Acceleration due to gravity at height and depth

On the earth's surface, acceleration due to gravity is $g$.

• At a height $h$ above the surface: $g_h = g\left(\frac{R}{R+h}\right)^2$ where $R$ is the radius of the earth.

  Since $R+h > R$, we have $g_h < g$ so gravity decreases as we go up.

• At a depth $d$ below the surface: $g_d = g\left(1 - \frac{d}{R}\right)$ for depth inside the earth (assuming uniform density).

  Since $d>0$, $g_d < g$ so gravity also decreases as we go down.

Therefore, Statement I is correct.

Check whether gravity is same at equal height and depth

Statement II says that gravity is same at height $h$ and depth $d$ if $h=d$.

Let $h=d=x$.

Then, $g_h = g\left(\frac{R}{R+x}\right)^2$ and $g_d = g\left(1 - \frac{x}{R}\right)$

These are not exactly equal in general.

For small $x$: $g_h \approx g\left(1 - \frac{2x}{R}\right)$ while $g_d = g\left(1 - \frac{x}{R}\right)$

Clearly, $g_h \neq g_d$

Hence, Statement II is incorrect.

Evaluate the options

• A: Statement I is correct but statement II is incorrect ✅
• B: Both correct ❌
• C: Statement I incorrect but statement II correct ❌
• D: Both incorrect ❌

Final answer

The correct option is: $\boxed{A}$

Comparison with stored answer

Stored correct answer: $A$

My derived answer also is $A$, so they agree.