The Question is from neet-physics gravitation gravitational-field-gravitational-potential, and it was created by the author VENKATESWARARAO

Question

A binary star system consists of two stars of masses $M$ and $2M$ separated by a distance $d$. A small asteroid of mass $m$ is placed on the line joining the two stars at a distance $x$ from the star of mass $M$.  If the net gravitational force on the asteroid is zero, what is the ratio $x/d$?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [A]
## CONCEPT FRAMEWORK
	**Primary Concept**: Gravitational Force
	**Related Topics**: Newton's Law of Universal Gravitation, Equilibrium
	**Prerequisites**: Understanding of gravitational force between two masses, force equilibrium
## SOLUTION PATHWAY
**Given Information**:
	Mass of star 1: $M$
	Mass of star 2: $2M$
	Distance between stars: $d$
	Distance of asteroid from star 1: $x$
	Mass of asteroid: $m$
	Net gravitational force on asteroid is zero
**Method & Steps**:
	1. Set up the forces on the asteroid.
		Reasoning: The asteroid experiences gravitational force from both stars. For the net force to be zero, these forces must be equal and opposite.
		Formula/Rule: $F = G\frac{m_1m_2}{r^2}$, where $F$ is the gravitational force, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the distance between them.
	2. Write the expressions for gravitational force of each star on the asteroid.
		Working: Force due to star of mass $M$: $F_1 = G\frac{Mm}{x^2}$. Force due to star of mass $2M$: $F_2 = G\frac{2Mm}{(d-x)^2}$.
		Key Point: The distance of the asteroid from the star of mass $2M$ is $(d-x)$.
	3. Equate the two forces and solve for $x$.
		Working: $G\frac{Mm}{x^2} = G\frac{2Mm}{(d-x)^2}$. Simplifying, we get $\frac{1}{x^2} = \frac{2}{(d-x)^2}$. Taking the square root of both sides, we get $\frac{1}{x} = \frac{\sqrt{2}}{d-x}$.
	4. Rearrange and solve for the ratio $x/d$.
		Working: $d - x = \sqrt{2}x$, which implies $d = x(1 + \sqrt{2})$, therefore $x/d = \frac{1}{1 + \sqrt{2}}$.
		Result: $\frac{x}{d} = \frac{1}{1 + \sqrt{2}}$
		Verification: The derived ratio satisfies the condition for zero net gravitational force.
## OPTION ANALYSIS
[A] ✓: This is the correct ratio derived by equating the gravitational forces and solving for $x/d$.
[B] :x: This ratio is incorrect, not derived from the correct force balance.
[C] :x: This ratio is incorrect, not derived from the correct force balance.
[D] :x: This ratio is incorrect, not derived from the correct force balance.
## LEARNING AIDS
	**Common Mistakes**:
		1. [Error 1]: Incorrectly assigning distances. Ensure the distance from the asteroid to the $2M$ star is $(d-x)$, not $x$.
		2. [Error 2]: Forgetting to take the square root after equating the forces. Remember to simplify the equation before solving for $x$.
	**Quick Tips**:
		1. [Helpful shortcut]:  When forces are equal, you can often equate the expressions directly and simplify.
		2. [Memory aid]: Draw a diagram to clearly visualize the distances and forces involved.

VENKATESWARARAO · Answer

The correct Option is "$\frac{1}{1 + \sqrt{2}}$"

VENKATESWARARAO · Answer

$\frac{1}{2 + \sqrt{2}}$

VENKATESWARARAO · Answer

$\frac{2}{1 + \sqrt{2}}$

VENKATESWARARAO · Answer

$\frac{\sqrt{2}}{1 + \sqrt{2}}$

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