Acceleration due to gravity above and below Earth's surface

Stand on a bathroom scale at sea level and it reads a certain weight. Carry that same scale to the top of Mount Everest, or imagine lowering it down the world's deepest mine shaft, and the reading changes — not because your mass changed, but because $g$ , the acceleration due to gravity, is not a fixed number stamped on the planet. It is a field, strongest at the surface and weaker in both directions you can move away from it: up into the sky, or down into the crust. The two ways of leaving the surface turn out to obey completely different laws, and that contrast is the whole point of this topic. Climb higher and you are simply getting farther from a fixed lump of mass — the familiar inverse-square law does the work, and $g$ falls off roughly as $1 - 2h/R$ . Dig deeper and something stranger happens: the shell of Earth above you stops pulling on you altogether, so only the shrinking sphere of mass beneath your feet counts, and $g$ falls off linearly, as $1 - d/R$ , all the way to zero at the very centre of the planet — the one place on Earth where you would be perfectly, eerily weightless while still surrounded by rock in every direction. This article builds both formulas from the surface value $g = GM/R^2$ that you already know, keeps every approximation honest about when it is allowed to be used, and works through two numbers you can hold onto: gravity at 32 km up (about where high-altitude balloons fly) and the depth at which gravity has dropped to half. By the end you should be able to derive both expressions from scratch, explain in one sentence why the two graphs of $g$ look so different on either side of the surface, and avoid the handful of traps that quietly cost marks in board exams.

Key takeaways

Above the surface, at height $h$ : $g_h = \dfrac{GM}{(R+h)^2} = g\left(1+\dfrac{h}{R}\right)^{-2} \approx g\left(1-\dfrac{2h}{R}\right)$ for $h \ll R$ — $g$ decreases because you are farther from a fixed mass.
Below the surface, at depth $d$ : $g_d = g\left(1-\dfrac{d}{R}\right)$ — $g$ decreases because the attracting mass itself shrinks as the outer shell stops contributing.
The height formula is a quadratic-in- $h/R$ approximation valid only for $h \ll R$ ; the depth formula is exact for a uniform-density Earth, with no small-quantity restriction on $d$ .
For the same small distance from the surface, altitude costs you gravity twice as fast as depth does — the $2h/R$ factor versus the $d/R$ factor.
$g$ is zero exactly at the Earth's centre ( $d = R$ ), not at its maximum there — a very common misconception.
Worked numbers to anchor intuition: at $h = 32\ \text{km}$ , $g$ drops from $9.8$ to $9.702\ \text{m/s}^2$ ; $g$ falls to half its surface value at $d = 3200\ \text{km}$ , exactly halfway to the centre.
Key idea: g falls off as the inverse square of distance above the surface (because the pulling mass is fixed) but falls off linearly below the surface (because the pulling mass itself shrinks), reaching zero only at the Earth's centre.

The core insight

Moving up, $g$ shrinks by the inverse-square law acting on a fixed mass ( $g_h \approx g(1-2h/R)$ ); moving down, $g$ shrinks linearly because only the shell inside your radius still pulls on you ( $g_d = g(1-d/R)$ ), vanishing at the centre.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Physics chapter on Gravitation, directly extending the surface relation $g = GM/R^2$ that you meet earlier in the same chapter. The prescribed scope asks you to know both variation laws — with altitude, $g_h = g(1-2h/R)$ for $h \ll R$ , and with depth, $g_d = g(1-d/R)$ — and to be able to derive each from the universal law of gravitation, $F = Gm_1m_2/r^2$ . It is a favourite two-mark and derivation question because it tests whether you actually understand why gravity changes, not just that it does; the exact height formula and the shell argument behind the depth formula are both fair game, but anything about a non-uniform (realistic) Earth density profile or about gravitational field theory is outside this chapter's scope.

The physical picture

Start from the surface formula you already trust: for a spherical Earth of mass $M$ and radius $R$ , treated as a point mass at its centre, the acceleration due to gravity at the surface is

g = \frac{GM}{R^2}

where: $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ is the universal gravitational constant, $M$ is Earth's mass, and $R \approx 6400\ \text{km}$ is its radius. Everything below is this same formula applied at a different distance from the centre, or with a different amount of mass doing the pulling.

Moving up, above the surface. At a height $h$ above the surface, you are now a distance $(R+h)$ from Earth's centre, and the entire mass $M$ still pulls on you (nothing has been left behind), so:

g_h = \frac{GM}{(R+h)^2}

where: $h$ is the height above the surface. Dividing through by $g = GM/R^2$ gives the compact form $g_h = g\left(1+\dfrac{h}{R}\right)^{-2}$ — the whole effect is that the denominator $(R+h)^2$ has grown, so the fraction shrinks. This is pure inverse-square behaviour: same mass, greater distance, weaker pull.

Moving down, below the surface. At a depth $d$ below the surface, you are a distance $(R-d)$ from Earth's centre — but now something changes besides the distance. A famous result about spherical shells (true for gravity, and used again for electric fields around charged shells) says that a uniform spherical shell exerts zero net gravitational force on any point strictly inside it: the pull from the near side of the shell and the pull from the far side cancel exactly. So of the whole Earth, only the inner sphere of radius $(R-d)$ — the mass "beneath" you — contributes any net pull; the shell of thickness $d$ above you contributes nothing. If the Earth has uniform density $\rho$ , that inner sphere has mass $M' = \tfrac{4}{3}\pi(R-d)^3\rho$ , giving:

g_d = \frac{GM'}{(R-d)^2} = \frac{4}{3}\pi G\rho(R-d)

where: $\rho$ is Earth's (assumed uniform) density and $(R-d)$ is the shrinking radius of the sphere still pulling on you. Because this expression is directly proportional to $(R-d)$ , and the surface value is $g = \tfrac{4}{3}\pi G\rho R$ , dividing the two gives the clean linear law $g_d = g(1-d/R)$ .

Figure

Two graphs of g against distance from Earths centre, one line falling off as inverse square above the surface and a straight line falling off linearly to zero below the surface, meeting at the surface value g

Key contrast. Above the surface, $g$ falls because you are moving farther from a fixed total mass — the inverse-square law bites, so the fall-off is quadratic in the small quantity $h/R$ (the $-2h/R$ term). Below the surface, $g$ falls because the attracting mass itself is shrinking as the shell above you drops out — the fall-off is exactly linear in $d/R$ . This is why the two curves either side of the surface look different: a gently curving fall on the outside, a straight-line fall on the inside, meeting smoothly at the surface value $g$ .

Derivation AThe small-height approximation for g above the surface

The goal is to show how the exact inverse-square expression $g_h = GM/(R+h)^2$ , which is awkward to use by hand for small everyday heights, collapses into the simple linear approximation $g_h \approx g(1-2h/R)$ that the syllabus actually expects you to derive and apply, and to be precise about exactly when that collapse is valid — because using it outside its range of validity is the single most common exam error on this topic.

Step 1. Write the exact expression and factor out $R$ from the denominator. Starting from $g_h = \dfrac{GM}{(R+h)^2}$ , factor $R^2$ out of $(R+h)^2 = R^2\left(1+\dfrac{h}{R}\right)^2$ , so that $g_h = \dfrac{GM}{R^2}\left(1+\dfrac{h}{R}\right)^{-2} = g\left(1+\dfrac{h}{R}\right)^{-2}$ , using $g = GM/R^2$ . The whole height-dependence is now packed into the single dimensionless ratio $h/R$ .

Step 2. Apply the binomial approximation. For any small $x$ , $(1+x)^n \approx 1+nx$ when higher powers of $x$ can be neglected. Here $x = h/R$ and $n = -2$ , so $\left(1+\dfrac{h}{R}\right)^{-2} \approx 1 - \dfrac{2h}{R}$ , keeping only the first-order term and discarding terms in $(h/R)^2$ and smaller.

Step 3. Substitute back to get the working formula. $g_h \approx g\left(1-\dfrac{2h}{R}\right)$ . This says: for every step you take upward that is a small fraction of Earth's radius, $g$ drops by twice that fraction of its surface value — the factor of $2$ is the signature of the inverse-square law linearised around the surface.

Step 4. State the validity condition honestly. The approximation is only trustworthy when $h \ll R$ (say $h/R$ no more than a few per cent), because that is exactly when the dropped terms in $(h/R)^2$ are negligible next to the kept term in $h/R$ . For heights that are a sizeable fraction of $R$ — high-altitude satellites, for instance — you must fall back on the exact form $g_h = GM/(R+h)^2$ ; plugging a large $h$ into the linear approximation can even predict a nonsensical negative $g$ . Consistency check: for $h = 0$ , both the exact and approximate expressions correctly give $g_h = g$ ; as $h \to \infty$ the exact form correctly tends to zero while the linear approximation would (wrongly) go negative — confirming the approximation is a small- $h$ tool only.

Worked examples

Board level - gravity at balloon altitude

Find the acceleration due to gravity at a height of 32 km above the Earth's surface. Take $g = 9.8\ \text{m/s}^2$ and $R_E = 6400\ \text{km}$ .

Step 1. Check that the approximation applies. Here $h = 32\ \text{km}$ and $R = 6400\ \text{km}$ , so $h/R = 32/6400 = 0.005$ — well under $1\%$ , comfortably satisfying $h \ll R$ . The linear approximation $g_h \approx g(1-2h/R)$ is safe to use.

Step 2. Substitute into the formula. $g_h = 9.8\left(1-\dfrac{2\times32}{6400}\right) = 9.8\left(1-\dfrac{64}{6400}\right) = 9.8(1-0.01)$ .

Step 3. Finish the arithmetic. $g_h = 9.8\times0.99 = \mathbf{9.702\ \text{m/s}^2}$ . So at 32 km altitude — roughly where a high-altitude weather balloon flies — gravity has weakened only slightly, from $9.8$ to about $9.70\ \text{m/s}^2$ , a drop of exactly $1\%$ , matching the $1\%$ value of $h/R$ we checked in Step 1.

Competitive level - depth for half gravity, then compare with an equal altitude

At what depth below the Earth's surface does the acceleration due to gravity reduce to 50% of its surface value? Take $R_E = 6400\ \text{km}$ . Then find what height above the surface would produce the same 50% reduction using the exact (not approximate) height formula, and compare the two distances.

Step 1. Solve the depth part first, since it is exact and simplest. Using $g_d = g(1-d/R)$ , set $g_d = 0.5g$ : $0.5g = g\left(1-\dfrac{d}{R}\right) \implies 0.5 = 1-\dfrac{d}{R} \implies \dfrac{d}{R} = 0.5$ , so $d = 0.5\times6400 = \mathbf{3200\ \text{km}}$ — exactly halfway to the centre.

Step 2. Set up the height part using the exact formula, since a height that halves $g$ is nowhere near $h \ll R$ , so the linear approximation would be invalid here. From $g_h = \dfrac{GM}{(R+h)^2} = g\left(\dfrac{R}{R+h}\right)^2$ , set $g_h = 0.5g$ : $0.5 = \left(\dfrac{R}{R+h}\right)^2$ .

Step 3. Solve for $h$ . Taking the square root, $\dfrac{R}{R+h} = \sqrt{0.5} = 0.7071$ , so $R+h = \dfrac{R}{0.7071} = 1.4142R$ , giving $h = 0.4142R = 0.4142\times6400 \approx \mathbf{2651\ \text{km}}$ .

Step 4. Compare the two answers. Halving $g$ needs a depth of $3200\ \text{km}$ but only a height of about $2651\ \text{km}$ — you have to travel noticeably farther into the Earth than away from it to achieve the same 50% drop, because the linear depth law falls more gently once $d/R$ stops being small, while the inverse-square height law keeps accelerating its own fall-off as $h$ grows. This is exactly why the simple "depth falls twice as slowly as height" rule of thumb from the common-traps section only holds for small $h$ and $d$ — it breaks down once you check it against the exact formula at large distances, as this problem does. Timing note: in a competitive setting, solve the depth part first (30 seconds, one substitution) and bank those marks; the height part needs a square root and is worth spending the remaining time on — do not let the exact formula's algebra eat into time you need for other questions.

Common traps

Applying the height formula $g_h = g(1-2h/R)$ when $h$ is comparable to or larger than $R$ — this approximation is only valid for $h \ll R$ ; for larger heights, use the exact form $g_h = GM/(R+h)^2$ , as the competitive example above shows.
Assuming $g$ decreases at the same rate with height and with depth — for small distances, height gives a factor of $2h/R$ while depth gives a factor of only $d/R$ , so $g$ falls off twice as fast with altitude as with depth for the same small distance from the surface.
Forgetting that the depth formula assumes uniform density of the Earth — real Earth density increases toward the core, so the actual variation deviates from this simple linear law; the syllabus wants you to know this is an idealisation.
Believing gravity is strongest at the Earth's centre — it is actually zero at the centre, since the pull from surrounding mass cancels out symmetrically in all directions, not maximal from being "deepest inside."
Mixing up which radius shrinks in the depth derivation — it is $(R-d)$ , the distance from the centre, not $d$ itself, that appears in $g_d = GM'/(R-d)^2$ before it simplifies to the linear form.

You should be able to…

Derive the approximate expression $g_h \approx g(1-2h/R)$ for height $h$ from the exact inverse-square relation, and state the condition $h \ll R$ under which it is valid.
Derive the expression $g_d = g(1-d/R)$ for depth $d$ , explaining the shell-theorem role of the enclosed mass and the uniform-density assumption.
Explain in one sentence why $g$ falls off quadratically (in $h/R$ ) above the surface but linearly (in $d/R$ ) below it.
State that $g$ is zero exactly at the Earth's centre, and be able to justify why using the depth formula.
Solve numerical problems for $g$ at a given altitude or depth, choosing correctly between the approximate and exact height formulas based on whether $h \ll R$ holds.

Ask the tutor: Want a worked comparison of how far you'd have to go up versus down to cut gravity by 10%, or help checking which formula applies to a tricky altitude-versus-depth numerical?

Why does gravity fall off differently above and below the surface — shouldn't it be the same formula both ways?

It would be the same formula if the mass doing the pulling stayed fixed in both cases, but it doesn't. Above the surface, the entire Earth is still below you and still pulls on you as one lump — only your distance from that fixed lump changes, so the ordinary inverse-square law $g \propto 1/r^2$ applies directly, giving the $g(1-2h/R)$ approximation. Below the surface, the situation is genuinely different: the shell of Earth now sitting above you contributes zero net pull (its near and far sides cancel), so the mass that counts is only the smaller sphere beneath you, and that mass itself shrinks as you go deeper. Two different physical situations, so two different laws — one curved (inverse-square), one straight-line (linear).

Is it true that if I dug a tunnel through the Earth I would float weightless once I passed the centre?

At the exact centre, yes — momentarily — because $g_d = g(1-d/R)$ gives zero when $d=R$ . But you would not stay weightless: as you continued past the centre toward the far side, the (idealised uniform-density) formula predicts $g$ growing again in the opposite direction, pulling you back toward the centre. In an idealised frictionless tunnel this actually produces a beautiful back-and-forth oscillation through the Earth, a topic sometimes explored beyond this syllabus using simple harmonic motion — but the CBSE Class 11 scope only asks you to know that $g$ is zero at the centre, not to analyse the full oscillation.

Why is the height formula only an approximation but the depth formula exact?

The height formula $g_h \approx g(1-2h/R)$ is obtained by throwing away small terms in a binomial expansion — a deliberate simplification of the exact $g_h = GM/(R+h)^2$ , done purely to make hand calculation easier when $h \ll R$ . The depth formula $g_d = g(1-d/R)$ , by contrast, is not an approximation of anything more complicated; given the assumption of uniform density, it falls straight out of the shell theorem with no terms discarded. Its only "approximation" is the physical assumption of uniform density itself, which is a modelling choice about the real Earth, not a mathematical shortcut within the formula.

Does this topic actually come up in board exams, or is it more for competitive exams?

Both, and it is one of the more reliable topics to expect. CBSE board papers regularly ask for the derivation of either $g_h \approx g(1-2h/R)$ or $g_d = g(1-d/R)$ as a 2 or 3-mark question, plus a numerical using one of them — exactly the two worked examples above. Competitive exams like JEE and NEET go further, testing the exact height formula for large $h$ , comparisons between height and depth effects for the same numerical distance, and occasionally the shell-theorem reasoning behind the depth formula. Knowing both the approximate and exact forms, and when each applies, covers you for both.