PhysicsCore26 min read

Gravitational Fields

One law, an inverse square, and a potential that is negative everywhere

This topic appears in:

01

A field is a region where a force is felt

Definition

Gravitational field strength — The gravitational force exerted per unit mass at a point: g = F/m, measured in N kg⁻¹.

Two masses attract each other across empty space with nothing visibly connecting them. The field is the way physics describes that: a region in which a mass experiences a force, with a strength and direction at every point.

Gravitational field strength g is defined as the force per unit mass. That definition is what makes it useful — it strips out the mass of whatever happens to be sitting there and describes the field itself. It also explains the coincidence that g = 9.81 N kg⁻¹ and the acceleration of free fall is 9.81 m s⁻²: they are the same quantity, since F = mg and F = ma give a = g.

Newton's law of gravitation:F = G m₁ m₂ / r²G = 6.67 × 10⁻¹¹ N m² kg⁻²field of a point mass:g = G M / r²always attractive; r is measured from the CENTREthe force is proportional to the product of the masses, and to 1/r²
G
the gravitational constantthe same everywhere in the universe
r
separation of the centresnot the surface-to-surface distance
M
the mass creating the fieldthe small mass in it does not appear in g

r is measured from the centre

For a satellite 300 km above the Earth, r is the Earth's radius plus 300 km — about 6670 km, not 300 km. Using the height above the surface instead of the distance from the centre is the most frequent error in this topic, and because the law is an inverse square it produces an answer wrong by a factor of nearly 500.

02

Potential, and why it is always negative

Gravitational potential φ is the work done per unit mass in bringing a mass from infinity to that point. The zero is placed at infinity, because that is the only place where masses genuinely stop interacting.

Since gravity is always attractive, it does the work of pulling a mass inwards — nothing has to be supplied. So the work done on the mass is negative, and the potential is negative everywhere, approaching zero only at infinity. A negative potential is not an error; it is the whole structure of a gravitational well.

Field and potential follow different powers of r, which is the distinction that most needs holding onto: field goes as 1/r² and potential as 1/r. They are linked by calculus — the field is the negative gradient of the potential.

φ = − G M / r(J kg⁻¹, always negative)E_p = m φ = − G M m / r(the energy of a mass in the field)g = − dφ/drthe field is the potential gradientpotential is per unit mass; potential energy includes the mass
φ
gravitational potentialwork per unit mass from infinity, in J kg⁻¹
E_p
potential energyφ multiplied by the mass placed there, in J
zero at infinity
the chosen referencewhich forces every finite value to be negative

Switch to Both and drag r. Doubling the distance quarters the field but only halves the potential. The potential curve is the shallower of the two and lies entirely below the axis — that is the gravitational well.

03

Orbits: gravity supplying the centripetal force

A satellite in a circular orbit is accelerating, because its direction is constantly changing. The only force acting is gravity, so gravity must be exactly the centripetal force required. Setting those two expressions equal is the key step in almost every orbit question.

The mass of the satellite cancels immediately, which is why the orbital speed and period depend only on the radius and the mass of the central body. A heavy satellite and a light one at the same altitude orbit at the same speed, and astronauts float not because gravity is absent but because they and the station are falling together.

gravity = centripetal force:GMm/r² = mv²/rv = √(GM/r)the satellite mass cancelsT² = 4π²r³ / (GM)Kepler's third laworbital speed decreases with radius — higher orbits are slower
v
orbital speedindependent of the satellite mass
T
the orbital periodT² proportional to r³
geostationary
T = 24 hourswhich fixes r at about 42 000 km from the centre
Worked example

A satellite orbits the Earth at a height of 400 km. Take the Earth's mass as 5.97 × 10²⁴ kg and radius as 6.37 × 10⁶ m. Find its orbital speed and period.

  1. r = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m.The height must be added to the Earth's radius, since r is measured from the centre.
  2. v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.77 × 10⁶).From equating gravitational and centripetal force, with the satellite mass already cancelled.
  3. = √(5.882 × 10⁷) = 7670 m s⁻¹.About 7.7 km per second — the familiar low-Earth-orbit speed, which is a useful check.
  4. T = 2πr/v = 2π(6.77 × 10⁶)/7670.Period is the circumference divided by the speed.
  5. = 5546 s ≈ 92 minutes.Close to the 90 minutes quoted for the International Space Station, confirming the working.

v = 7.67 km s⁻¹, T ≈ 92 minutes

The four quantities, and how they scale

  1. Force F = GMm/r² — inverse square, and needs both masses.
  2. Field g = GM/r² — inverse square, force per unit mass.
  3. Potential φ = −GM/r — inverse first power, and negative.
  4. Potential energy E_p = −GMm/r — potential times the mass in the field.
  5. Escape velocity comes from ½mv² = GMm/r, giving v = √(2GM/r).
  6. A geostationary orbit needs T = 24 h, equatorial, and moving west to east.
04

Fields near a surface, and why g looks constant

The inverse square law says the field weakens with distance, yet every mechanics question treats g as a fixed 9.81. Both are correct, and the reason is a matter of scale.

The Earth's radius is about 6370 km. Climbing a 100 m building changes r by roughly one part in 64 000, and squaring that still leaves a change of about 0.003% — far below the precision of any school measurement. Over the small heights of ordinary mechanics the field is genuinely uniform for practical purposes.

Over larger distances the variation becomes impossible to ignore. At the altitude of the International Space Station g has fallen to about 8.7 N kg⁻¹, roughly 89% of its surface value, and at geostationary height it is under 0.25 N kg⁻¹. This is why a satellite question must always use the inverse square law rather than the constant value.

LocationDistance from centreg (N kg⁻¹)
Earth surface6370 km9.81
Top of Everest6379 km9.79
Space station6770 km8.69
Geostationary orbit42 200 km0.22
Moon distance384 000 km0.0027

Uniform field or radial field?

Near a surface the field lines are effectively parallel and evenly spaced — a uniform field, where E_p = mgh applies and potential energy rises linearly with height. Further out the lines converge radially on the centre, the field follows 1/r², and the energy must be found from −GMm/r instead. Using mgh for a satellite is a serious error, and knowing which model applies is often the first decision a question demands.

Practice questions

5 questions · 17 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Define gravitational field strength, and explain why it has the same numerical value as the acceleration of free fall.
Model answer

Gravitational field strength is the gravitational force per unit mass at a point, g = F/m. Since the gravitational force on a mass is F = mg and Newton's second law gives F = ma, equating them gives a = g — so a freely falling mass accelerates at the numerical value of the field strength.

Examiner tip. One mark for the definition as force per unit mass, one for the F = mg = ma argument.

SQ2[2 marks]
Explain why gravitational potential is always negative.
Model answer

Potential is defined as the work done per unit mass in bringing a mass from infinity, where the potential is defined as zero. Because gravity is always attractive, it does positive work pulling the mass inwards, so the work done on the mass is negative. Every finite distance therefore has a negative potential, rising to zero only at infinity.

Examiner tip. Both the zero-at-infinity convention and the attractive nature of gravity are needed.

SQ3[2 marks]
A planet's surface field strength is g. At a distance of three planetary radii from its centre, state the field strength and the potential as fractions of their surface values.
Model answer

Field follows an inverse square, so it is g/9. Potential follows an inverse first power, so it is one third of the surface value. The two scale differently, which is why they must never be treated interchangeably.

Examiner tip. The mark is for using the different powers — 1/r² for field and 1/r for potential.

Solved numericals

1 · 4 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Calculate the gravitational field strength at its surface and the escape velocity from it.
Full working
  1. g = GM/r² = (6.67 × 10⁻¹¹)(7.35 × 10²²)/(1.74 × 10⁶)²Using the surface radius as r.[1]
  2. g = 4.902 × 10¹² / 3.028 × 10¹² = 1.62 N kg⁻¹About one sixth of Earth's, which is the well-known figure and a good check.[1]
  3. Escape velocity: ½mv² = GMm/r, so v = √(2GM/r)The kinetic energy must equal the depth of the potential well.[1]
  4. v = √(2 × 4.902 × 10¹² / 1.74 × 10⁶) = √(5.635 × 10⁶) = 2370 m s⁻¹About 2.4 km s⁻¹, far less than Earth's 11.2 km s⁻¹.[1]

g = 1.62 N kg⁻¹, escape velocity 2.37 km s⁻¹

Exam questions

1 · 7 marks

Multi-part questions with a full mark scheme.

Q1[7 marks]
A geostationary satellite remains above a fixed point on the equator. Take M(Earth) = 5.97 × 10²⁴ kg.
(a) State two conditions, other than its period, for an orbit to be geostationary.
(b) Show that the orbital radius is about 4.2 × 10⁷ m.
(c) Find the height above the Earth's surface, given the Earth's radius is 6.37 × 10⁶ m.
(d) Explain why the mass of the satellite does not appear in the calculation.
Mark scheme
  1. (a) The orbit must be in the plane of the equatorAny other plane would make the satellite drift north and south.[1]
  2. and the satellite must travel west to east, in the same sense as the Earth's rotationOtherwise it would move relative to the ground even with a 24-hour period.[1]
  3. (b) T = 24 h = 86 400 s. Using T² = 4π²r³/(GM), r³ = GMT²/(4π²)Rearranging Kepler's third law for r.[1]
  4. r³ = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(86400)²/(4π²) = 7.53 × 10²²Careful with the square of the period.[1]
  5. r = 4.22 × 10⁷ m as requiredTaking the cube root.[1]
  6. (c) Height = 4.22 × 10⁷ − 6.37 × 10⁶ = 3.58 × 10⁷ mSubtracting the Earth's radius, since r was measured from the centre.[1]
  7. (d) Equating GMm/r² with mv²/r cancels the satellite mass m from both sides, so the orbit depends only on r and the Earth's mass.The mark is for identifying the cancellation, not merely asserting independence.[1]

(a) equatorial, west to east; (b) 4.22 × 10⁷ m; (c) 3.58 × 10⁷ m; (d) m cancels