Two planets have masses M and 16 M and their radii are a and 2a, respectively. The separation betwee

Q: Why do we only need to reach the neutral point instead of the surface of the smaller planet?

Because once the body crosses the neutral point, the gravitational pull of the smaller planet becomes stronger than that of the larger planet. It will naturally 'fall' towards the smaller planet without needing any extra initial push.

Q: How do we find the distance of the neutral point?

By equating the gravitational field intensities of both planets at a distance $x$. Set $\frac{GM_1}{x^2} = \frac{GM_2}{(d-x)^2}$ and solve for $x$.

Q: Why is the final kinetic energy taken as zero?

We are asked for the *minimum* firing speed. This implies the body should just barely make it to the neutral point. If it arrives with zero velocity, it has used exactly the minimum energy required.

Q: What happens to the potential energy from the smaller planet when the body is at the larger planet's surface?

It must be included! The total potential energy at any point is the algebraic sum of the potential energies due to *all* masses in the system. Don't forget the $-\frac{GMm}{8a}$ term.

Q: Does the mass of the fired body matter?

No. As you can see in the final energy equation, the mass $m$ cancels out from every term. The required speed depends only on the masses of the planets and the distances involved.

Question

Two planets have masses $M$ and $16 M$ and their radii are $a$ and $2a$, respectively. The separation between the centres of the planets is $10a$. A body of mass $m$ is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is

Options:

(A) $2\sqrt{\frac{GM}{a}}$

(B) $4\sqrt{\frac{GM}{a}}$

(C) $\frac{GM^2}{ma}$

(D) $\frac{3}{2}\sqrt{\frac{5GM}{a}}$

Sigma Percentile · Accepted Answer

The problem of launching a projectile from one celestial body to another is a classic application of the conservation of mechanical energy and the concept of gravitational fields. Let's embark on this journey to find the minimum firing speed required to send a mass from a larger planet to a smaller one.

### Analyzing the Setup

Imagine two planets suspended in space. The larger planet has a massive $16M$ and a radius of $2a$. The smaller planet has a mass of $M$ and a radius of $a$. The distance between their centers is a vast $10a$.

We are tasked with firing a body of mass $m$ from the surface of the larger planet directly towards the smaller one. The key question is: **What is the absolute minimum speed required for the body to reach the smaller planet?**

A common misconception is that the body must be fired with enough energy to reach the *surface* of the smaller planet. However, this ignores the gravitational pull of the smaller planet itself!

### The Neutral Point (Null Point)

As the body travels away from the larger planet, it is constantly pulled back by its massive gravity. But simultaneously, it is being pulled forward by the smaller planet. There exists a magical point in space between them where these two opposing gravitational forces perfectly cancel each other out. This is called the **neutral point** or null point.

If we can just give the body enough kinetic energy to reach this neutral point, it will have zero velocity exactly at that spot. But once it nudges even a millimeter past it, the gravitational pull of the smaller planet becomes dominant, and the body will "fall" the rest of the way to the smaller planet's surface.

Let's locate this neutral point, $P$. Let its distance from the center of the larger planet be $x$. At $P$, the gravitational fields are equal:
$$ \frac{G(16M)}{x^2} = \frac{GM}{(10a - x)^2} $$

Taking the square root of both sides simplifies things beautifully:
$$ \frac{4}{x} = \frac{1}{10a - x} $$

Cross-multiplying gives:
$$ 40a - 4x = x \implies 5x = 40a \implies x = 8a $$

So, the neutral point is located at a distance of $8a$ from the center of the larger planet (and $2a$ from the center of the smaller planet).

### The Master Equation: Conservation of Energy

Now, we apply the principle of conservation of mechanical energy. The total energy of the body at the surface of the larger planet must equal its total energy at the neutral point.

**Initial State (at the surface of the $16M$ planet):**
The body is at a distance of $2a$ from the center of the larger planet and $8a$ from the center of the smaller planet.
$$ K_i = \frac{1}{2}mu^2 $$
$$ U_i = -\frac{G(16M)m}{2a} - \frac{GMm}{8a} $$

**Final State (at the neutral point $P$):**
The body is at a distance of $8a$ from the center of the larger planet and $2a$ from the center of the smaller planet. Since we are looking for the *minimum* firing speed, the kinetic energy here will be zero.
$$ K_f = 0 $$
$$ U_f = -\frac{G(16M)m}{8a} - \frac{GMm}{2a} $$

Equating the initial and final total energies ($K_i + U_i = K_f + U_f$):
$$ \frac{1}{2}mu^2 - \frac{16GMm}{2a} - \frac{GMm}{8a} = 0 - \frac{16GMm}{8a} - \frac{GMm}{2a} $$

### Final Calculation

To make the algebra smooth, let's find a common denominator of $8a$ for all the potential energy terms:
$$ \frac{1}{2}mu^2 - \frac{64GMm}{8a} - \frac{GMm}{8a} = -\frac{16GMm}{8a} - \frac{4GMm}{8a} $$

Combining the terms:
$$ \frac{1}{2}mu^2 - \frac{65GMm}{8a} = -\frac{20GMm}{8a} $$

Now, isolate the kinetic energy term:
$$ \frac{1}{2}mu^2 = \frac{65GMm}{8a} - \frac{20GMm}{8a} $$
$$ \frac{1}{2}mu^2 = \frac{45GMm}{8a} $$

Notice how the mass of the body, $m$, cancels out from both sides. This means the required escape speed is independent of the mass of the projectile!
$$ u^2 = \frac{45GM}{4a} $$

Taking the square root gives us our final, elegant answer:
$$ u = \sqrt{\frac{45GM}{4a}} = \frac{3\sqrt{5}}{2}\sqrt{\frac{GM}{a}} $$

This is the exact minimum speed required to send the body on its interplanetary journey.

Select Answer:

Visualized Solution

$Visualizing the Two-Planet System$

$The Neutral Point (Null Point)$

$Locating the Neutral Point$

$Conservation of Mechanical Energy$

$Energy Equation Setup$

$Simplifying the Energy Equation$

$Calculating Minimum Speed$

$What if the speed was higher?$

The Sigma Insight: Gravitational Potential and Potential Energy

Analyzing the Setup

The Neutral Point (Null Point)

The Master Equation

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; } Visualizing the Two-Planet System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Neutral Point (Null Point)

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

.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; } Energy Equation Setup

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if the speed was higher?