The Question is from neet-physics gravitation gravitational-potential-energy, 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, the gravitational potential energy of the asteroid is:

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Gravitational Force and Potential Energy
	**Related Topics**: Newton's Law of Universal Gravitation, Superposition Principle
	**Prerequisites**: Understanding of gravitational force, potential energy, and solving quadratic equations.
## SOLUTION PATHWAY
**Given Information**:
	Masses of stars: $M$ and $2M$
	Distance between stars: $d$
	Mass of asteroid: $m$
	Distance of asteroid from mass $M$: $x$
	Net gravitational force on the asteroid is zero.
**Method & Steps**:
	1. Find the position 'x' where the net gravitational force is zero.
		Reasoning: The gravitational forces from the two stars on the asteroid must be equal and opposite for the net force to be zero.
		Formula/Rule: $F = G\frac{Mm}{r^2}$
	2. Set the gravitational forces equal and solve for x.
		Working: $G\frac{Mm}{x^2} = G\frac{2Mm}{(d-x)^2}$. Simplifying gives $\frac{1}{x^2} = \frac{2}{(d-x)^2}$. Taking the square root of both sides gives $\frac{1}{x} = \frac{\sqrt{2}}{d-x}$. Solving for x, we get $(d-x) = x\sqrt{2}$, so $x = \frac{d}{1+\sqrt{2}} = d(\sqrt{2}-1)$.
		Key Point: Only consider the physical solution where x is between 0 and d.
	3. Calculate the gravitational potential energy of the asteroid at this position.
		Working: $U = -\frac{GMm}{x} - \frac{G(2M)m}{d-x} = -GMm(\frac{1}{d(\sqrt{2}-1)} + \frac{2}{d-d(\sqrt{2}-1)}) = -GMm(\frac{1}{d(\sqrt{2}-1)} + \frac{2}{d(2-\sqrt{2})})$. Simplifying further: $U = -\frac{GMm}{d}(\frac{1}{\sqrt{2}-1} + \frac{2}{2-\sqrt{2}}) = -\frac{GMm}{d}(\frac{\sqrt{2}+1}{1} + \frac{2(2+\sqrt{2})}{2}) = -\frac{GMm}{d}(\sqrt{2}+1+2+\sqrt{2}) = -\frac{GMm}{d}(3+2\sqrt{2})$.
		Result: $U = -GMm\frac{(3+2\sqrt{2})}{d}$
		Verification: The units and sign are consistent with gravitational potential energy.
 ## OPTION ANALYSIS
[A] ✓: The calculated potential energy matches this option.
[B] :x: This option uses $3-2\sqrt{2}$ instead of $3+2\sqrt{2}$.
[C] :x: This option uses a factor of 2 before the entire expression and has the wrong coefficient for the square root term.
[D] :x: This option uses $2+\sqrt{2}$ instead of $3+2\sqrt{2}$.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrectly setting up the force equation or not considering the correct distance from each mass.
		2.[Error 2]: Making algebraic errors in solving for 'x' or in calculating the potential energy.
	**Quick Tips**:
		1.[Helpful shortcut]: Rationalize the denominators immediately after obtaining fractions to make the final addition easier.
		2.[Memory aid]: Remember that potential energy is always negative for a gravitational system.

VENKATESWARARAO · Answer

The correct Option is "-GMm(3+2√2)/d"

VENKATESWARARAO · Answer

-GMm(3-2√2)/d

VENKATESWARARAO · Answer

-2GMm(3+√2)/d

VENKATESWARARAO · Answer

-GMm(2+√2)/d

NEET Physics Gravitation Gravitational Potential Energy MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company