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

Question

The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from this platform is $fv$, where $v$ is its escape velocity from the surface of the Earth. The value of f is

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
	**Primary Concept**: Escape Velocity and Conservation of Energy
	**Related Topics**: Gravitational Potential Energy, Kinetic Energy
	**Prerequisites**: Definition of Escape Velocity, Formula for Gravitational Potential Energy
## SOLUTION PATHWAY
**Given Information**:
	* Earth's radius: R
	* Platform height from surface: R
	* Escape velocity from surface: v
	* Escape velocity from platform: fv
**Method & Steps**:
	1.  **Escape velocity at surface:** The escape velocity at the surface of the Earth is given by $v = \sqrt{\frac{2GM}{R}}$, where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
		Reasoning: This is the standard formula for escape velocity derived from energy conservation.
		Formula/Rule: $v = \sqrt{\frac{2GM}{R}}$
	2.  **Escape velocity at height R:**  At a height R from the surface, the distance from the center of the Earth is 2R. The escape velocity at this height, denoted as v', is given by $v' = \sqrt{\frac{2GM}{2R}}$
		Reasoning: The distance from the center of the Earth is 2R, so we substitute 2R for R in the escape velocity formula.
		Formula/Rule: $v' = \sqrt{\frac{2GM}{2R}}$
	3.  **Relate v' and v:**  We are given that $v' = fv$. Substitute the expressions for v and v'. We have  $fv = \sqrt{\frac{2GM}{2R}} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}} = \frac{1}{\sqrt{2}} v$.
		Working: Substituting and simplifying the expressions.
		Key Point: The escape velocity decreases as we move away from the Earth's center.
	4. **Find f:** From the relation $fv = \frac{1}{\sqrt{2}} v$, we get $f= \frac{1}{\sqrt{2}}$.
		Result: $f = \frac{1}{\sqrt{2}}$
		Verification: The value is consistent with the expected decrease in escape velocity at a higher altitude.
 ## OPTION ANALYSIS
[A] :x: Incorrect. This is not the correct ratio of escape velocities.
[B] :x: Incorrect. This is not the correct ratio of escape velocities.
[C] :x: Incorrect. This would be a higher escape velocity, which is not expected at a higher altitude.
[D] ✓: Correct. The escape velocity at height R is $\frac{1}{\sqrt{2}}$ times the escape velocity at the surface.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Using R instead of 2R for the distance from the center of the Earth at height R. [How to avoid]: Always consider the distance from the center of the Earth for gravitational calculations.
		2.[Error 2]: Incorrectly manipulating square roots. [How to avoid]: Be careful when taking square roots and simplifying expressions.
	**Quick Tips**:
		1.[Helpful shortcut]: Remember the inverse square relationship in gravitational potential and escape velocity.
		2.[Memory aid]:  The escape velocity decreases with height as the gravitational force weakens.

VENKATESWARARAO · Answer

$\frac{1}{3}$

VENKATESWARARAO · Answer

$\frac{1}{2}$

VENKATESWARARAO · Answer

$\sqrt 2 $

VENKATESWARARAO · Answer

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

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