A planet of radius R = 1/10 * (radius of earth) has the same mass density as earth. Scientists dig a

Q: Why does gravity vary linearly inside the planet? Shouldn't it follow the inverse-square law?

The inverse-square law $g = \frac{GM}{r^2}$ applies outside a spherical mass. Inside a uniform solid planet of radius $R$, at a distance $x$ from the center, the gravitational pull from the shell of mass outside $x$ cancels out completely (Shell Theorem). Only the mass inside the sphere of radius $x$ contributes to gravity, which scales as $x^3$. Dividing this mass by $x^2$ gives a linear relationship: $g(x) \propto x$.

Q: Why are the limits of integration from $4R/5$ to $R$ instead of $0$ to $R/5$?

The variable of integration $x$ represents the distance from the center of the planet, not the depth from the surface. Since the well is dug to a depth of $R/5$ from the surface (where $x = R$), the bottom of the well is at a distance of $R - R/5 = 4R/5$ from the center. Therefore, the wire spans from $x = 4R/5$ to $x = R$.

Q: How did we get $g \propto R$ for the planet's surface gravity?

The formula for surface gravity is $g = \frac{GM}{R^2}$. Since mass $M = \text{density} \times \text{volume} = \rho \times \frac{4}{3}\pi R^3$, substituting this gives $g = \frac{G \rho \frac{4}{3}\pi R^3}{R^2} = \frac{4}{3}\pi G \rho R$. Since the density $\rho$ is the same as Earth's, $g$ is directly proportional to $R$.

Q: What would happen to the tension at the bottom of the wire?

The tension at any point in the wire supports only the portion of the wire hanging below it. At the very bottom of the wire ($x = 4R/5$), there is no mass hanging below, so the tension is zero. The tension increases as you go up the wire, reaching its maximum value of $108 \text{ N}$ at the top ($x = R$).

Q: Can we solve this using the average value of gravity?

Yes! Because the gravity varies linearly with distance $x$, the average gravitational acceleration experienced by the wire is simply the arithmetic mean of gravity at its two ends: $g_{\text{avg}} = \frac{g(R) + g(4R/5)}{2} = \frac{g_s + 0.8 g_s}{2} = 0.9 g_s$. The total mass of the wire is $m = \lambda L = 10^{-3} \times (1.2 \times 10^5) = 120 \text{ kg}$. The total force is $F = m \cdot g_{\text{avg}} = 120 \times (0.9 \times 1) = 108 \text{ N}$.

Question

A planet of radius $R = \frac{1}{10} 	imes 	ext{(radius of earth)}$ has the same mass density as earth. Scientists dig a well of depth $\frac{R}{5}$ on it and lower a wire of the same length and of linear mass density $10^{-3} 	ext{ kg m}^{-1}$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of earth $= 6 	imes 10^6 	ext{ m}$ and the acceleration due to gravity of earth is $10 	ext{ ms}^{-2}$)

Options:

(A) $96 	ext{ N}$

(B) $108 	ext{ N}$

(C) $120 	ext{ N}$

(D) $150 	ext{ N}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on the surface of a compact, dense planet.

This planet is a miniature version of Earth—it shares the exact same mass density $ho$, but its radius $R$ is scaled down to a mere tenth of Earth's radius $R_e$.

$$
R = \frac{R_e}{10}
$$

On this planet, scientists have dug a narrow, radial well of depth $R/5$ directly into the crust.

They lower a uniform wire of the same length, $L = R/5$, into this well.

Our task is to find the upward holding force $F$ that a person must exert at the top of the wire to keep it suspended in equilibrium, ensuring it doesn't touch the bottom or sides of the well.

---

### The Gravity Profile of the Planet

Before we can calculate the force on the wire, we must understand how gravity behaves on and inside this planet.

First, let's find the acceleration due to gravity at the planet's surface, $g_s$.

For any spherical body of mass $M$ and radius $R$, the surface gravity is given by Newton's law of gravitation:

$$
g_s = \frac{GM}{R^2}
$$

Since the mass can be written in terms of its uniform density $ho$ as $M = \frac{4}{3}\pi R^3 ho$, we can substitute this to find:

$$
g_s = \frac{G \left(\frac{4}{3}\pi R^3 hoight)}{R^2} = \frac{4}{3}\pi G ho R
$$

This reveals a beautiful linear relationship: for a constant density, the surface gravity is directly proportional to the radius of the planet ($g_s \propto R$).

Using this scaling law, we can easily relate the planet's surface gravity to Earth's surface gravity $g_e$:

$$
g_s = g_e 	imes \frac{R}{R_e} = 10 	imes \frac{1}{10} = 1 	ext{ ms}^{-2}
$$

Now, what about the gravity *inside* the planet?

As we descend into the well, we are going below the surface.

Inside a uniform solid sphere, the gravitational field at a distance $x$ from the center ($x < R$) is due only to the mass contained within the sphere of radius $x$.

This leads to a linear variation of gravity with distance from the center:

$$
g(x) = g_s \left(\frac{x}{R}ight)
$$

As we go deeper (smaller $x$), the local acceleration due to gravity decreases linearly, reaching zero at the very center of the planet.

---

### Setting up the Integration

Because the gravitational acceleration $g(x)$ varies continuously along the length of the wire, we cannot simply multiply the total mass of the wire by a single value of gravity.

Instead, we must use calculus to sum up the forces acting on every tiny segment of the wire.

Let's consider an infinitesimal element of the wire of length $dx$ located at a distance $x$ from the center of the planet.

The mass $dm$ of this tiny element is:

$$
dm = \lambda \, dx
$$

where $\lambda = 10^{-3} 	ext{ kg m}^{-1}$ is the linear mass density of the wire.

The downward gravitational force $dF$ acting on this element is:

$$
dF = dm \cdot g(x) = (\lambda \, dx) \cdot \left(g_s \frac{x}{R}ight) = \frac{\lambda g_s}{R} x \, dx
$$

To find the total downward force $F$ on the wire, we integrate $dF$ over the entire span of the wire.

Since the well is dug to a depth of $R/5$, the wire extends from a distance of $x_{	ext{bottom}} = R - \frac{R}{5} = \frac{4R}{5}$ to $x_{	ext{top}} = R$ from the center of the planet.

Thus, our limits of integration are from $\frac{4R}{5}$ to $R$:

$$
F = \int_{4R/5}^{R} \frac{\lambda g_s}{R} x \, dx
$$

---

### Performing the Integration

Let's pull the constants out of the integral and integrate $x$:

$$
F = \frac{\lambda g_s}{R} \int_{4R/5}^{R} x \, dx = \frac{\lambda g_s}{R} \left[ \frac{x^2}{2} ight]_{4R/5}^{R}
$$

$$
F = \frac{\lambda g_s}{2R} \left( R^2 - \left(\frac{4R}{5}ight)^2 ight)
$$

Simplifying the term inside the brackets:

$$
R^2 - \frac{16R^2}{25} = \frac{9R^2}{25}
$$

Substituting this back into our equation:

$$
F = \frac{\lambda g_s}{2R} \left( \frac{9R^2}{25} ight) = \frac{9}{50} \lambda g_s R
$$

This is our master formula! It elegantly combines the physical dimensions of the planet and the properties of the wire.

---

### The Final Calculation

Now, let's substitute our numerical values into the master formula:

- Linear mass density, $\lambda = 10^{-3} 	ext{ kg m}^{-1}$
- Surface gravity of the planet, $g_s = 1 	ext{ ms}^{-2}$
- Radius of the planet, $R = \frac{R_e}{10} = \frac{6 	imes 10^6 	ext{ m}}{10} = 6 	imes 10^5 	ext{ m}$

Plugging these in:

$$
F = \frac{9}{50} 	imes 10^{-3} 	imes 1 	imes (6 	imes 10^5)
$$

$$
F = \frac{9}{50} 	imes 600
$$

$$
F = 9 	imes 12 = 108 	ext{ N}
$$

Thus, the force applied at the top of the wire by the person holding it in place is exactly **$108 	ext{ N}$**, which corresponds to **Option (b)**.

Select Answer:

Visualized Solution

Visualizing the Setup

Relating Planet's Gravity to Earth's Gravity

Calculating Surface Gravity $g_{s}$

Gravity Variation with Depth

Force on an Infinitesimal Element

Setting up the Integration Limits

Integrating the Force Equation

Simplifying the Algebraic Expression

Substituting Numerical Values

Calculating the Final Force

The Sigma Insight: Acceleration due to Gravity and its Variation

Analyzing the Setup

The Gravity Profile of the Planet

Setting up the Integration

Performing the Integration

The Final Calculation

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