Gravitational Fields and Orbital Mechanics

Newton's Law of Universal Gravitation

Every object with mass exerts an attractive gravitational force on every other object with mass. Newton quantified this observation in his Law of Universal Gravitation:

F = GMm / r²

where:

• F = gravitational force (N)
• G = Universal Gravitational Constant = 6.674 × 10⁻¹¹ N m² kg⁻²
• M = mass of the larger body (kg)
• m = mass of the smaller body (kg)
• r = centre-to-centre separation between the two bodies (m)

Several features are worth noting for HSC purposes:

• The force is always attractive — gravitational repulsion does not exist.
• The force obeys an inverse-square law: doubling the separation reduces F by a factor of 4; tripling it reduces F by a factor of 9.
• The force acts equally on both bodies (Newton's Third Law pair): Earth pulls the Moon with the same magnitude force as the Moon pulls Earth.
• r is measured from centre to centre, not surface to surface. For a satellite orbiting at height h above Earth's surface (radius R_E), use r = R_E + h.

Worked Example 1 — Force between Earth and Moon

Given: M_Earth = 5.972 × 10²⁴ kg, M_Moon = 7.342 × 10²² kg, r = 3.844 × 10⁸ m.

Step 1 — Write the formula: F = GMm / r²

Step 2 — Substitute values:

F = (6.674 × 10⁻¹¹) × (5.972 × 10²⁴) × (7.342 × 10²²) / (3.844 × 10⁸)²

Step 3 — Numerator: 6.674 × 5.972 × 7.342 = 6.674 × 43.82 ≈ 292.4; combine powers: 10⁻¹¹ × 10²⁴ × 10²² = 10³⁵; numerator ≈ 2.924 × 10³⁷

Step 4 — Denominator: (3.844)² = 14.776; (10⁸)² = 10¹⁶; denominator ≈ 1.478 × 10¹⁷

Step 5 — Divide: F ≈ 2.924 × 10³⁷ / 1.478 × 10¹⁷ ≈ 1.978 × 10²⁰ N

Result: F ≈ 1.98 × 10²⁰ N (consistent with the accepted value of ~1.98 × 10²⁰ N). Units check: N m² kg⁻² × kg × kg / m² = N. ✓

Gravitational Field Strength

Rather than always computing forces between pairs of masses, physicists use the concept of a gravitational field. A field exists at every point in space around a massive object; it describes the force that would act on a unit mass placed at that point.

The gravitational field strength g at a distance r from the centre of mass M is:

g = GM / r²

Units: N kg⁻¹ (equivalent to m s⁻²). At Earth's surface, g ≈ 9.8 N kg⁻¹.

Key relationships:

• The force on a mass m placed in the field is F = mg, which — when expanded — returns F = GMm/r².
• g decreases with the square of distance from the centre: moving from Earth's surface to twice Earth's radius reduces g to one-quarter.
• g is a vector directed towards the centre of the source mass.

Worked Example 2 — Field strength at altitude

The International Space Station orbits at approximately h = 4.00 × 10⁵ m above Earth's surface. Calculate the gravitational field strength at that altitude.

Given: G = 6.674 × 10⁻¹¹ N m² kg⁻², M_E = 5.972 × 10²⁴ kg, R_E = 6.371 × 10⁶ m.

Step 1 — Find orbital radius: r = R_E + h = 6.371 × 10⁶ + 4.00 × 10⁵ = 6.771 × 10⁶ m

Step 2 — Apply formula: g = GM/r² = (6.674 × 10⁻¹¹ × 5.972 × 10²⁴) / (6.771 × 10⁶)²

Step 3 — Numerator: 6.674 × 5.972 = 39.85; combine powers: 10⁻¹¹⁺²⁴ = 10¹³; numerator ≈ 3.985 × 10¹⁴

Step 4 — Denominator: (6.771)² = 45.85; (10⁶)² = 10¹²; denominator ≈ 4.585 × 10¹³

Step 5 — g ≈ 3.985 × 10¹⁴ / 4.585 × 10¹³ ≈ 8.69 N kg⁻¹

Result: g ≈ 8.69 N kg⁻¹ at ISS altitude. This is about 89% of the surface value — astronauts are not weightless because gravity is absent; they are in continuous free-fall around Earth (apparent weightlessness).