Distance between the centres of two stars is 10a. The masses of these stars are M and 16M and their

Q: Why do we only need to reach the neutral point $P$? Why not all the way to the surface of the smaller star?

Because once the projectile crosses the neutral point $P$, the gravitational pull of the smaller star becomes stronger than that of the larger star. The net force will then point towards the smaller star, pulling the projectile to its surface automatically without requiring any extra initial kinetic energy.

Q: Why is the potential energy negative? What does it physically represent?

Gravitational potential energy is negative because gravity is an attractive force. A negative potential energy indicates a bound state, meaning energy must be supplied to the system to separate the masses to infinity.

Q: How did we get the distance of the neutral point as $2a$ and $8a$?

By setting the gravitational field intensities of both stars equal to each other: $GM/r_1^2 = G(16M)/r_2^2$. Taking the square root gives $r_2 = 4r_1$. Since the total distance between their centres is $10a$, we solve $r_1 + 4r_1 = 10a$ to get $r_1 = 2a$ and $r_2 = 8a$.

Q: Does the mass of the projected particle $m$ affect the minimum escape velocity?

No, the mass of the projectile $m$ cancels out on both sides of the energy conservation equation. This is a beautiful consequence of the equivalence principle: all masses accelerate at the same rate in a given gravitational field.

Q: What would happen if the initial velocity was slightly less than $v_{\min}$?

If the launch speed is less than $v_{\min}$, the projectile will not have enough kinetic energy to reach the neutral point $P$. It will stop somewhere before $P$ and then be pulled back to the surface of the larger star.

Question

Distance between the centres of two stars is $10a$. The masses of these stars are $M$ and $16M$ and their radii $a$ and $2a$ respectively. A body of mass $m$ is fired straight from the surface of the larger star towards the surface of the smaller star. What should be its minimum initial speed to reach the surface of the smaller star? Obtain the expression in terms of $G, M$ and $a$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on the surface of a massive star, looking out across the dark expanse of space toward another, smaller star.

This is not just a standard physics problem; it is a cosmic tug-of-war.

We have two celestial bodies:
- A smaller star of mass $M$ and radius $a$.
- A larger star of mass $16M$ and radius $2a$.
- Their centres are separated by a distance of $10a$.

Our goal is to launch a small projectile of mass $m$ from the surface of the larger star so that it successfully reaches the surface of the smaller star.

What is the absolute minimum speed required for this journey?

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### The Concept of the Neutral Point

As the projectile travels from the larger star to the smaller star, it experiences two opposing gravitational forces:
- The gravitational pull of the larger star, pulling it back.
- The gravitational pull of the smaller star, pulling it forward.

Because gravity is an inverse-square law, the pull of the larger star dominates when the projectile is close to it.

However, as the projectile moves further away, the pull of the larger star weakens, and the pull of the smaller star strengthens.

At some unique point $P$ between the two stars, these two forces must perfectly balance. This is the **neutral point** (or zero-gravity point).

$$\frac{GM}{r_1^2} = \frac{G(16M)}{r_2^2}$$

Where:
- $r_1$ is the distance from the centre of the smaller star ($C_1$).
- $r_2$ is the distance from the centre of the larger star ($C_2$).

Simplifying this equation by taking the square root on both sides:

$$\frac{r_2}{r_1} = 4 \implies r_2 = 4r_1$$

Since the total distance between the centres is $10a$:

$$r_1 + r_2 = 10a \implies r_1 + 4r_1 = 10a \implies 5r_1 = 10a$$

This gives us the exact location of the neutral point:
- $r_1 = 2a$
- $r_2 = 8a$

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### The Physics of Minimum Launch Speed

Why is the neutral point so critical?

If we launch the projectile with just enough speed to reach the neutral point $P$, it will arrive there with virtually zero velocity.

Once it crosses $P$ by even an infinitesimal distance, the gravitational pull of the smaller star becomes stronger than that of the larger star.

From that point onward, the smaller star's gravity will naturally pull the projectile down to its surface without requiring any additional energy!

Therefore, the **minimum launch speed** $v_{\min}$ is the speed required to take the projectile from the surface of the larger star to the neutral point $P$ with zero final velocity.

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### Applying Conservation of Energy

Let's apply the principle of **Conservation of Mechanical Energy** between the initial launch position (surface of the larger star) and the neutral point $P$.

#### 1. Initial State (at the surface of the larger star):
- Distance from $C_2$ is $r_{2i} = 2a$ (the radius of the larger star).
- Distance from $C_1$ is $r_{1i} = 10a - 2a = 8a$.

The initial potential energy $U_i$ is:

$$U_i = -\frac{GMm}{8a} - \frac{16GMm}{2a}$$

To simplify, let's find a common denominator:

$$U_i = -\frac{GMm}{8a} - \frac{64GMm}{8a} = -\frac{65GMm}{8a}$$

The initial kinetic energy is:

$$K_i = \frac{1}{2}mv_{\min}^2$$

#### 2. Final State (at the neutral point $P$):
- Distance from $C_1$ is $r_1 = 2a$.
- Distance from $C_2$ is $r_2 = 8a$.

The potential energy at $P$ is:

$$U_P = -\frac{GMm}{2a} - \frac{16GMm}{8a}$$

Simplifying this expression:

$$U_P = -\frac{GMm}{2a} - \frac{2GMm}{a} = -\frac{GMm}{2a} - \frac{4GMm}{2a} = -\frac{5GMm}{2a}$$

To make comparison easier, let's write it with a denominator of $8a$:

$$U_P = -\frac{20GMm}{8a}$$

At the minimum launch speed, the kinetic energy at $P$ is zero:

$$K_P = 0$$

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### Solving for $v_{\min}$

Equating the total mechanical energy of the two states:

$$K_i + U_i = K_P + U_P$$

$$\frac{1}{2}mv_{\min}^2 - \frac{65GMm}{8a} = 0 - \frac{20GMm}{8a}$$

$$\frac{1}{2}mv_{\min}^2 = \frac{65GMm}{8a} - \frac{20GMm}{8a}$$

$$\frac{1}{2}mv_{\min}^2 = \frac{45GMm}{8a}$$

We can cancel the mass of the projectile $m$ from both sides:

$$\frac{1}{2}v_{\min}^2 = \frac{45GM}{8a}$$

$$v_{\min}^2 = \frac{90GM}{8a} = \frac{45GM}{4a}$$

Taking the square root to find the final velocity:

$$v_{\min} = \sqrt{\frac{45GM}{4a}} = \sqrt{\frac{9 	imes 5 	imes GM}{4a}}$$

$$v_{\min} = \frac{3\sqrt{5}}{2} \sqrt{\frac{GM}{a}}$$

This is our final, elegant result! It shows that the minimum launch speed depends purely on the mass of the smaller star $M$, the scaling parameter $a$, and the universal gravitational constant $G$.

Visualized Solution

Understanding the Two-Star System

The Concept of the Neutral Point $P$

Locating the Neutral Point $P$

Solving for $r_{1}$ and $r_{2}$

Analyzing the Initial Position

Applying Conservation of Mechanical Energy

Calculating Initial Potential Energy $U_{i}$

Simplifying $U_{i}$

Calculating Potential Energy at $P$

Simplifying $U_{P}$

Setting up the Energy Equation

Solving for $v_{m i n}$

Finding the Final Expression

The Way Forward: What if the Direction is Reversed?

The Sigma Insight: Gravitational Potential and Potential Energy

Analyzing the Setup

The Concept of the Neutral Point

The Physics of Minimum Launch Speed

Applying Conservation of Energy

# 1

# 2

Solving for $v_{m i n}$

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Visualized Solution

Understanding the Two-Star System

The Concept of the Neutral Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Locating the Neutral Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2​

Analyzing the Initial Position

Applying Conservation of Mechanical Energy

Calculating Initial Potential Energy .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ui​

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

Calculating Potential Energy at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

Setting up the Energy Equation

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

Finding the Final Expression

The Way Forward: What if the Direction is Reversed?

The Sigma Insight: Gravitational Potential and Potential Energy

Analyzing the Setup

The Concept of the Neutral Point

The Physics of Minimum Launch Speed

Applying Conservation of Energy

# 1

# 2

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

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy required to move a body of mass m from an orbit of radius 2R to 3R is

The Concept of the Neutral Point $P$

Locating the Neutral Point $P$

Solving for $r_{1}$ and $r_{2}$

Calculating Initial Potential Energy $U_{i}$

Simplifying $U_{i}$

Calculating Potential Energy at $P$

Simplifying $U_{P}$

Solving for $v_{m i n}$

Solving for $v_{m i n}$

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