Gravitational Fields

SPEC 3.7.1, 3.7.2.1, 3.7.2.2 • FIELDS AND THEIR CONSEQUENCES

Understanding force fields, Newton's law of gravitation, and gravitational field strength

What is a Force Field?

Definition

A force field is a region in which a body experiences a non-contact force.

Force fields are one of the great unifying concepts in physics. They allow us to describe how objects can exert forces on each other without physical contact.

Key Concept

A force field can be represented as a vector - it has both magnitude and direction. The direction must be determined by inspection of the field.

Force fields arise from three fundamental interactions:

Mass - creates gravitational fields

Static charge - creates electric fields

Moving charges - creates magnetic fields

Types of Force Fields

Created by mass

Always attractive

Relatively weak force

Created by charge

Opposite charges attract

Like charges repel

Key Difference

Masses always attract, but charges can attract (opposite) or repel (like). This is the most commonly tested distinction.

Comparing Gravitational and Electric Fields

FeatureGravitationalElectric

Force lawInverse-squareInverse-square

Field linesYes - radial patternYes - radial pattern

Potential conceptYesYes

Equipotential surfacesYesYes

Nature of forceAlways attractiveAttractive or repulsive

Remember

Both fields follow inverse-square laws: the force decreases with the square of the distance from the source.

Newton's Law of Gravitation

Newton's Law

F = \frac{G m _{1} m _{2}}{r ^{2}}

SymbolQuantityUnit

G

Gravitational constant = 6.67 × 10⁻¹¹ — a universal constant that determines the strength of gravitational attraction between massesN m² kg⁻²

r

Distance between centres of massesm

Inverse-Square Law

Forces are equal and opposite (Newton's 3rd Law). If distance doubles, force reduces to ¼. If distance triples, force becomes ¹⁄₉.

01Worked Example

Calculate the gravitational force between the Earth and the Moon.

Mass of Earth = 5.97 × 10²⁴ kg

Mass of Moon = 7.35 × 10²² kg

Distance between centres = 3.84 × 10⁸ m

1

Write the equation:

F = \frac{G m _{1} m _{2}}{r ^{2}}

2

Substitute values:

F = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24} \times 7.35 \times 1 0 ^{22}}{( 3.84 \times 1 0 ^{8} ) ^{2}}

3

Calculate numerator:

6.67 \times 5.97 \times 7.35 \times 1 0^{- 11 + 24 + 22} = 2.93 \times 1 0^{37}

4

Calculate denominator:

(3.84 \times 1 0^{8})^{2} = 1.47 \times 1 0^{17}

5

Divide:

F = \frac{2.93 \times 1 0 ^{37}}{1.47 \times 1 0 ^{17}} = 1.99 \times 1 0^{20}

N

Answer: F = 2.0 × 10²⁰ N

Worked Example02

Two lead spheres, each of mass 50 kg, are placed with their centres 0.40 m apart. Calculate the gravitational force between them.

1

Identify values:

m_{1} = m_{2} = 50

kg,

r = 0.40

m

2

Apply formula:

F = \frac{G m _{1} m _{2}}{r ^{2}} = \frac{6.67 \times 1 0 ^{- 11} \times 50 \times 50}{0.4 0 ^{2}}

3

Calculate:

F = \frac{6.67 \times 1 0 ^{- 11} \times 2500}{0.16}

4

F = \frac{1.67 \times 1 0 ^{- 7}}{0.16} = 1.04 \times 1 0^{- 6}

N

Answer: F = 1.0 × 10⁻⁶ N (about 1 μN)

This tiny force shows why we don't notice gravitational attraction between everyday objects - it's incredibly weak compared to Earth's pull on them.

03Worked Example

At what distance from Earth's centre would the gravitational force on a 1000 kg satellite be 5000 N?

Mass of Earth = 5.97 × 10²⁴ kg

1

Rearrange

F = \frac{G m _{1} m _{2}}{r ^{2}}

to find

r

:

r^{2} = \frac{G m _{1} m _{2}}{F}

2

Substitute:

r^{2} = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24} \times 1000}{5000}

3

Calculate:

r^{2} = \frac{3.98 \times 1 0 ^{17}}{5000} = 7.96 \times 1 0^{13}

4

Take square root:

r = 7.96 \times 1 0^{13} = 8.92 \times 1 0^{6}

m

Answer: r = 8.9 × 10⁶ m (about 8900 km from Earth's centre)

Gravitational Field Strength

Definition

Gravitational field strength (g) is the force per unit mass experienced by a small test mass placed in the field.

g = \frac{F}{m}

g tells us the force experienced per kilogram of test mass at that point in the field

Important

The units N kg⁻¹ and m s⁻² are equivalent. Near Earth's surface,

g \approx 9.81

N kg⁻¹ (or 9.81 m s⁻²).

Field Strength in a Radial Field

For a point mass (or uniform sphere), we can derive an expression for

g

at distance

r

from the centre:

Derivation

Starting with:

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

and

g = \frac{F}{m}

Substituting:

g = \frac{1}{m} \times \frac{GM m}{r ^{2}} = \frac{GM}{r ^{2}}

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

Critical Point

Notice that

g

does not depend on the test mass

m

. The field strength at any point is a property of the field alone, determined only by the source mass

M

and distance

r

.

Remember

This is why all objects fall with the same acceleration in a gravitational field (ignoring air resistance) -

g

is independent of mass.

Gravitational Field Lines

Radial Field

(around a planet)

Uniform Field

(near Earth's surface)

Field Line Rules

Field lines point in the direction of the force on a mass

For gravity, lines always point toward the mass

Closer lines = stronger field

Lines never cross

04Worked Example

Calculate the gravitational field strength at Earth's surface.

Mass of Earth = 5.97 × 10²⁴ kg

Radius of Earth = 6.37 × 10⁶ m

1

Use:

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

2

At the surface,

r = R_{E a r t h} = 6.37 \times 1 0^{6}

m

3

Substitute:

g = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24}}{( 6.37 \times 1 0 ^{6} ) ^{2}}

4

Calculate:

g = \frac{3.98 \times 1 0 ^{14}}{4.06 \times 1 0 ^{13}} = 9.81

N kg⁻¹

Answer: g = 9.81 N kg⁻¹ (or 9.81 m s⁻²)

05Worked Example

Calculate the gravitational field strength at an altitude of 400 km above Earth's surface (the altitude of the ISS).

Mass of Earth = 5.97 × 10²⁴ kg

Radius of Earth = 6.37 × 10⁶ m

1

Find total distance from Earth's centre:

r = 6.37 \times 1 0^{6} + 400 \times 1 0^{3}

2

r = 6.37 \times 1 0^{6} + 0.40 \times 1 0^{6} = 6.77 \times 1 0^{6}

m

3

Apply:

g = \frac{GM}{r ^{2}} = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24}}{( 6.77 \times 1 0 ^{6} ) ^{2}}

4

Calculate:

g = \frac{3.98 \times 1 0 ^{14}}{4.58 \times 1 0 ^{13}} = 8.69

N kg⁻¹

Answer: g = 8.7 N kg⁻¹

At ISS altitude, g is still about 89% of the surface value. Astronauts float because they're in continuous free fall, not because gravity is weak!

Worked Example06

The gravitational field strength on Mars is 3.7 N kg⁻¹. Mars has a radius of 3.39 × 10⁶ m. Calculate the mass of Mars.

1

Rearrange

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

to find M:

M = \frac{g r ^{2}}{G}

2

Substitute:

M = \frac{3.7 \times ( 3.39 \times 1 0 ^{6} ) ^{2}}{6.67 \times 1 0 ^{- 11}}

3

Calculate numerator:

3.7 \times 1.15 \times 1 0^{13} = 4.26 \times 1 0^{13}

4

Divide:

M = \frac{4.26 \times 1 0 ^{13}}{6.67 \times 1 0 ^{- 11}} = 6.4 \times 1 0^{23}

kg

Answer: M = 6.4 × 10²³ kg

07Worked Example

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

Radius of Earth = 6.37 × 10⁶ m

1

At surface:

g_{0} = \frac{GM}{R ^{2}}

. At altitude h:

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

2

We need

g = \frac{1}{2} g_{0}

, so:

\frac{GM}{( R + h ) ^{2}} = \frac{1}{2} \times \frac{GM}{R ^{2}}

3

Simplify:

\frac{R ^{2}}{( R + h ) ^{2}} = \frac{1}{2}

, so

(R + h)^{2} = 2 R^{2}

4

Take square root:

R + h = R 2

, so

h = R (2 - 1)

5

Calculate:

h = 6.37 \times 1 0^{6} \times (1.414 - 1) = 6.37 \times 1 0^{6} \times 0.414

6

h = 2.64 \times 1 0^{6}

m = 2640 km

Answer: h = 2640 km (about 0.41 Earth radii)

Practice

Question 1: A satellite of mass 800 kg orbits Earth at an altitude of 200 km. Calculate the gravitational force on the satellite.

Mass of Earth = 5.97 × 10²⁴ kg

Radius of Earth = 6.37 × 10⁶ m

1

Distance from Earth's centre:

r = 6.37 \times 1 0^{6} + 200 \times 1 0^{3} = 6.57 \times 1 0^{6}

m

2

Apply:

F = \frac{G m _{1} m _{2}}{r ^{2}} = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24} \times 800}{( 6.57 \times 1 0 ^{6} ) ^{2}}

3

Calculate:

F = \frac{3.18 \times 1 0 ^{17}}{4.32 \times 1 0 ^{13}} = 7370

N

Answer: F = 7.4 kN

Practice

Question 2: Calculate the gravitational field strength at the surface of Mars.

Mass of Mars = 6.42 × 10²³ kg

Radius of Mars = 3.39 × 10⁶ m

1

Apply:

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

2

Substitute:

g = \frac{6.67 \times 1 0 ^{- 11} \times 6.42 \times 1 0 ^{23}}{( 3.39 \times 1 0 ^{6} ) ^{2}}

3

Calculate:

g = \frac{4.28 \times 1 0 ^{13}}{1.15 \times 1 0 ^{13}} = 3.72

N kg⁻¹

Answer: g = 3.7 N kg⁻¹ (about 38% of Earth's surface gravity)

Practice

Question 3: At what height above Earth's surface does the gravitational field strength drop to 4.9 N kg⁻¹?

Mass of Earth = 5.97 × 10²⁴ kg

Radius of Earth = 6.37 × 10⁶ m

1

Rearrange

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

to find r:

r = \frac{GM}{g}

2

Substitute:

r = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24}}{4.9}

3

Calculate:

r = 8.12 \times 1 0^{13} = 9.01 \times 1 0^{6}

m

4

Find height:

h = r - R = 9.01 \times 1 0^{6} - 6.37 \times 1 0^{6} = 2.64 \times 1 0^{6}

m

Answer: h = 2640 km

Note: 4.9 N kg⁻¹ is half of 9.8 N kg⁻¹, confirming our earlier result that g halves at about 2640 km altitude.

Summary

Key Formulas

FormulaDescription

F = \frac{G m _{1} m _{2}}{r ^{2}}

Newton's law of gravitation

g = \frac{F}{m}

Definition of field strength

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

Field strength in radial field

Key Exam Points

A force field is a region where a body experiences a non-contact force

Gravitational forces are always attractive; electric forces can attract or repel

Both follow inverse-square laws

Distance r is measured between centres of mass

Field strength g is independent of test mass

G = 6.67 × 10⁻¹¹ N m² kg⁻² (given in data booklet)