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Question

If the radius of Earth were doubled, keeping its mass constant, the escape velocity from Earth would:

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Escape Velocity
	**Related Topics**: Gravitational Potential Energy, Kinetic Energy, Conservation of Energy
	**Prerequisites**: Understanding of escape velocity and its formula.
## SOLUTION PATHWAY
**Given Information**:
	* The mass of Earth is constant.
	* The radius of Earth is doubled.
**Method & Steps**:
	1. Define the escape velocity formula.
		Reasoning: The escape velocity is the minimum speed an object needs to escape the gravitational pull of a celestial body.
		Formula/Rule: $v_e = \sqrt{\frac{2GM}{r}}$, where $v_e$ is the escape velocity, G is the gravitational constant, M is the mass of the celestial body, and r is its radius.
	2. Analyze the effect of doubling the radius.
		Working: Let $v_{e1}$ be the initial escape velocity with radius $r_1 = r$, and $v_{e2}$ be the new escape velocity with radius $r_2 = 2r$.  $v_{e1} = \sqrt{\frac{2GM}{r}}$ and $v_{e2} = \sqrt{\frac{2GM}{2r}}$.
		Key Point: The mass M and gravitational constant G remain constant.
	3. Find the ratio of the new escape velocity to the original.
		Working: $\frac{v_{e2}}{v_{e1}} = \frac{\sqrt{\frac{2GM}{2r}}}{\sqrt{\frac{2GM}{r}}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
	4. Conclude the change in escape velocity.
		Result: The new escape velocity is $\frac{1}{\sqrt{2}}$ times the original escape velocity. Therefore, the escape velocity decreases by a factor of $\sqrt{2}$.
		Verification: The result aligns with the inverse square root relationship between escape velocity and radius.
## OPTION ANALYSIS
[A] :x:  The escape velocity changes when the radius changes, so it does not remain unchanged.
[B] :x:  The escape velocity decreases, not increases.
[C] ✓:  The correct calculation shows that the escape velocity decreases by a factor of $\sqrt{2}$.
[D] :x:  The escape velocity decreases, not increases.
## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting the square root in the escape velocity formula. Always remember the square root is present in the formula.
		2.[Error 2]:  Confusing the effect of radius and mass on escape velocity. In this case, only the radius changes.
	**Quick Tips**:
		1.[Helpful shortcut]: The escape velocity is inversely proportional to the square root of the radius, when the mass is constant.  $v_e \propto \frac{1}{\sqrt{r}}$
		2.[Memory aid]:  Remember the formula for escape velocity $v_e = \sqrt{\frac{2GM}{r}}$ to easily derive the relationship.

P.NARASIMHA · Answer

Remain unchanged

P.NARASIMHA · Answer

Increase by a factor of 2

P.NARASIMHA · Answer

The correct Option is "Decrease by a factor of √2"

P.NARASIMHA · Answer

Increase by a factor of √2

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