The Question is from neet-physics gravitation , and it was created by the author ABDUL JILANI BASHA

Question

At what height above the Earth's surface is the acceleration due to gravity half of its value at the surface? (R is the radius of Earth)

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option C
## CONCEPT FRAMEWORK
**Primary Concept**: Variation of Gravity with Height
**Related Topics**: Gravitational Force, Acceleration due to Gravity
**Prerequisites**: Newton's Law of Universal Gravitation, Understanding of acceleration due to gravity
## SOLUTION PATHWAY
**Given Information**:
* Acceleration due to gravity at the surface = g
* Acceleration due to gravity at height h = g/2
* Radius of Earth = R
**Method & Steps**:
1.  Write the formula for acceleration due to gravity at a height h above the Earth's surface.
    Reasoning: We need to relate the acceleration due to gravity at a certain height to its value at the surface.
    Formula/Rule: $g_h = \frac{GM}{(R+h)^2}$, where G is the gravitational constant, M is the mass of the Earth, R is the radius of the Earth, and h is the height above the surface.
2. Write the formula for acceleration due to gravity at the surface.
    Reasoning: We need the reference value of gravity at the surface to compare with.
    Formula/Rule: $g = \frac{GM}{R^2}$
3.  Set up the given condition that acceleration at height h is half of the surface value.
    Reasoning: We are given the condition that $g_h = g/2$.
    Working:  $\frac{GM}{(R+h)^2} = \frac{1}{2} \frac{GM}{R^2}$
4. Solve for h.
    Working: $\frac{1}{(R+h)^2} = \frac{1}{2R^2}$ ; $(R+h)^2 = 2R^2$ ; $R+h = \sqrt{2}R$ ; $h = \sqrt{2}R - R$ ; $h = (\sqrt{2} - 1)R$
    Key Point: We have isolated the height h.
5. State the final result.
    Result: The height at which the acceleration due to gravity is half its value at the surface is $(\sqrt{2} - 1)R$.
    Verification: The derived formula matches the correct option.
## OPTION ANALYSIS
[A] :x: Incorrect. This would correspond to the square root of the acceleration being halved.
[B] :x: Incorrect. At a height of R, the gravity is not half of its surface value. It is 1/4th of the surface value, according to the inverse square law.
[C] ✓: Correct. This matches the derived result.
[D] :x: Incorrect. At a height of 2R, the gravity is 1/9th of its surface value, according to the inverse square law.
## LEARNING AIDS
**Common Mistakes**:
1. **Forgetting to Square the Radius**: Using $(R+h)$ instead of $(R+h)^2$ in the gravity formula.
How to avoid: Always remember that gravity follows an inverse square law.
2. **Confusing Radius and Height**: Using R+h incorrectly in the calculation. 
How to avoid: Distinguish between the radius of the Earth and the height above the surface.
**Quick Tips**:
1. **Inverse Square Law**: Remember that the acceleration due to gravity decreases with the square of the distance from the center of the Earth.
2. **Relationship between g and height**: When we go up in height, the value of g decreases.

ABDUL JILANI BASHA · Answer

R/2

ABDUL JILANI BASHA · Answer

R

ABDUL JILANI BASHA · Answer

The correct Option is "$(\sqrt{2} - 1)R$"

ABDUL JILANI BASHA · Answer

2R

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