Edexcel IAL Physics revision

Edexcel IAL Physics revision · A2 — Mechanics & fields

Gravitational fields

Newton's inverse-square law, applied from lab bench to geostationary orbit. The algebra is short; the marks concentrate around one negative sign — gravitational potential — and one satellite: the geostationary one.

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What the syllabus demands

  • Use Newton's law of gravitation: F = GMm/r²
  • Define gravitational field strength g = F/m; use g = GM/r²
  • Define gravitational potential; use φ = −GM/r
  • Analyse circular orbits: equate gravitational and centripetal force
  • Describe geostationary orbits and derive orbital speed and period relationships

Definitions that earn marks

Clear definitions to practise — check your course mark scheme

Newton's law of gravitation
The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation: F = GMm ÷ r².
Gravitational field strength
The gravitational force per unit mass at a point: g = F ÷ m. Units: N/kg.
Gravitational potential (φ)
The work done per unit mass in bringing a small test mass from infinity to the point: φ = −GM ÷ r. Negative because gravity is attractive and the potential at infinity is defined as zero.
Geostationary orbit
An equatorial orbit with a period of 24 hours in the direction of the Earth's rotation, so the satellite remains above the same point on the equator.

The equations

Gravitational forceF = G M m ÷ r² · N
Field strengthg = G M ÷ r² · N/kg
Potentialφ = −G M ÷ r · J/kg
Orbit conditionG M m ÷ r² = m v² ÷ r

More equations to practise: the Edexcel IAL formula sheet.

Where the marks die

Common mistakes to check

  1. 01

    Dropping the minus sign on potential. φ = −GM/r: potential is always negative, rises towards zero at infinity, and 'higher potential' means less negative. Escape questions live and die on this sign.

  2. 02

    Using surface g for orbital altitude. g falls as 1/r² — and r is measured from the centre of the Earth, not from the surface. Adding the altitude to the radius is a required step.

  3. 03

    Geostationary answers missing a condition. Three are needed: period 24 hours, orbit above the equator, moving in the same direction as the Earth's rotation.

  4. 04

    Confusing field strength (vector, N/kg, −dφ/dr) with potential (scalar, J/kg). Field strength is the gradient of potential — where potential flattens, the field weakens.

One worked example, done properly

Question

Calculate the orbital speed of a satellite 400 km above Earth's surface. (M = 6.0 × 10²⁴ kg, R = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹)

Method

  1. 1.r = R + h = 6.4 × 10⁶ + 4.0 × 10⁵ = 6.8 × 10⁶ m.
  2. 2.Gravity provides centripetal force: GMm/r² = mv²/r, so v = √(GM/r).
  3. 3.v = √(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ ÷ 6.8 × 10⁶).

v ≈ 7.7 × 10³ m/s (7.7 km/s)

Fit these topics into your free physics revision plan

Common questions

Asked, answered.

Why is gravitational potential negative?

Potential at infinity is defined as zero, and gravity is attractive — so work must be done against gravity to move a mass out to infinity. Every point closer than infinity therefore has less than zero potential.

How do you find the speed of an orbiting satellite?

Set the gravitational force equal to the required centripetal force: GMm/r² = mv²/r. The satellite mass cancels, giving v = √(GM/r) — the speed depends only on the central mass and orbital radius.

Why must geostationary satellites orbit over the equator?

The orbit's centre must be the Earth's centre, so the orbital plane passes through it. Only an equatorial orbit lets the satellite stay above the same ground point; any inclined orbit would trace a daily figure-of-eight over the surface.

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