The masses and radii of the Earth and Moon are (m1, R1) and (m2, R2), respectively. Their centres ar

Q: Why is the final energy taken as zero?

To find the *minimum* escape velocity, we assume the particle just barely reaches infinity. At infinity, potential energy is zero, and if it just barely reaches there, its kinetic energy is also zero.

Q: Why do we add the potential energies from both masses?

Gravitational potential energy is a scalar quantity. The total potential energy of a system is the algebraic sum of the potential energies due to each individual mass.

Q: Does the mass of the particle $m$ matter?

No, as you can see in the derivation, the mass $m$ cancels out from both sides of the energy conservation equation. Escape velocity is independent of the mass of the escaping body.

Q: What if the particle was not at the midpoint?

If it were at a distance $x$ from $m_1$, the distances would be $x$ and $r-x$. The potential energy would be $-\frac{Gm_1m}{x} - \frac{Gm_2m}{r-x}$. The rest of the process remains the same.

Q: Why is the potential energy negative?

Gravitational force is always attractive. We define potential energy to be zero at infinity. As a mass moves from infinity towards another mass, the gravitational field does positive work, meaning the potential energy decreases from zero, becoming negative.

Question

The masses and radii of the Earth and Moon are $(m_1, R_1)$ and $(m_2, R_2)$, respectively. Their centres are at a distance $r$ apart. Find the minimum escape velocity for a particle of mass $m$ to be projected from the middle of these two masses.

Options:

(A) $v = \frac{1}{2}\sqrt{\frac{4G(m_1 + m_2)}{r}}$

(B) $v = \sqrt{\frac{4G(m_1 + m_2)}{r}}$

(C) $v = \frac{1}{2}\sqrt{\frac{2G(m_1 + m_2)}{r}}$

(D) $v = \frac{\sqrt{2G(m_1 + m_2)}}{r}$

Sigma Percentile · Accepted Answer

### The Cosmic Setup
Imagine a vast expanse of space where two massive celestial bodies, say the Earth and the Moon, are locked in a gravitational dance. Let their masses be $m_1$ and $m_2$, and the distance separating their centers be $r$. Now, picture a tiny particle of mass $m$ placed exactly at the midpoint between them. Because it's right in the middle, its distance from both the Earth and the Moon is exactly $\frac{r}{2}$.

Our mission is to launch this particle so that it completely escapes the gravitational clutches of *both* planets.

### The Physics of Escaping
What does it mean to "escape"? In physics, escaping a gravitational field means reaching an infinite distance away. At infinity, the gravitational pull becomes zero, and consequently, the gravitational potential energy is also zero.

To find the *minimum* escape velocity, we assume that the particle uses up all its kinetic energy just to reach infinity. Therefore, upon reaching infinity, its kinetic energy drops to zero. This implies that the total mechanical energy (Kinetic + Potential) at infinity is exactly zero.

### The Master Equation: Conservation of Energy
We can solve this elegantly using the Principle of Conservation of Mechanical Energy. The total energy of the particle at the launch point must equal its total energy at infinity.

$$E_{	ext{initial}} = E_{	ext{final}} = 0$$

The initial energy consists of two parts:
1.  **Kinetic Energy ($K_i$):** The energy we provide by launching it with velocity $v$. So, $K_i = \frac{1}{2}mv^2$.
2.  **Potential Energy ($U_i$):** Here is the crucial part. The particle is in the gravitational field of *both* masses. Therefore, its total potential energy is the sum of the potential energies due to each mass.
    $$U_i = -\frac{Gm_1m}{r/2} - \frac{Gm_2m}{r/2}$$

Setting the sum of initial kinetic and potential energies to zero, we get:

$$\frac{1}{2}mv^2 - \frac{Gm_1m}{r/2} - \frac{Gm_2m}{r/2} = 0$$

### The Final Calculation
Let's simplify this equation. First, we can bring the $2$ from the denominator up to the numerator:

$$\frac{1}{2}mv^2 - \frac{2Gm_1m}{r} - \frac{2Gm_2m}{r} = 0$$

Now, move the potential energy terms to the right side of the equation:

$$\frac{1}{2}mv^2 = \frac{2Gm_1m}{r} + \frac{2Gm_2m}{r}$$

Notice something beautiful? The mass of the particle, $m$, appears in every term. We can cancel it out! This proves a fundamental truth: **escape velocity is independent of the mass of the escaping object.**

$$\frac{1}{2}v^2 = \frac{2G}{r}(m_1 + m_2)$$

Finally, multiply both sides by $2$ and take the square root to isolate $v$:

$$v^2 = \frac{4G}{r}(m_1 + m_2)$$

$$v = \sqrt{\frac{4G(m_1 + m_2)}{r}}$$

And there we have it! The minimum escape velocity required to break free from this dual-planet system.

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Visualized Solution

$Visualizing the Setup$

$Concept of Escape Velocity$

$Conservation of Mechanical Energy$

$Setting up the Equation$

$Solving for v$

$Final Calculation$

$Conclusion$

The Sigma Insight: Escape Speed and Motion of Satellites

The Cosmic Setup Imagine a vast expanse of space where two massive celestial bodies, say the Earth and the Moon, are locked in a gravitational dance

The Physics of Escaping What does it mean to "escape"? In physics, escaping a gravitational field means reaching an infinite distance away

The Master Equation

The Final Calculation Let's simplify this equation

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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; } Concept of Escape Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Mechanical Energy

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

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

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

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