Suppose two planets (spherical in shape) of radii R and 2R, but mass M and 9M respectively have a ce

Q: Why do we only need to reach the neutral point?

Once the satellite crosses the neutral point, the gravitational pull of the larger planet becomes stronger than that of the smaller planet. It will automatically pull the satellite in, so we don't need to provide any extra energy for that part of the journey.

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

We are looking for the *minimum* speed required. This means the satellite should just barely reach the neutral point. If it reaches there with zero velocity, it has exactly the minimum energy needed. Any more energy, and it would cross it with some speed left over.

Q: Why do we consider the potential energy of the second planet when the satellite is on the first planet?

Gravity has an infinite range! Even though the satellite is on the surface of the first planet, the second planet is still pulling on it. This pull contributes to the total potential energy of the system and cannot be ignored.

Q: What happens if the satellite crosses the neutral point?

If it crosses the neutral point, it enters the dominant gravitational field of the second planet. It will then accelerate towards the second planet and crash into its surface.

Q: Can we use equations of motion ($v^2 = u^2 + 2as$) here?

Absolutely not! The gravitational force (and therefore acceleration) changes continuously as the distance changes. Equations of motion are only valid for *constant* acceleration. We must use energy conservation.

Question

Suppose two planets (spherical in shape) of radii $R$ and $2R$, but mass $M$ and $9M$ respectively have a centre to centre separation $8R$ as shown in the figure. A satellite of mass $m$ is projected from the surface of the planet of mass $M$ directly towards the centre of the second planet. The minimum speed $v$ required for the satellite to reach the surface of the second planet is $\sqrt{\frac{aGM}{7R}}$, then the value of $a$ is …………… .
[Take, the two planets are fixed in their position]

Sigma Percentile · Accepted Answer

### The Cosmic Setup: Two Giants and a Tiny Satellite

Imagine you are standing on the surface of a planet, looking across the vast emptiness of space at a massive neighboring giant. Your mission is to launch a satellite from your planet so that it reaches the giant.

At first glance, you might think you need to provide enough energy to shoot the satellite all the way to the surface of the second planet. But physics offers us a beautiful shortcut! You don't need to throw it all the way; you just need to throw it past the **neutral point**.

### The Magic of the Neutral Point

The neutral point is the exact location in space where the gravitational pull of the first planet is perfectly balanced by the gravitational pull of the second planet. Once the satellite crosses this invisible boundary, the giant planet's gravity takes over and pulls the satellite in for the rest of the journey.

Let's find this magical point $P$. Suppose it is at a distance $x$ from the center of the first planet (mass $M$). The distance between the centers of the two planets is $8R$. Therefore, the distance from $P$ to the second planet (mass $9M$) is $(8R - x)$.

Equating the gravitational fields at $P$:

$$ \frac{GM}{x^2} = \frac{G(9M)}{(8R - x)^2} $$

Taking the square root of both sides makes the algebra incredibly elegant:

$$ \frac{1}{x} = \frac{3}{8R - x} $$

Cross-multiplying gives us $8R - x = 3x$, which simplifies to $4x = 8R$, and finally, **$x = 2R$**.

So, the neutral point is located at a distance of $2R$ from the center of the first planet.

### The Energy Equation: A Tale of Two States

To find the minimum launch speed $v$, we must ensure the satellite just barely reaches the neutral point. This means its kinetic energy at point $P$ will be exactly zero. We apply the **Conservation of Mechanical Energy** between the launch point (the surface of the first planet) and the neutral point $P$.

**State 1: At the surface of the first planet**
The satellite is at a distance $R$ from the center of the first planet and a distance of $(8R - R) = 7R$ from the center of the second planet. The initial energy $E_i$ is the sum of its kinetic energy and the potential energies due to *both* planets:

$$ E_i = \frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{G(9M)m}{7R} $$

**State 2: At the neutral point $P$**
The satellite is at a distance $2R$ from the first planet and $(8R - 2R) = 6R$ from the second planet. Its kinetic energy is zero. The final energy $E_f$ is:

$$ E_f = 0 - \frac{GMm}{2R} - \frac{G(9M)m}{6R} $$

Let's simplify $E_f$:

$$ E_f = - \frac{GMm}{2R} - \frac{3GMm}{2R} = - \frac{4GMm}{2R} = - \frac{2GMm}{R} $$

### The Final Mathematical Symphony

Now, we equate the initial and final energies ($E_i = E_f$):

$$ \frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{9GMm}{7R} = - \frac{2GMm}{R} $$

To isolate the kinetic energy term, we move the potential energy terms to the right side:

$$ \frac{1}{2}mv^2 = \frac{GMm}{R} + \frac{9GMm}{7R} - \frac{2GMm}{R} $$

$$ \frac{1}{2}mv^2 = \frac{9GMm}{7R} - \frac{GMm}{R} $$

Finding a common denominator:

$$ \frac{1}{2}mv^2 = \frac{9GMm - 7GMm}{7R} = \frac{2GMm}{7R} $$

Multiplying both sides by 2 and dividing by $m$:

$$ v^2 = \frac{4GM}{7R} $$

Taking the square root, we find the minimum required speed:

$$ v = \sqrt{\frac{4GM}{7R}} $$

The problem states that this minimum speed is $\sqrt{\frac{aGM}{7R}}$. By directly comparing our result with the given expression, it is crystal clear that **$a = 4$**.

This problem is a beautiful demonstration of how energy conservation and gravitational fields intertwine to govern the mechanics of the cosmos!

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Finding the Neutral Point P$

$Solving for x$

$Conservation of Energy$

$Setting up the Energy Equation$

$Simplifying Final Energy$

$Equating Energies$

$Calculating Minimum Speed$

$Finding the Value of a$

$The Way Forward$

The Sigma Insight: Gravitational Potential and Potential Energy

The Cosmic Setup

The Magic of the Neutral Point

The Energy Equation

The Final Mathematical Symphony

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Enter Numerical Value:

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; } Finding the Neutral Point P

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

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

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

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

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

.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; } Finding the Value of a

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