Gravitational field strength g — Physics

Before this

This leads to

Satellite orbits

Where does 9.8 come from?

You have used $g = 9.8$ m/s² in almost every mechanics problem. You have used it for falling stones, for weights and for the tension in ropes. But nobody measured 9.8 by dropping every possible object. So why is it 9.8, and not 5 or 20? Is it the same on a mountain top? On the Moon? At the centre of the Earth?

The last lecture gave the law of gravitation, $F = \frac{G m _{1} m _{2}}{r ^{2}}$ . It is a law about two masses. It also promised a different point of view: the pull that one mass, the Earth, gives to each kilogram placed near it. That point of view is called the gravitational field. It answers every one of the questions above, and it shows that 9.8 is not a magic number. It is what the law of gravitation gives for the Earth's mass and size.

A field: the pull per kilogram

Imagine placing a small object near the Earth, and measuring the gravitational force on it. Then replace it with an object of twice the mass. The force doubles. Use three times the mass, and the force triples. The force always depends on what you put there.

But one thing does not change: the force on each kilogram. Divide the force by the mass, and you get the same number every time. That number belongs to the place, not to the object. It tells you how strongly gravity acts at that point in space, whatever you put there.

Gravitational field strength

A gravitational field is a region where a mass feels a gravitational force. The gravitational field strength $g$ at a point is the force per unit mass on a small object placed there.

$g = \frac{F}{m}$ , measured in N/kg.

The object must be small, so that its own pull does not disturb the field it is measuring. Physicists call it a "test mass". The field also has a direction: the direction of the force on the test mass, which is towards the centre of the Earth. So $g$ is a vector.

Field lines show the direction of the force on a test mass: towards the Earth's centre, from every side. Where the lines are closer together, near the surface, the field is stronger.

The picture also connects to the last lecture. There, the inverse square law came from lines spreading over a larger and larger sphere. These are the same lines. Near the surface they are crowded together, and the field is strong. Far away they are spread out, and the field is weak.

Why N/kg and m/s² are the same unit

You may have noticed that $g$ is sometimes written as 9.8 N/kg and sometimes as 9.8 m/s². These are not two different quantities that happen to share a number. They are the same unit, written two ways.

A newton is defined by Newton's second law: 1 N is the force that gives 1 kg an acceleration of 1 m/s². So 1 N = 1 kg m/s². Divide by a kilogram:

1 N/kg = \frac{1 kg m/s ^{2}}{1 kg} = 1 m/s^{2}

This is also a statement about physics, not only about units. If gravity is the only force on an object, the resultant force is $F = m g$ . Newton's second law then gives $a = \frac{F}{m} = g$ . So the field strength at a point equals the acceleration of anything falling freely there. The 9.8 N/kg of the field and the 9.8 m/s² of a falling stone are the same fact, seen from two sides.

From the law of gravitation to g = GM/r²

Now combine the definition with the law of gravitation. A planet of mass $M$ pulls a test mass $m$ at distance $r$ from its centre with a force

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

Divide by $m$ to get the force per kilogram:

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

Look at what just happened. The test mass $m$ has cancelled. It is gone from the answer. This is why a feather and a hammer fall together on the Moon, where there is no air. The field gives every kilogram the same pull, so it gives every object the same acceleration.

Field strength of a planet

Outside a planet of mass $M$ , at distance $r$ from its centre:

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

$g$ depends only on the planet's mass and on the distance from its centre. It does not depend on the mass of the object placed there.

The cancelling of $m$ is the most important idea in this lecture, and it is also where students most often go wrong. Some quantities in a gravity problem belong to the place. Others depend on the object you put there. Sort them below.

Sorter — a property of the place, or of the object?

Imagine replacing a 1 kg ball with a 5 kg ball at the same place, with no air. Decide whether each quantity would change. Every card explains itself when you place it.

Replacing a 1 kg ball with a 5 kg ball at the same place, with no air resistance:

Changes with the object's mass: the gravitational force on the ball; its weight; the reading on a scale it stands on; the upward force needed to hold it still; the force of the ball on the Earth.
The same for any object at that place: the field strength g; the ball's acceleration when dropped; the time it takes to fall 2 m; the direction of the pull.

The pattern is simple. The field strength and the acceleration of free fall belong to the place. Every force, including weight, is $m g$ , so it belongs to the object at that place.

The 9.8, calculated

Worked Example 1 — g at the Earth's surface

The Earth's mass is $5.97 \times 1 0^{24}$ kg and its radius is $6.37 \times 1 0^{6}$ m. Find $g$ at the surface.

At the surface, $r$ is the Earth's radius. Handle the numbers and the powers of ten separately. On top: $6.67 \times 5.97 \approx 39.8$ , and $1 0^{- 11} \times 1 0^{24} = 1 0^{13}$ , so the top is about $3.98 \times 1 0^{14}$ . Underneath: $6.3 7^{2} \approx 40.6$ , and $(1 0^{6})^{2} = 1 0^{12}$ , so the bottom is about $4.06 \times 1 0^{13}$ .

g = \frac{GM}{R ^{2}} \approx \frac{3.98 \times 1 0 ^{14}}{4.06 \times 1 0 ^{13}} \approx 9.8 N/kg

There it is. The number you have used for years comes from the Earth's mass and radius, put into the law of gravitation. A heavier Earth of the same size would give a larger $g$ . A larger Earth with the same mass would give a smaller one.

Stop and think

Your mass is 60 kg. You travel to the Moon. Does your mass change? Does your weight change?

Show the answer

Your mass does not change. It is the amount of matter in you, and it is 60 kg everywhere. Your weight does change, because weight is a force, $W = m g$ , and $g$ depends on the place. On the Earth, $W = 60 \times 9.8 = 588$ N. On the Moon, where $g \approx 1.6$ N/kg, $W \approx 60 \times 1.6 = 96$ N. A bathroom scale measures the force you push on it. So on the Moon it would show about one sixth of the usual reading, even though you are made of exactly the same matter.

How g changes with distance

Above the surface, $g = \frac{GM}{r ^{2}}$ is an inverse square law, exactly like the force in the last lecture. At twice the distance from the centre, $g$ is a quarter as large. At three times, a ninth.

Near the surface, the change is small, because a mountain is tiny compared with the Earth. The summit of Mount Everest is about 8.85 km above sea level. That increases $r$ from 6370 km to about 6379 km, and

(\frac{6370}{6379})^{2} \approx 0.997

So $g$ on the summit is about 0.3% smaller, roughly 9.79 instead of 9.81 N/kg. That is small, but sensitive instruments called gravimeters can easily measure it.

Even at sea level, $g$ is not exactly the same everywhere. It is about 9.780 N/kg at the equator and about 9.832 N/kg at the poles. Two effects cause this. The Earth spins, and it bulges at the equator, so a point on the equator is a little farther from the centre. Measured values at the surface range from about 9.764 N/kg to about 9.834 N/kg. The lowest is on Nevado Huascarán, a mountain in Peru. The highest is at the surface of the Arctic Ocean. The value 9.80665 m/s², called "standard gravity", is an agreed reference number, not a measurement at one particular place. In this course, Upper Secondary work uses 9.8.

Inside the Earth

What happens if you go down, into a deep mine? The formula $g = \frac{GM}{r ^{2}}$ seems to say that $g$ gets larger as $r$ gets smaller. But that formula is for points outside the planet. Inside, something else happens.

Newton proved a result called the shell theorem. A uniform spherical shell of mass pulls on an object outside it as if all its mass were at the centre. On an object inside it, the pulls from all parts of the shell cancel exactly. So when you are inside the Earth, only the mass closer to the centre than you pulls on you. The layers above your head pull in all directions and cancel.

Now imagine a model Earth with the same density everywhere. Halfway to the centre, the sphere beneath you has half the radius, so it has $(\frac{1}{2})^{3} = \frac{1}{8}$ of the mass. But you are also twice as close to its centre, so its pull per kilogram is multiplied by $2^{2} = 4$ . Together: $\frac{1}{8} \times 4 = \frac{1}{2}$ . In this uniform model, $g$ inside falls in proportion to $r$ , down to zero at the centre.

Use the graph builder to read $g$ at any distance from the centre, inside and outside.

Graph builder — g from the centre of the Earth to far out in space

Move the point along the graph. Read g at each distance from the Earth's centre. Then use the presets to compare the centre, a point halfway down, the surface, the top of Everest and the Space Station.

Distance from the Earth's centre: 6370 km

Values of g from g = GM/r² outside the Earth (M = 5.97 × 10²⁴ kg, R = 6370 km), and from a uniform-Earth model inside, where g is proportional to r.

In the uniform-Earth model, g rises in a straight line from zero at the centre to about 9.81 N/kg at the surface. Outside, it falls as 1/r².

Place	Distance from centre (km)	g (N/kg)
centre (model)	0	0
halfway down (model)	3185	4.91
surface	6370	9.81
top of Everest	6379	9.79
Space Station	6770	8.69
2 Earth radii	12 740	2.45

What did you see? In the model, $g$ is largest at the surface. Going up, it falls slowly at first, by only 0.3% on Everest, and then as $1/ r^{2}$ . Going down, it falls in a straight line to zero at the centre. At the centre, the Earth pulls you equally in every direction, so the forces cancel.

The real Earth is not uniform. Its core is much denser than its outer rocks. So much mass is concentrated near the centre that $g$ in the real Earth actually rises a little as you go down through the rocks. It reaches a maximum of about 10.7 N/kg just outside the outer core. Only inside the core does it fall towards zero. So the straight line inside is a model, and a useful one, but it is not a measurement.

That is it. That is the whole idea of a field: one number at each place, set by how much mass is pulling and how far away it is.

g on other worlds

The same formula works for every planet and moon. Here are surface values from NASA's planetary fact sheet.

Body	Mass (× 10²⁴ kg)	Diameter (km)	g at the surface (N/kg)
Mercury	0.330	4879	3.7
Venus	4.87	12 104	8.9
Earth	5.97	12 756	9.8
Moon	0.073	3475	1.6
Mars	0.642	6792	3.7
Jupiter	1898	142 984	23.1

Look at Mercury and Mars. Mars has almost twice Mercury's mass, yet the two have the same $g$ . The reason is the radius. Mars is also wider, and the $r^{2}$ underneath cancels the extra mass. Jupiter has more than 300 times the Earth's mass, but its $g$ is only about 2.4 times as large, because Jupiter is also about 11 times as wide. Jupiter has no solid surface, so its value is given at a chosen level in its atmosphere.

Worked Example 2 — g at the height of the Space Station

The Space Station orbits about 400 km above the surface. Find $g$ there.

First turn the height into a distance from the centre: $r = 6.37 \times 1 0^{6} + 0.40 \times 1 0^{6} = 6.77 \times 1 0^{6}$ m.

The top of the fraction is the same as before, $GM \approx 3.98 \times 1 0^{14}$ . Underneath: $6.7 7^{2} \approx 45.8$ , so $r^{2} \approx 4.58 \times 1 0^{13}$ . So $g \approx 3.98 \times 1 0^{14} \div 4.58 \times 1 0^{13} \approx 8.7$ N/kg.

Check with a ratio: $(\frac{6.37}{6.77})^{2} \approx 0.885$ , and $0.885 \times 9.81 \approx 8.7$ ✓. This matches the last lecture's astronaut, who felt 89% of their usual pull. The astronauts float because they and the station are falling freely together, not because the field is missing.

Worked Example 3 — g on Mars, from its mass and size

Use the table to find $g$ at the surface of Mars, and the weight there of a 75 kg astronaut.

The radius is half the diameter: $6792 \div 2 = 3396$ km $= 3.396 \times 1 0^{6}$ m. The mass is $0.642 \times 1 0^{24} = 6.42 \times 1 0^{23}$ kg.

g = \frac{6.67 \times 1 0 ^{- 11} \times 6.42 \times 1 0 ^{23}}{( 3.396 \times 1 0 ^{6} ) ^{2}} \approx 3.7 N/kg

This matches NASA's value. The astronaut's weight is $W = m g \approx 75 \times 3.71 \approx 278$ N, compared with $75 \times 9.8 = 735$ N on the Earth. The mass is still 75 kg.

Worked Example 4 — Find a planet's mass from its g

A space probe lands on a planet of radius $3.2 \times 1 0^{6}$ m. It measures $g = 5.2$ N/kg at the surface. Find the planet's mass.

Rearrange $g = \frac{GM}{R ^{2}}$ : multiply both sides by $R^{2}$ and divide by $G$ , so $M = \frac{g R ^{2}}{G}$ .

On top: $5.2 \times (3.2 \times 1 0^{6})^{2} = 5.2 \times 10.24 \times 1 0^{12} \approx 53.2 \times 1 0^{12}$ . Divide by $6.67 \times 1 0^{- 11}$ : $\frac{53.2}{6.67} \approx 7.98$ , and $\frac{1 0 ^{12}}{1 0 ^{- 11}} = 1 0^{23}$ . So $M \approx 8.0 \times 1 0^{23}$ kg.

Check that the answer is sensible: it is a little more than Mars, $6.4 \times 1 0^{23}$ kg, for a planet a little smaller than Mars, so its $g$ should be a little higher than 3.7 N/kg. It is: 5.2 N/kg. This is how astronomers "weigh" planets and moons. Measure $g$ , or the motion of something orbiting it, and the law of gravitation gives the mass.

Three classic mistakes

Trap 1 — Saying weight in kilograms

"She weighs 60 kg" is everyday language, but in physics it mixes two quantities. Mass is in kilograms and does not change from place to place. Weight is a force, $W = m g$ , in newtons, and it changes with $g$ . Write "her mass is 60 kg" and "her weight is 588 N".

Trap 2 — Thinking heavier objects have a larger g

A heavier object feels a larger force, but it also has more mass to accelerate. The two effects cancel exactly: $g = \frac{F}{m}$ is the same for every object at that place. Without air, a heavy ball and a light ball fall together.

Trap 3 — Using g = GM/r² inside a planet

The formula $g = \frac{GM}{r ^{2}}$ , with the planet's whole mass, holds only outside it. Inside, the layers farther out than you pull equally in all directions and cancel, so only the mass nearer the centre counts. Using the full mass at a small $r$ would predict an enormous $g$ near the centre, when the true value is zero.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Use $G = 6.67 \times 1 0^{- 11}$ N m²/kg².

The Moon's mass is $7.35 \times 1 0^{22}$ kg and its radius is $1.74 \times 1 0^{6}$ m. Find $g$ at its surface, and the weight there of a 60 kg person.

Hint 1
Use $g = \frac{GM}{R ^{2}}$ with the Moon's values.

Hint 2
$GM \approx 4.90 \times 1 0^{12}$ , and $R^{2} \approx 3.03 \times 1 0^{12}$ .

Full answer
$g \approx \frac{4.90 \times 1 0 ^{12}}{3.03 \times 1 0 ^{12}} \approx 1.6$ N/kg, which matches NASA's 1.6. The weight is $60 \times 1.62 \approx 97$ N, about a sixth of the 588 N on the Earth. The mass is still 60 kg.
A planet has twice the Earth's mass and twice its radius. What is $g$ at its surface?

Hint 1
You do not need any numbers except 9.8. Write one factor for each change.

Hint 2
Doubling $M$ gives × 2. Doubling $R$ gives × $\frac{1}{4}$ .

Full answer
$g$ is multiplied by $2 \times \frac{1}{4} = \frac{1}{2}$ , so $g \approx 9.8 \div 2 = 4.9$ N/kg. A bigger planet does not always have stronger gravity at its surface. Size works against mass.
At what height above the Earth's surface is $g$ half its surface value? Give your answer in km, using $R = 6370$ km.

Hint 1
$g$ is proportional to $\frac{1}{r ^{2}}$ . For $g$ to halve, $r^{2}$ must double.

Hint 2
So $r = 2 \times R$ . Then subtract $R$ to get the height.

Full answer
$r = 2 \times 6370 \approx 9009$ km, so the height is $9009 - 6370 \approx 2640$ km. Check: $(\frac{6370}{9009})^{2} \approx 0.50$ ✓. Notice the trap: halving $g$ does not need double the distance. Doubling the distance would make $g$ a quarter.
For a uniform planet of density $ρ$ and radius $R$ , the mass is $M = \frac{4}{3} π R^{3} ρ$ . (a) Show that $g = \frac{4}{3} π GρR$ at the surface. (b) The Earth's average density is about 5510 kg/m³. Check that this gives 9.8 N/kg. (c) Mars's average density is about 3910 kg/m³ and its radius is 3396 km. Explain, using (a), why Mars's $g$ is so much smaller than the Earth's.

Hint 1
Substitute $M = \frac{4}{3} π R^{3} ρ$ into $g = \frac{GM}{R ^{2}}$ , and cancel $R^{2}$ .

Hint 2
For (c), compare the two factors that change: $ρ$ and $R$ .

Full answer
(a) $g = \frac{G \times \frac{4}{3} π R ^{3} ρ}{R ^{2}} = \frac{4}{3} π GρR$ . (b) $\frac{4}{3} \times 3.14 \times 6.67 \times 1 0^{- 11} \times 5510 \times 6.37 \times 1 0^{6} \approx 9.8$ N/kg ✓. (c) Mars's density is $3910 \div 5510 \approx 0.71$ of the Earth's, and its radius is $3396 \div 6370 \approx 0.53$ of the Earth's. So its $g$ is about $0.71 \times 0.53 \approx 0.38$ of the Earth's: $0.38 \times 9.81 \approx 3.7$ N/kg, as NASA gives. For planets made of similar material, $g$ grows in proportion to the radius, so a smaller planet has weaker surface gravity.

The bigger idea

The field idea changes how we think about forces. The law of gravitation describes two objects pulling each other across empty space. The field description says something different: the Earth gives every point around it a value of $g$ , whether or not anything is there. An object placed at a point simply responds to the field at that point.

This way of thinking runs through the rest of physics. Electric charges create electric fields, and magnets and currents create magnetic fields. In each case, one object sets up a field, and another responds to the field where it is. The next lecture uses the gravitational field to explain how satellites stay in orbit: the field supplies exactly the centripetal acceleration that circular motion needs.

Key concepts in this lesson

Gravitational field strength $g$ is the gravitational force per unit mass on a small test mass, $g = \frac{F}{m}$ , in N/kg.
One N/kg is the same unit as one m/s², and the field strength at a point equals the acceleration of free fall there.
Outside a planet, $g = \frac{GM}{r ^{2}}$ , and the mass of the object placed in the field cancels out.
The Earth's mass and radius give $g \approx 9.8$ N/kg at the surface, and $g$ falls as $1/ r^{2}$ with distance from the centre.
Mass stays the same everywhere, while weight $W = m g$ changes with the field strength of the place.
Inside a planet only the mass nearer the centre pulls, so in a uniform model $g$ falls to zero at the centre.