The height in terms of radius of the earth (R), at which the acceleration due to gravity becomes g/9, where g is acceleration due to gravity on earth's surface, is _______.

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q30 – JEE-MAIN 2026

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Home Physics GravitationQuestion #187

The height in terms of radius of the earth (

R

), at which the acceleration due to gravity becomes

\frac{g}{9}

, where

g

is acceleration due to gravity on earth's surface, is _______.

Clue 1 of 3: Gravitational Acceleration Variation

Recall how the acceleration due to gravity changes with altitude above the Earth's surface.

🧠 Full Solution Path

Step 1: Identify the relevant formula for gravitational acceleration.✦ Active

The acceleration due to gravity at a height $h$ above the Earth's surface, $g^{'}$ , is related to the acceleration due to gravity on the surface, $g$ , by the formula:

g^{'} = g {(\frac{R}{R + h})}^{2}

where $R$ is the radius of the Earth.

Step 2: Substitute the given condition and simplify.○ Expand

We are given that the acceleration due to gravity becomes $\frac{g}{9}$ . So, we set $g^{'} = \frac{g}{9}$ :

\frac{g}{9} = g {(\frac{R}{R + h})}^{2}

Dividing both sides by $g$ gives:

\frac{1}{9} = {(\frac{R}{R + h})}^{2}

Step 3: Solve for the height

h

.○ Expand

Taking the square root of both sides:

\sqrt{\frac{1}{9}} = \frac{R}{R + h}

\frac{1}{3} = \frac{R}{R + h}

Cross-multiplying yields:

R + h = 3 R

Subtracting $R$ from both sides gives the height $h$ :

h = 3 R - R = 2 R

Thus, the height is $2 R$ .

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Physics Question 30 – JEE-MAIN 2026

Clue 1 of 3: Gravitational Acceleration Variation

🧠 Full Solution Path