Gravitational Field Strength of a Point Mass

A2 · 9 min

Combining the definition of gravitational field strength with Newton's law of gravitation gives a formula for the field around any planet or star: $g = GM/r^2$ . It tells you how strong gravity is at the top of a mountain, at the height of the space station, or on the surface of Mars, and it explains why we can treat $g$ as constant in everyday problems even though it is not really constant at all. Paper 4 asks for the derivation, for calculations, and for explanations of why $g$ is approximately constant near the surface.

Deriving $g = GM/r^2$

The syllabus requires this derivation. It takes two lines, and each line is a mark.

Field strength due to a point mass

Place a small test mass $m$ at distance $r$ from a point mass $M$ . By Newton's law of gravitation the force on it is

F = \frac{GMm}{r^{2}}

Gravitational field strength is defined as force per unit mass, $g = F/m$ , so

g = \frac{F}{m} = \frac{GMm}{r^{2}m} = \frac{GM}{r^{2}}

Key result

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

$M$ : the mass creating the field (kg)
$r$ : distance from the centre of the mass (m), with $r \ge R$ for a sphere of radius $R$
$g$ : field strength in $\text{N kg}^{-1}$ , directed towards $M$

The field strength depends only on the mass creating the field and the distance from it. It does not depend on the test mass at all: a feather and a hammer at the same point experience the same field strength (and, without air resistance, the same acceleration).

How $g$ varies with distance

Outside a planet, $g \propto 1/r^2$ : an inverse-square law.

At the surface ( $r = R$ ), $g = g_0 = GM/R^2$ .
At $r = 2R$ , $g = g_0/4$ .
At $r = 3R$ , $g = g_0/9$ .

The graph below shows $g$ (in units of the surface value $g_0$ ) against $r$ (in units of the planet's radius $R$ ).

y = min(x, 1/x^2) (1, 0) -- (1, 1)

For $r > R$ the curve is $g = g_0(R/r)^2$ : it falls steeply at first, then more and more gently, approaching zero but never reaching it. Gravity has infinite range.

Tip

The straight line from the origin to $r = R$ shows the field inside a planet of uniform density, where $g \propto r$ (only the mass closer to the centre than you contributes). This is beyond the syllabus, but it explains why $g$ is greatest at the surface and why the graph you may be shown has a peak at $r = R$ .

A useful shortcut follows from $g_0 = GM/R^2$ , which gives $GM = g_0R^2$ . Then at any distance $r$ :

g = g_0\left(\frac{R}{r}\right)^{2}

This lets you solve problems with only $g_0$ and $R$ , without $G$ or $M$ .

Why $g$ is approximately constant near the Earth's surface

If $g$ depends on $r$ , why do we use $g = 9.81\ \text{N kg}^{-1}$ for everything from dropping a ball to firing a projectile?

At a height $h$ above the surface, $r = R + h$ :

\frac{g}{g_0} = \left(\frac{R}{R + h}\right)^{2}

The Earth's radius is $6.37 \times 10^{6}\ \text{m}$ . For everyday heights $h \ll R$ , so $R + h \approx R$ and the ratio is very close to $1$ :

Height above surface	$g$ / $\text{N kg}^{-1}$	Percentage of surface value
$0$	$9.81$	$100\%$
$1\ \text{km}$	$9.81$	$99.97\%$
$8.85\ \text{km}$ (top of Everest)	$9.79$	$99.7\%$
$400\ \text{km}$ (space station)	$8.69$	$88.5\%$
$3.58 \times 10^{4}\ \text{km}$ (geostationary)	$0.224$	$2.3\%$

Key result

Near the Earth's surface, changes in height are very small compared with the radius of the Earth ( $\Delta h \ll R$ ), so the distance $r$ from the centre, and hence $g = GM/r^2$ , hardly changes. The field lines are almost parallel over a small region: the field is approximately uniform.

Field strength is a vector

Where two or more masses contribute, the resultant field strength is the vector sum of the individual fields. Along the line joining two masses, the fields point in opposite directions, so subtract them. Somewhere between two masses there is always a point where the resultant field is zero (closer to the smaller mass).

Field strength and density

For a uniform sphere of density $\rho$ , $M = \tfrac{4}{3}\pi R^3\rho$ , so the surface field strength is

g_0 = \frac{GM}{R^{2}} = \frac{4}{3}\pi G\rho R

Planets of the same density have surface field strengths proportional to their radii. This rearranges to find the mean density of a planet from measurements of $g_0$ and $R$ .

Field strength problems

Identify which mass creates the field; $M$ is that mass.
Find $r$ from the centre of that mass.
Use $g = GM/r^2$ directly, or $g = g_0(R/r)^2$ if only surface data are given.
For two masses, calculate each field separately, decide its direction, then add as vectors.
For comparisons, use ratios: $g \propto M/r^2$ .

Worked examples

Field strength on Mars

Mars has mass $6.42 \times 10^{23}\ \text{kg}$ and radius $3.39 \times 10^{6}\ \text{m}$ . Calculate the gravitational field strength at its surface and the weight there of a $75\ \text{kg}$ astronaut.

Solution

g = \frac{GM}{R^{2}} = \frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{(3.39 \times 10^{6})^{2}} = 3.73\ \text{N kg}^{-1}

Weight $= mg = 75 \times 3.73 = 280\ \text{N}$ .

Field strength at the space station

Calculate the gravitational field strength at the International Space Station, $400\ \text{km}$ above the Earth's surface. Use $g_0 = 9.81\ \text{N kg}^{-1}$ and $R = 6.37 \times 10^{6}\ \text{m}$ .

Solution

g = g_0\left(\frac{R}{R + h}\right)^{2} = 9.81 \times \left(\frac{6.37 \times 10^{6}}{6.77 \times 10^{6}}\right)^{2} = 8.69\ \text{N kg}^{-1}

Gravity at the space station is still nearly $90\%$ of its surface value. The astronauts float because they are in free fall, not because gravity is absent.

Height for a given field strength

At what height above the Earth's surface is the gravitational field strength half its surface value?

Solution

\frac{g}{g_0} = \left(\frac{R}{r}\right)^{2} = \frac{1}{2} \quad\Rightarrow\quad r = \sqrt{2}\,R = 1.414R

Height $h = r - R = 0.414R = 0.414 \times 6.37 \times 10^{6} = 2.64 \times 10^{6}\ \text{m}$ .

Comparing planets

Planet X has three times the mass of the Earth and twice its radius. Calculate the gravitational field strength at its surface.

Solution

$g \propto M/R^2$ :

\frac{g_X}{g_E} = \frac{3}{2^{2}} = 0.75 \quad\Rightarrow\quad g_X = 0.75 \times 9.81 = 7.36\ \text{N kg}^{-1}

Resultant field between the Earth and the Moon

A spacecraft is $1.0 \times 10^{8}\ \text{m}$ from the centre of the Earth on the line to the Moon. The Earth–Moon distance is $3.84 \times 10^{8}\ \text{m}$ ; $M_E = 5.97 \times 10^{24}\ \text{kg}$ , $M_M = 7.35 \times 10^{22}\ \text{kg}$ . Calculate the resultant gravitational field strength at the spacecraft.

Solution

Field due to the Earth (towards the Earth):

g_E = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(1.0 \times 10^{8})^{2}} = 3.98 \times 10^{-2}\ \text{N kg}^{-1}

Field due to the Moon (towards the Moon), at distance $2.84 \times 10^{8}\ \text{m}$ :

g_M = \frac{6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{(2.84 \times 10^{8})^{2}} = 6.08 \times 10^{-5}\ \text{N kg}^{-1}

They are in opposite directions, so the resultant is $3.98 \times 10^{-2} - 0.006 \times 10^{-2} = 3.98 \times 10^{-2}\ \text{N kg}^{-1}$ towards the Earth. Here the Moon's field is negligible; it only becomes comparable very close to the Moon.

Density of Mars

Use the surface field strength of Mars, $3.73\ \text{N kg}^{-1}$ , and its radius, $3.39 \times 10^{6}\ \text{m}$ , to calculate its mean density. Compare with the Earth's mean density of $5.5 \times 10^{3}\ \text{kg m}^{-3}$ .

Solution

\rho = \frac{3g_0}{4\pi GR} = \frac{3 \times 3.73}{4\pi \times 6.67 \times 10^{-11} \times 3.39 \times 10^{6}} = 3.9 \times 10^{3}\ \text{kg m}^{-3}

Mars is less dense than the Earth, suggesting a smaller iron core in proportion to its size.

Watch out

Confusing $g$ and $G$ . $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ is a universal constant. $g$ is the field strength at a particular place, and changes from place to place.

Watch out

Using the wrong mass. In $g = GM/r^2$ , $M$ is the mass that creates the field, not the mass placed in it. The field of the Earth at the Moon's position uses the Earth's mass.

Watch out

Saying " $g$ is constant near the Earth because gravity is uniform". That is circular. The reason is that the change in height is negligible compared with the Earth's radius, so $r$ (and therefore $GM/r^2$ ) is almost unchanged.

Exam tip

"Derive $g = GM/r^2$ " (2 marks): write $F = GMm/r^2$ and $g = F/m$ , then combine. State that $m$ is a small test mass.
"Explain why $g$ is approximately constant near the Earth's surface" (1–2 marks): $h$ is much smaller than $R$ , so $r$ is approximately equal to $R$ (and the field lines are approximately parallel).
Sketch graphs of $g$ against $r$ should start at $r = R$ (not at the origin), show a curve decreasing ever more slowly, and not touch the $r$ axis.
When a graph of $g$ against $1/r^2$ is given, it is a straight line through the origin with gradient $GM$ .

Summary

Derivation: $F = GMm/r^2$ and $g = F/m$ give $g = GM/r^2$ .
$g$ depends on the mass creating the field and the distance from its centre, not on the test mass.
Outside a sphere $g \propto 1/r^2$ ; at the surface $g_0 = GM/R^2$ , so $GM = g_0R^2$ and $g = g_0(R/r)^2$ .
Near the surface $g$ is approximately constant because height changes are tiny compared with $R$ .
Field strength is a vector; fields from several masses add as vectors.
$g_0 = \tfrac{4}{3}\pi G\rho R$ for a uniform planet.

Practice questions

Question

Derive the expression $g = GM/r^2$ , stating any assumption about the test mass.
Calculate the gravitational field strength at the surface of the Moon ( $M = 7.35 \times 10^{22}\ \text{kg}$ , $R = 1.74 \times 10^{6}\ \text{m}$ ) and at the surface of Jupiter ( $M = 1.90 \times 10^{27}\ \text{kg}$ , $R = 7.15 \times 10^{7}\ \text{m}$ ).
Calculate the height above the Earth's surface at which $g = 6.0\ \text{N kg}^{-1}$ .
Calculate the gravitational field strength of the Earth at the radius of a geostationary orbit, $4.22 \times 10^{7}\ \text{m}$ , and show that it equals the centripetal acceleration of a geostationary satellite.
Explain why a mass falling $20\ \text{m}$ near the Earth's surface can be treated as having constant acceleration.
Planet P has the same density as the Earth but half its radius. Find the field strength at its surface.
Sketch a graph showing how $g$ varies with distance $r$ from the centre of a planet of radius $R$ , for $r$ from $R$ to $4R$ . Mark the values at $R$ , $2R$ and $4R$ in terms of $g_0$ .
A star of mass $M$ and a planet of mass $M/400$ are a distance $d$ apart. (a) Find, in terms of $d$ , the distance from the star of the point between them where the gravitational field strength is zero. (b) Explain why there is no such point beyond the planet on the line joining their centres.

Answers

$F = GMm/r^2$ on a small test mass $m$ (small so it does not disturb the field); $g = F/m = GM/r^2$ .
Moon: $g = 6.67 \times 10^{-11} \times 7.35 \times 10^{22}/(1.74 \times 10^{6})^2 = 1.62\ \text{N kg}^{-1}$ . Jupiter: $6.67 \times 10^{-11} \times 1.90 \times 10^{27}/(7.15 \times 10^{7})^2 = 24.8\ \text{N kg}^{-1}$ .
$r = R\sqrt{g_0/g} = 6.37 \times 10^{6} \times \sqrt{9.81/6.0} = 8.15 \times 10^{6}\ \text{m}$ , so $h = 1.78 \times 10^{6}\ \text{m}$ .
$g = 6.67 \times 10^{-11} \times 5.97 \times 10^{24}/(4.22 \times 10^{7})^2 = 0.224\ \text{N kg}^{-1}$ . Centripetal acceleration $r\omega^2 = 4.22 \times 10^{7} \times (2\pi/86\,400)^2 = 0.223\ \text{m s}^{-2}$ , equal within rounding: gravity provides the centripetal acceleration.
$20\ \text{m}$ is negligible compared with $R = 6.37 \times 10^{6}\ \text{m}$ , so $r$ and hence $g = GM/r^2$ change by a negligible amount (about $0.0006\%$ ).
$g_0 \propto \rho R$ , so $g_P = 9.81/2 = 4.9\ \text{N kg}^{-1}$ .
A curve starting at $g_0$ at $r = R$ , falling to $g_0/4$ at $2R$ and $g_0/16$ at $4R$ , decreasing ever less steeply and never reaching zero.
(a) $\dfrac{GM}{x^2} = \dfrac{G(M/400)}{(d - x)^2} \Rightarrow \dfrac{x}{d - x} = 20 \Rightarrow x = \dfrac{20d}{21} = 0.952d$ from the star. (b) Beyond the planet both fields point in the same direction (back towards the star and the planet), so they add and cannot cancel.

Deriving g=GM/r2g = GM/r^2g=GM/r2

How ggg varies with distance

Why ggg is approximately constant near the Earth's surface

Field strength is a vector

Field strength and density

Worked examples

Practice questions

Console

Deriving $g = GM/r^2$

How $g$ varies with distance

Why $g$ is approximately constant near the Earth's surface