?

In calculus, the local maxima or minima of a continuous function occur where its first derivative is zero. Setting $\frac{dE}{dx} = 0$ allows us to find the critical points.

Q: What happens if $x > R$?

If $x > R$, the point is outside the planet. The enclosed mass would just be the total mass of the planet, which is a constant. The field would then simply decrease as $\frac{1}{x^2}$.

Q: Why did we ignore the negative root when taking the square root?

The variable $x$ represents a radial distance from the center of the planet, which is a physical length. Therefore, it must be a positive value.

Question

The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $\rho(r) = \rho_0 \left(1 - \frac{r^2}{R^2}\right)$. Then, the gravitational field is maximum at

Options:

(A) $r = \frac{\sqrt{3}}{4} R$

(B) $r = R$

(C) $r = \sqrt{\frac{5}{9}} R$

(D) $r = \frac{1}{\sqrt{3}} R$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a planet where the mass isn't distributed evenly. Instead of being uniform, the planet is denser at the core and gradually becomes less dense as you move towards the surface. This is mathematically described by the variable density function $ho(r) = ho_0 \left(1 - \frac{r^2}{R^2}ight)$.

Our goal is to find the exact distance $x$ from the center where the gravitational pull (or field) is the absolute strongest. To do this, we need to understand how much mass is pulling on an object at any given distance $x$.

### The Master Equation
According to Newton's shell theorem (or Gauss's law for gravitation), the gravitational field $E$ at a distance $x$ inside a spherically symmetric body depends **only** on the mass enclosed within that radius $x$. The mass outside this radius exerts no net gravitational force.

The formula is:
$$E = \frac{G M_{	ext{enclosed}}}{x^2}$$

Because the density is variable, we cannot simply multiply the total volume by a constant density. We must build the enclosed mass layer by layer. Imagine an infinitely thin spherical shell of radius $r$ and thickness $dr$. The volume of this shell is $dV = 4\pi r^2 dr$. The mass of this tiny shell is $dM = ho(r) dV$.

### Integrating the Mass
To find the total enclosed mass $M_{	ext{enclosed}}$ up to our distance $x$, we integrate these elemental masses from the center ($r=0$) to $x$:
$$M_{	ext{enclosed}} = \int_0^x ho_0 \left(1 - \frac{r^2}{R^2}ight) 4\pi r^2 dr$$

Let's expand and solve this integral:
$$M_{	ext{enclosed}} = 4\pi ho_0 \int_0^x \left(r^2 - \frac{r^4}{R^2}ight) dr$$
$$M_{	ext{enclosed}} = 4\pi ho_0 \left[ \frac{r^3}{3} - \frac{r^5}{5R^2} ight]_0^x$$
$$M_{	ext{enclosed}} = 4\pi ho_0 \left( \frac{x^3}{3} - \frac{x^5}{5R^2} ight)$$

### Finding the Maximum Field
Now, we substitute this enclosed mass back into our gravitational field equation:
$$E = \frac{G}{x^2} \left[ 4\pi ho_0 \left( \frac{x^3}{3} - \frac{x^5}{5R^2} ight) ight]$$
$$E = 4\pi G ho_0 \left( \frac{x}{3} - \frac{x^3}{5R^2} ight)$$

We now have a function for the gravitational field $E$ in terms of $x$. To find where this field is maximum, we turn to calculus. The maximum value occurs where the derivative of the function with respect to $x$ is zero.
$$\frac{dE}{dx} = 0$$

Differentiating our expression for $E$:
$$\frac{d}{dx} \left[ 4\pi G ho_0 \left( \frac{x}{3} - \frac{x^3}{5R^2} ight) ight] = 0$$
$$4\pi G ho_0 \left( \frac{1}{3} - \frac{3x^2}{5R^2} ight) = 0$$

Since the constants cannot be zero, the term inside the parenthesis must be zero:
$$\frac{1}{3} = \frac{3x^2}{5R^2}$$

### Final Calculation
Solving for $x^2$:
$$x^2 = \frac{5R^2}{9}$$

Taking the square root (and keeping only the positive value since distance cannot be negative), we get:
**$$x = \sqrt{\frac{5}{9}} R$$**

This is the exact distance from the center where the gravitational field reaches its peak intensity!

Animated Solution for Physics - Gravitation: The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $ρ (r) = ρ_{0} (1 - \frac{r ^{2}}{R ^{2}})$ . Then, the gravitational field is maximum at

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Gravitational Field Formula$

$Elemental Mass$

$Setting up the Integral$

$Evaluating the Integral$

$Field Expression$

$Condition for Maximum Field$

$Differentiating the Field$

$Final Answer$

The Sigma Insight: Gravitational Force

Analyzing the Setup Imagine a planet where the mass isn't distributed evenly

The Master Equation According to Newton's shell theorem (or Gauss's law for gravitation), the gravitational field $E$ at a distance $x$ inside a spherically symmetric body depends only on the mass enclosed within that radius $x$

Integrating the Mass

Finding the Maximum Field

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gravitational Field Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elemental Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Integral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating the Integral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Field Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Maximum Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differentiating the Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer