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Question

Assume that the acceleration due to gravity on the surface of the moon is $0.2$ times the acceleration due to gravity on the surface of the earth. If ${R_e}$ is the maximum range of a projectile on the earth’s surface, what is the maximum range on the surface of the moon for the same velocity of projection

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
**Primary Concept**: Projectile Motion and Range
**Related Topics**: Kinematics, Acceleration due to gravity
**Prerequisites**: Understanding of projectile motion equations, especially the range formula.
## SOLUTION PATHWAY
**Given Information**:
* Acceleration due to gravity on the moon ($g_m$) = 0.2 * acceleration due to gravity on Earth ($g_e$)
* Range on Earth = $R_e$
* Same initial velocity for both cases
**Method & Steps**:
1.  **Recall the range formula for a projectile:** The maximum range of a projectile is given by $R = \frac{v^2 \sin(2	heta)}{g}$, where $v$ is the initial velocity, $	heta$ is the launch angle, and $g$ is the acceleration due to gravity. For maximum range, $	heta = 45^{\circ}$, and $\sin(2	heta) = 1$.  Therefore, $R = \frac{v^2}{g}$.
    Reasoning:  This formula relates range to velocity and gravity.
    Formula/Rule:  $R = \frac{v^2 \sin(2	heta)}{g}$
2.  **Express the range on Earth:** $R_e = \frac{v^2}{g_e}$
    Working: Substituting the given value of $\sin(2	heta)$ for maximum range
    Key Point: We are given the range on earth, so substituting g as $g_e$
3.  **Express the range on the moon:** $R_m = \frac{v^2}{g_m}$
    Working: Substituting g as $g_m$
    Key Point:  The initial velocity is the same.
4.  **Relate $g_m$ to $g_e$:** $g_m = 0.2 g_e$
   Reasoning: Given in the problem statement.
5.  **Substitute $g_m$ into the range on the moon equation:** $R_m = \frac{v^2}{0.2 g_e} = \frac{1}{0.2} \frac{v^2}{g_e} = 5 \frac{v^2}{g_e}$
    Working: Substituting $g_m$ in $R_m$ equation
6.  **Substitute $R_e$ into the equation for $R_m$:** Since $R_e = \frac{v^2}{g_e}$, $R_m = 5 R_e$
    Result: $R_m = 5R_e$
    Verification: The derived relation matches the correct answer.
## OPTION ANALYSIS
[A] :x: Incorrect. The range on the moon is larger, not smaller than the range on Earth.
[B] :x: Incorrect. The range is not twice the Earth's range.
[C] :x: Incorrect. The range is not half of the Earth's range.
[D] ✓: Correct. The range on the moon is 5 times the Earth's range.
## LEARNING AIDS
**Common Mistakes**:
1. **Incorrect formula**: Using incorrect projectile motion equations.
   How to avoid: Memorize or derive the correct range formula.
2. **Incorrect substitution**: Incorrectly substituting values for gravity.
   How to avoid: Double-check that the correct gravity is used for each location.
**Quick Tips**:
1. **Range formula**: Remember the range formula and how it relates to gravity.
2. **Direct proportionality**: The range is inversely proportional to gravity when velocity is constant.

P.NARASIMHA · Answer

$0.2\;{R_e}$

P.NARASIMHA · Answer

$2\;{R_e}$

P.NARASIMHA · Answer

$0.5\;{R_e}$

P.NARASIMHA · Answer

The correct Option is "$5\;{R_e}$"

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