There is a crater of depth R/100 on the surface of the moon (radius R). A projectile is fired vertic

Q: Why is the potential inside the crater not just $-GM/r$?

The formula $-GM/r$ only applies to points outside or on the surface of a spherically symmetric mass distribution. Inside the body, the mass outside the radius $r$ also contributes to the potential in a different way, leading to the quadratic potential formula $V(r) = -\\frac{GM}{2R^3}(3R^2 - r^2)$.

Q: Why is the launch velocity equal to the escape velocity from the surface, even though we are launching from inside a crater?

The question specifies that the projectile is fired with a velocity 'which is equal to the escape velocity $v$ from the surface of the moon'. This is a given constraint of the problem, meaning we must use the surface escape velocity value $v_e = \\sqrt{\\frac{2GM}{R}}$ as our initial velocity, regardless of the launch location.

Q: Does the projectile escape the Moon's gravity completely?

No! If it were launched from the surface with escape velocity, it would escape to infinity. However, because it is launched from inside a crater (deeper in the potential well) with the surface escape velocity, its initial energy is slightly less than what is required to escape to infinity. Therefore, it reaches a finite maximum height $h$ and falls back.

Q: How did we get $r = 0.99R$ for the launch point?

The crater has a depth of $R/100 = 0.01R$. Since the crater is dug into the surface of the Moon, the distance of the bottom of the crater from the center of the Moon is $R - 0.01R = 0.99R$.

Q: Can we use the approximation $g' = g(1 - 2h/R)$ or similar gravity variation formulas here?

Absolutely not! Those approximations are only valid for very small heights ($h \\ll R$). Here, the maximum height reached is around $99.5R$, which is massive compared to the radius of the Moon. We must use exact gravitational potential energy formulas.

Question

There is a crater of depth $\frac{R}{100}$ on the surface of the moon (radius $R$). A projectile is fired vertically upward from the crater with velocity, which is equal to the escape velocity $v$ from the surface of the moon. Find the maximum height attained by the projectile.

Sigma Percentile · Accepted Answer

### The Cosmic Launchpad: Setting the Scene

Imagine standing at the bottom of a deep, silent crater on the surface of the Moon.

This is not just any crater; its depth is a significant fraction of the Moon's radius, specifically $d = \frac{R}{100}$.

Because you are standing at the bottom of this crater, you are physically closer to the center of the Moon than someone standing on the flat lunar plains.

Your distance from the center of the Moon is:

$$r_A = R - \frac{R}{100} = 0.99R$$

Now, you launch a projectile vertically upwards with a speed equal to the escape velocity from the Moon's surface.

This setup presents a beautiful paradox: if you launch something with escape velocity, it should escape to infinity, right?

But wait! The escape velocity is calculated for a launch from the **surface**.

Since you are launching from **inside** a crater, you are deeper within the Moon's gravitational potential well.

Will the projectile still escape, or will the extra gravitational grip pull it back to a finite height?

Let's find out!

### The Physics of Escape Velocity

To analyze this problem, we must first write down the expression for the escape velocity from the surface of the Moon.

The escape velocity $v_e$ is the minimum speed needed for a body to escape to infinity with zero kinetic energy remaining.

By conserving energy from the surface to infinity, we find:

$$v_e = \sqrt{\frac{2GM}{R}}$$

where $G$ is the universal gravitational constant, $M$ is the mass of the Moon, and $R$ is its radius.

The problem states that our projectile is launched with this exact speed:

$$v_A = v_e = \sqrt{\frac{2GM}{R}}$$

### The Energy Conservation Principle

Since gravity is a conservative force, the total mechanical energy of the projectile remains constant throughout its flight.

Let $A$ be the launch point at the bottom of the crater, and $B$ be the highest point reached.

At point $B$, the projectile momentarily stops before falling back down, so its velocity $v_B = 0$.

Applying the conservation of mechanical energy between $A$ and $B$:

$$K_A + U_A = K_B + U_B$$

$$\frac{1}{2} m v_A^2 + m V_A = 0 + m V_B$$

where $m$ is the mass of the projectile, and $V_A$ and $V_B$ are the gravitational potentials at points $A$ and $B$, respectively.

Dividing the entire equation by $m$ simplifies it to:

$$\frac{1}{2} v_A^2 + V_A = V_B$$

Now, substitute the launch velocity $v_A = \sqrt{\frac{2GM}{R}}$ into this equation:

$$\frac{1}{2} \left(\sqrt{\frac{2GM}{R}}\right)^2 + V_A = V_B$$

$$\frac{GM}{R} + V_A = V_B$$

Rearranging this gives us our master equation:

$$\frac{GM}{R} = V_B - V_A$$

This elegant equation tells us that the kinetic energy per unit mass supplied at launch must equal the potential difference between the highest point and the launch point.

### Navigating the Gravitational Potential Well

To proceed, we need to find the exact expressions for the gravitational potentials $V_A$ and $V_B$.

Point $B$ is at a height $h$ above the surface of the Moon, which means its distance from the center is $r_B = R + h$.

Since this point lies outside the Moon, we can treat the Moon as a point mass concentrated at its center.

The potential at $B$ is simply:

$$V_B = -\frac{GM}{R + h}$$

Point $A$, however, lies inside the Moon's spherical boundary at a distance $r_A = 0.99R$ from the center.

We cannot use the simple $-\frac{GM}{r}$ formula here because some of the Moon's mass lies at a radius greater than $r_A$.

Instead, we must use the formula for the gravitational potential inside a uniform solid sphere of mass $M$ and radius $R$:

$$V(r) = -\frac{GM}{2R^3} \left(3R^2 - r^2\right)$$

Substituting $r = 0.99R$ into this formula:

$$V_A = -\frac{GM}{2R^3} \left[3R^2 - (0.99R)^2\right]$$

$$V_A = -\frac{GM}{2R^3} \left[3R^2 - 0.9801R^2\right]$$

$$V_A = -\frac{GM}{2R} \left[3 - 0.9801\right]$$

$$V_A = -\frac{GM}{2R} \left[2.0199\right]$$

$$V_A = -1.00995 \frac{GM}{R}$$

Notice that this potential is slightly more negative than the surface potential ($-1.0 \frac{GM}{R}$), which confirms that the projectile starts deeper in the potential well.

### The Mathematical Climax: Solving for Height

Now, let's substitute $V_A$ and $V_B$ back into our master energy equation:

$$\frac{GM}{R} = V_B - V_A$$

$$\frac{GM}{R} = -\frac{GM}{R + h} - \left(-1.00995 \frac{GM}{R}\right)$$

$$\frac{GM}{R} = -\frac{GM}{R + h} + 1.00995 \frac{GM}{R}$$

Dividing both sides by $GM$ to eliminate the constants:

$$\frac{1}{R} = -\frac{1}{R + h} + \frac{1.00995}{R}$$

Now, isolate the term containing $h$:

$$\frac{1}{R + h} = \frac{1.00995}{R} - \frac{1}{R}$$

$$\frac{1}{R + h} = \frac{0.00995}{R}$$

Taking the reciprocal of both sides:

$$R + h = \frac{R}{0.00995}$$

$$R + h \approx 100.5025 R$$

$$h \approx 99.5025 R \approx 99.5 R$$

### Conceptual Takeaways and the Way Forward

What a spectacular result!

Because the projectile was launched from just a tiny bit deeper inside the Moon ($0.01R$ below the surface), it failed to escape to infinity.

Instead, it reached a massive but finite height of **$99.5 R$** before gravity finally brought its upward journey to a halt.

This problem beautifully illustrates how sensitive escape trajectories are to the initial potential energy of the launch site.

If you want to explore further, try calculating the maximum height if the crater was dug to a depth of $10\%$ of the Moon's radius ($d = 0.1R$).

You will find that the maximum height drops drastically, showing just how deep and powerful gravitational wells truly are!

Visualized Solution

Visualizing the Setup

Understanding Escape Velocity

Conservation of Mechanical Energy

Substituting Launch Speed

Potential at the Highest Point $B$

Potential Inside the Crater (Point $A$ )

Simplifying $V_{A}$

Substituting Potentials into Energy Equation

Isolating the Height Term

Calculating the Maximum Height

What if the Crater was Deeper?

The Sigma Insight: Gravitational Potential and Potential Energy

The Cosmic Launchpad

The Physics of Escape Velocity

The Energy Conservation Principle

Navigating the Gravitational Potential Well

The Mathematical Climax

Conceptual Takeaways and the Way Forward

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