The masses and radii of the Earth and the Moon are M1, R1 and M2, R2 respectively. Their centres are

Q: Why can we ignore the radii of the Earth and the Moon in this calculation?

Since the particle is projected from the midway point in space between the Earth and the Moon, it is already outside both bodies. Therefore, we only care about the distance from the particle to their centers of mass, which is $d/2$. The individual radii $R_1$ and $R_2$ do not affect the potential energy at this midway point.

Q: What does 'escape to infinity' actually mean physically?

Physically, escaping to infinity means the particle reaches a region of space so far away that the gravitational pull of both the Earth and the Moon becomes negligible (effectively zero), and the particle has no tendency to fall back toward them.

Q: Why is the escape velocity independent of the mass of the projected particle?

The gravitational force (and thus the potential energy) is directly proportional to the mass of the particle $m$. Since the kinetic energy is also directly proportional to $m$, the mass term cancels out when we equate kinetic energy to potential energy. Thus, the required velocity is independent of the particle's mass.

Q: What if the particle was projected with a speed greater than this minimum speed?

If projected with a speed $v > v_e$, the particle will still escape to infinity, but it will arrive there with some leftover kinetic energy. Its final speed at infinity (terminal speed) can be found using $v_{\infty} = \sqrt{v^2 - v_e^2}$.

Q: Can we use this same method if the Earth and Moon were moving?

If the Earth and Moon are in orbit around their common center of mass, the system becomes a dynamic three-body problem. We would then need to account for the kinetic energy of the Earth and Moon, and potentially use a rotating coordinate system, which introduces centrifugal and Coriolis terms.

Question

The masses and radii of the Earth and the Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are a distance $d$ apart. The minimum speed with which a particle of mass $m$ should be projected from a point midway between the two centres so as to escape to infinity is _________.

Sigma Percentile · Accepted Answer

### Introduction to the Gravitational Trap

Imagine standing on a cosmic bridge suspended exactly halfway between the **Earth** and the **Moon**.

From this unique vantage point, you feel the gravitational tug of two massive worlds pulling you in opposite directions.

If you drop a particle here, it hangs in a delicate, unstable equilibrium.

But what if you wanted to launch this particle so that it escapes this system entirely and journeys into the deep, dark abyss of interstellar space?

This is the classic problem of finding the **escape velocity** from a multi-body gravitational field.

Let's dive into the physics of how we can break free from this double gravitational trap.

### Analyzing the Setup

Let the Earth have a mass of $M_1$ and the Moon have a mass of $M_2$.

Their centers are separated by a distance $d$.

We place a test particle of mass $m$ exactly midway between them.

This means the distance from the particle to the center of the Earth is:

$$r_1 = \frac{d}{2}$$

And the distance to the center of the Moon is also:

$$r_2 = \frac{d}{2}$$

To find the minimum speed $v_e$ required to project this particle to infinity, we must look at the total energy of the system.

### The Master Equation: Conservation of Energy

Gravity is a **conservative force**.

This means we can use the **Law of Conservation of Mechanical Energy** to solve our problem.

Let's write down the total mechanical energy of the particle at any point:

$$E = K + U$$

Where $K$ is the kinetic energy and $U$ is the gravitational potential energy.

When the particle "escapes to infinity," it reaches a point infinitely far away where the gravitational influence of both the Earth and the Moon drops to zero.

Thus, the potential energy at infinity is:

$$U_f = 0$$

For the **minimum** launch speed, the particle should just barely reach infinity, meaning its kinetic energy when it gets there will also be zero:

$$K_f = 0$$

Therefore, the total mechanical energy at infinity is:

$$E_f = K_f + U_f = 0$$

By the conservation of energy, the initial mechanical energy $E_i$ at the launch point must also be zero:

$$E_i = K_i + U_i = 0$$

This is our master equation!

### Calculating the Gravitational Potential Energy

Gravitational potential energy is a scalar quantity, so we can simply add the potential energy contributions from both the Earth and the Moon.

The potential energy due to the Earth is:

$$U_1 = -\frac{GM_1 m}{d/2} = -\frac{2GM_1 m}{d}$$

Similarly, the potential energy due to the Moon is:

$$U_2 = -\frac{GM_2 m}{d/2} = -\frac{2GM_2 m}{d}$$

Adding these together gives the total initial potential energy $U_i$:

$$U_i = U_1 + U_2 = -\frac{2Gm}{d}(M_1 + M_2)$$

This negative energy represents the **gravitational well** or trap the particle is stuck in.

To escape, we must supply an equal and opposite amount of positive energy, known as the **binding energy**:

$$	ext{Binding Energy} = -U_i = \frac{2Gm}{d}(M_1 + M_2)$$

### Finding the Escape Velocity

We supply this binding energy in the form of initial kinetic energy $K_i$:

$$K_i = \frac{1}{2}mv_e^2$$

Setting $K_i$ equal to the binding energy:

$$\frac{1}{2}mv_e^2 = \frac{2Gm}{d}(M_1 + M_2)$$

Notice something beautiful here: the mass of our test particle, $m$, appears on both sides of the equation.

We can cancel it out completely!

$$\frac{1}{2}v_e^2 = \frac{2G}{d}(M_1 + M_2)$$

This tells us that whether we are launching a tiny electron or a massive spaceship, the escape velocity remains exactly the same.

Now, multiplying both sides by $2$:

$$v_e^2 = \frac{4G}{d}(M_1 + M_2)$$

Taking the square root of both sides, we arrive at our final elegant result:

$$v_e = 2\sqrt{\frac{G(M_1 + M_2)}{d}}$$

This is the minimum speed required to launch the particle to infinity from the midway point.

Visualized Solution

Visualizing the Setup

The Principle of Conservation of Energy

Potential Energy due to Earth

Potential Energy due to Moon

Total Initial Potential Energy

Understanding Binding Energy

Setting up Kinetic Energy

Cancelling the Particle Mass

Solving for Escape Velocity

The Way Forward

The Sigma Insight: Gravitational Potential and Potential Energy

Introduction to the Gravitational Trap

Analyzing the Setup

The Master Equation

Calculating the Gravitational Potential Energy

Finding the Escape Velocity

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