g at height h: g' = g(1-2h/R). g at depth d: g' = g(1-d/R).

Variation of g with Height and Depth

Question

A satellite is orbiting at a height $h = 3200$ km above Earth’s surface. At what depth below Earth’s surface would the value of $g$ be the same as at this height? Take $R_E = 6400$ km.

Solution — Step by Step

For $h \ll R$ , the formula is:

g_h = g\left(1 - \frac{2h}{R}\right)

Plug in $h = 3200$ km and $R = 6400$ km:

g_h = g\left(1 - \frac{2 \times 3200}{6400}\right) = g\left(1 - 1\right) = 0

Wait — that gives zero, which means our approximation broke down. At $h = R/2$ , we’re no longer in the “small height” regime. Let’s use the exact formula instead.

The exact expression comes directly from Newton’s law:

g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2}

With $h = 3200$ km and $R = 6400$ km, we get $h/R = 1/2$ :

g_h = \frac{g}{\left(\frac{3}{2}\right)^2} = \frac{g}{\frac{9}{4}} = \frac{4g}{9}

The formula for $g$ at depth $d$ below the surface is:

g_d = g\left(1 - \frac{d}{R}\right)

This one is always linear — there’s no “approximate vs exact” issue here because it’s derived from a uniform sphere model, not an inverse-square law at a distance.

Set $g_d = g_h$ :

g\left(1 - \frac{d}{R}\right) = \frac{4g}{9}

1 - \frac{d}{R} = \frac{4}{9}

\frac{d}{R} = 1 - \frac{4}{9} = \frac{5}{9}

d = \frac{5}{9} \times 6400 \approx \mathbf{3556 \text{ km}}

Why This Works

The key physics here is that the two variation laws come from completely different physical models. The height formula ( $g \propto 1/r^2$ ) treats Earth as a point mass — as you go higher, all of Earth’s mass is still pulling you, just from farther away. The depth formula ( $g \propto r$ ) comes from the shell theorem: mass in the shell above you contributes zero net force, so only the mass of the sphere below counts.

At depth $d$ , the effective mass pulling you is $M \cdot (R-d)^3/R^3$ , and the distance is $(R-d)$ . When you combine these, the distance squared in the denominator cancels partially with the numerator, leaving a clean linear relationship. This is why $g$ at depth decreases slowly compared to $g$ at height — at the surface, going down even 1000 km barely changes $g$ , while going up 1000 km has a much larger effect.

This question tests whether you remember which formula to use. Board exams often give small heights (use approximate), while JEE loves giving $h = R/2$ or $h = R$ precisely to catch students using the wrong formula.

Alternative Method

You can work backwards from the depth side. The depth formula $g_d = g(1 - d/R)$ tells us $g_d$ varies linearly from $g$ at $d=0$ to $0$ at $d=R$ (the centre).

So we need $g_d = 4g/9$ , which means:

\frac{g_d}{g} = \frac{4}{9} \approx 0.444

On a linear scale from 1 (surface) to 0 (centre), the fraction $4/9$ of the way corresponds to depth $d = (1 - 4/9)R = 5R/9$ .

This “fraction of $g$ ” approach is faster in MCQs where you don’t need the numerical answer.

Common Mistake

The most common error is using $g_h = g(1 - 2h/R)$ when $h$ is large. This approximation is valid only when $h \ll R$ (typically $h < 100$ km for board-level accuracy). In this problem, $h = R/2$ , so the approximate formula gives a completely wrong answer ( $g_h = 0$ ). Always check the ratio $h/R$ first — if it’s not negligible, switch to the exact formula $g/(1 + h/R)^2$ .

For quick MCQ comparison: at height $h = R$ (one full Earth radius above surface), $g$ drops to $g/4$ . At depth $d = R/2$ (halfway to the centre), $g$ drops to $g/2$ . These reference points are worth remembering — JEE Main has directly tested both in recent years.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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