Physics · Ch 8 — Gravitation

Acceleration Due to Gravity Below and Above the Surface of Earth

8.6

Acceleration Due to Gravity Below and Above the Surface of Earth

Variation of Acceleration Due to Gravity with Height and Depth

The acceleration due to gravity $g$ is not a universal constant. Its value changes as we move away from the Earth's surface — either upward into the atmosphere or downward into the Earth's interior. The reason is simple: gravity depends on distance from the Earth's centre, and the effective mass that pulls you changes when you go inside the Earth.

We treat the Earth as a uniform sphere of radius $R$ and mass $M$ , with its centre at $O$ . On the surface, the acceleration due to gravity is

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

where $G = 6.67 \times 10^{-11} \ \text{N m}^2 \ \text{kg}^{-2}$ is the universal gravitational constant.

1. Variation with Height (Above the Surface)

Consider a point at a height $h$ above the Earth's surface. Its distance from the Earth's centre is $r = R + h$ . The acceleration due to gravity at this height, call it $g_h$ , is given directly by Newton's law of gravitation:

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

We want to compare $g_h$ with the surface value $g$ . Divide the two expressions:

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

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

This is the exact formula. For practical calculations, especially when $h$ is small compared to $R$ (which is about 6400 km), we can expand it using the binomial theorem.

Write $\frac{R}{R+h} = \frac{1}{1 + h/R}$ . Then

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

For $h \ll R$ , expand:

\left(1 + \frac{h}{R}\right)^{-2} \approx 1 - \frac{2h}{R} + \frac{3h^2}{R^2} - \cdots

Neglecting terms of order $(h/R)^2$ and higher, we get the approximate formula:

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

Watch out

This approximation is valid only when $h \ll R$ . For heights comparable to $R$ (e.g., geostationary orbit at $h \approx 36000$ km), you must use the exact formula $g_h = g R^2/(R+h)^2$ .

Key result: As height increases, $g$ decreases. At a height $h = R$ (one Earth radius above the surface), $g_h = g/4$ .

2. Variation with Depth (Below the Surface)

Now consider a point at a depth $d$ below the Earth's surface. Its distance from the centre is $r = R - d$ . The crucial new idea is that the gravitational force at this point comes only from the mass of the Earth that lies inside the sphere of radius $r$ . The mass outside this sphere (the spherical shell between $r$ and $R$ ) exerts zero net gravitational force on a particle inside it — this is a consequence of the shell theorem.

Let $M'$ be the mass of the Earth enclosed within radius $r$ . Assuming uniform density $\rho$ ,

\rho = \frac{M}{\frac{4}{3}\pi R^3} = \frac{M'}{\frac{4}{3}\pi r^3}

Hence

M' = M \left( \frac{r^3}{R^3} \right) = M \left( \frac{R-d}{R} \right)^3

The acceleration due to gravity at depth $d$ , call it $g_d$ , is then

g_d = \frac{G M'}{r^2} = \frac{G}{r^2} \left[ M \left( \frac{r^3}{R^3} \right) \right] = \frac{GM r}{R^3}

Substitute $r = R - d$ :

g_d = \frac{GM (R-d)}{R^3} = \frac{GM}{R^2} \left(1 - \frac{d}{R}\right)

But $\frac{GM}{R^2} = g$ , the surface value. Therefore

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

Important

Inside the Earth, $g$ decreases linearly with depth. At the centre ( $d = R$ ), $g_d = 0$ . This makes physical sense: at the centre, the mass of the Earth pulls equally in all directions, so the net gravitational force is zero.

Note

The linear decrease assumes uniform density. The real Earth has a denser core, so the actual variation is not perfectly linear — but the qualitative trend (decrease toward the centre) holds.

3. Comparison: Above vs. Below

Location	Distance from centre	Formula for $g$	Behaviour
Surface	$R$	$g = GM/R^2$	Reference value
Height $h$	$R+h$	$g_h = g R^2/(R+h)^2$	Decreases as $1/r^2$
Depth $d$	$R-d$	$g_d = g (1 - d/R)$	Decreases linearly

Figure 7.8.ag at a height h above the surface of the earth.

Fig. 7.8.a — g at a height h above the surface of the earth.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

The figure shows a cross-section of the Earth drawn as a shaded disk. From the centre of the disk to a point on its lower-right edge, a straight line is drawn and labelled $R_E$ — this is the Earth’s radius. On the opposite side, at the top of the disk, a vertical arrow rises from the surface straight upward. The arrow is labelled $h$ , and at its tip sits a small dot representing a point mass (the object whose weight we are studying). The arrow makes it clear that $h$ is measured from the surface, not from the centre.

The physical idea is straightforward: as you move away from the Earth’s surface, the gravitational force weakens. The figure isolates the effect of altitude by showing the object at a height $h$ above ground, with the Earth’s full radius $R_E$ drawn for comparison. The key question the diagram helps answer is: how does the acceleration due to gravity $g$ change when you are no longer on the surface?

The textbook uses this geometry to derive the formula for $g$ at a height $h$ . At the surface, the distance from the Earth’s centre is $R_E$ , and the acceleration is

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

where $G$ is the universal gravitational constant and $M$ is the mass of the Earth. At a height $h$ above the surface, the distance from the centre becomes $R_E + h$ . The acceleration due to gravity at that height, call it $g_h$ , is therefore

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

Dividing the two expressions gives a compact relation:

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

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

This is the central result tied to Fig. 7.8.a. Every symbol is defined: $g$ is the surface value ( $9.8\ \text{m/s}^2$ ), $R_E$ is the Earth’s radius (about $6.4 \times 10^6\ \text{m}$ ), and $h$ is the height above the surface. The formula shows that $g_h$ is always less than $g$ , and the decrease becomes significant only when $h$ is a substantial fraction of $R_E$ .

Watch out

A common mistake is to use $h$ as the distance from the centre. The figure’s arrow starts at the surface, not at the centre — $h$ is the altitude above ground. The full distance from the centre is $R_E + h$ , not $h$ alone. …

Figure 7.8.bg at a depth d. In this case only the smaller sphere of radius (RE-d) contributes to g.

Fig. 7.8.b — g at a depth d. In this case only the smaller sphere of radius (RE-d) contributes to g.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

The figure shows a cross-section of the Earth as a perfect sphere of radius $R_E$ and total mass $M_E$ . A dashed circle of radius $R_E - d$ is drawn inside it, concentric with the outer surface. The region between the two circles — a spherical shell of thickness $d$ — is shaded or marked with a small double-arrow labelled $d$ at the top. The inner sphere (radius $R_E - d$ ) is labelled $M_s$ , and the outer shell is the part of the Earth that lies above a point at depth $d$ .

The physical idea is straightforward: if you go a distance $d$ below the Earth's surface, the gravitational acceleration you feel is not due to the whole Earth. The shell of material above you (thickness $d$ ) exerts zero net gravitational pull on you — a result from Newton's shell theorem. Only the mass of the smaller sphere beneath you, of radius $R_E - d$ , contributes. That inner sphere has mass $M_s$ , which is less than $M_E$ because the shell has been removed.

The textbook uses this picture to derive the formula for $g$ at depth $d$ . Assuming uniform density $\rho$ , the mass of the inner sphere is proportional to its volume:

M_s = \rho \cdot \frac{4}{3}\pi (R_E - d)^3

while the Earth's total mass is $M_E = \rho \cdot \frac{4}{3}\pi R_E^3$ .

The acceleration due to gravity at depth $d$ is then:

g_d = \frac{G M_s}{(R_E - d)^2}

Substituting $M_s$ from above and simplifying gives the key result:

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

where $g = \frac{G M_E}{R_E^2}$ is the acceleration at the surface. Each symbol: $g_d$ is $g$ at depth $d$ , $g$ is surface gravity, $d$ is depth below surface, $R_E$ is Earth's radius. …

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