A rocket has to be launched from earth in such a way that it never returns. If E is the minimum ener

Q: Why do we use the formula $E = \frac{GMm}{R}$ instead of $\frac{1}{2}mv^2$?

Both are correct! The escape energy is the kinetic energy $\frac{1}{2}mv^2$ required to overcome the gravitational potential energy $-\frac{GMm}{R}$. To just escape, the total energy must be zero, so $\frac{1}{2}mv^2 - \frac{GMm}{R} = 0$, which means the required kinetic energy $E$ is exactly $\frac{GMm}{R}$.

Q: Why did we take the cube root of the volume ratio?

The volume of a sphere is $\frac{4}{3}\pi R^3$. If $V_e = 64 V_m$, then $R_e^3 = 64 R_m^3$. To find the relationship between the radii $R_e$ and $R_m$, we must take the cube root of both sides, and the cube root of 64 is 4.

Q: What if the densities were not equal?

If the densities were different, you couldn't simply say the mass ratio equals the volume ratio. You would have to use the formula $M = \text{Volume} \times \text{Density}$ and plug in the specific density values to find the true mass ratio.

Q: Does the mass of the rocket matter?

For the *ratio* of the escape energies, the mass of the rocket cancels out, so it doesn't matter. However, the absolute amount of energy required definitely depends on how heavy the rocket is!

Q: Why is the escape energy inversely proportional to the radius?

The closer you are to the center of a massive body, the stronger its gravitational pull. A smaller radius means you are starting deeper within the planet's 'gravity well', requiring more energy to climb out.

Question

A rocket has to be launched from earth in such a way that it never returns. If $E$ is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have, if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon.

Options:

(A) $\frac{E}{64}$

(B) $\frac{E}{16}$

(C) $\frac{E}{32}$

(D) $\frac{E}{4}$

Sigma Percentile · Accepted Answer

### The Cosmic Launchpad

Imagine you are standing on a launchpad, looking up at a massive rocket. Your mission? To send this rocket so far away that it never, ever returns to Earth. To achieve this, you need to give it a very specific amount of energy, known as the **escape energy**.

Now, imagine taking that exact same rocket and transporting it to the Moon. The Moon is smaller and less massive than the Earth. Naturally, you'd expect it to be easier to launch the rocket from there. But exactly *how much* easier? That is the beautiful puzzle we are going to solve today.

### The Master Equation

Before we crunch any numbers, we need our master tool. What is escape energy? It is the minimum kinetic energy required to completely overcome the gravitational potential energy that binds the rocket to the planet.

Mathematically, the escape energy $E$ is given by:
$E = \frac{GMm}{R}$

Here, $G$ is the universal gravitational constant, $M$ is the mass of the planet, $m$ is the mass of the rocket, and $R$ is the radius of the planet. Notice how the energy depends directly on the planet's mass and inversely on its radius.

### Decoding the Geometry

The problem gives us a crucial clue: the Earth's volume $V_e$ is 64 times the Moon's volume $V_m$.
$V_e = 64 V_m$

We know that planets are roughly spherical, and the volume of a sphere is $\frac{4}{3}\pi R^3$. This means volume is proportional to the cube of the radius.
$\frac{4}{3}\pi R_e^3 = 64 	imes \frac{4}{3}\pi R_m^3$

If we take the cube root of both sides, we find a simple, elegant relationship between their radii:
$R_e = 4 R_m$

The Earth is exactly four times wider than the Moon!

### The Mass Connection

Next, we need to figure out the masses. The problem states that the density of the Earth $ho_e$ and the Moon $ho_m$ are equal.
$ho_e = ho_m$

Density is simply mass divided by volume. If their densities are the same, then the ratio of their masses must be exactly equal to the ratio of their volumes.
$\frac{M_e}{V_e} = \frac{M_m}{V_m}$

Since the Earth has 64 times the volume, it must also have 64 times the mass!
$M_e = 64 M_m$

### The Final Calculation

Now we have everything we need. Let's set up a ratio of the escape energy from Earth $E_e$ to the escape energy from the Moon $E_m$.

$\frac{E_e}{E_m} = \frac{\frac{GM_e m}{R_e}}{\frac{GM_m m}{R_m}}$

The $G$ and the rocket's mass $m$ cancel out beautifully. We are left with:
$\frac{E_e}{E_m} = \left(\frac{M_e}{M_m}ight) 	imes \left(\frac{R_m}{R_e}ight)$

Let's plug in the ratios we discovered earlier. The mass ratio is 64, and the inverse radius ratio is $\frac{1}{4}$.
$\frac{E_e}{E_m} = 64 	imes \frac{1}{4} = 16$

This tells us that escaping the Earth requires 16 times more energy than escaping the Moon. Since the problem defines the Earth's escape energy as $E$, we can easily find the Moon's escape energy:
$E_m = \frac{E}{16}$

And there we have it! By breaking down the problem into geometry and mass, we found our answer with pure logical elegance.

Select Answer:

Visualized Solution

$E_{esc a p e}$

$E = \frac{GM m}{R}$

$V_{e} = 64 V_{m}$

$ρ_{e} = ρ_{m}$

$\frac{E _{e}}{E _{m}} = ?$

$\frac{E _{e}}{E _{m}} = 16$

$E_{m} = \frac{E}{16}$

\text{Conclusion}

The Sigma Insight: Escape Speed and Motion of Satellites

The Cosmic Launchpad

The Master Equation

Decoding the Geometry

The Mass Connection

The 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; } Eescape​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=RGMm​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ve​=64Vm​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρe​=ρm​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Em​Ee​​=?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Em​Ee​​=16

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Em​=16E​