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Physics 3% exam weight

Gravitation

Part of the MDCAT study roadmap. Physics topic phy-20 of Physics.

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Gravitation

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

  • Newton’s law of universal gravitation: every two point masses attract each other with force F = G m₁ m₂ / r², where G = 6.674 × 10⁻¹¹ N·m²/kg² and r is the centre-to-centre distance.
  • Gravitational field strength at distance r from mass M: g = GM / r². At Earth’s surface g₀ ≈ 9.8 m/s².
  • Escape velocity vₑ = √(2GM/r); orbital velocity v₀ = √(GM/r). Escape is exactly √2 times orbital at the same radius.
  • Kepler’s third law: T² = (4π² / GM) a³ — period squared scales with semi-major axis cubed.
  • MDCAT traps: sign of gravitational PE (U = −GmM/r, always negative for bound systems), and substituting (R + h) instead of r when a satellite is at altitude h.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Newton’s Law and the Gravitational Constant

Two masses m₁ and m₂ separated by centre-to-centre distance r pull on each other with F = G m₁ m₂ / r², directed along the line joining their centres. G is universal — the same on Earth, on Mars, or between galaxies — and has been measured by the Cavendish torsion-balance experiment. Because the force obeys an inverse-square law, doubling r quarters the force; tripling r reduces it to one-ninth.

Field Strength, Potential, and Potential Energy

Gravitational field strength g at a point is force per unit test mass, g = GM/r². Near Earth’s surface this gives g₀ = GM/R² ≈ 9.8 m/s². Gravitational potential V = −GM/r is the work done per unit mass to bring a test mass from infinity to point r; potential is always negative and approaches 0 at infinity. The associated potential energy for a two-body system is U = −Gm₁m₂/r.

Orbits, Escape, and Kepler’s Laws

A satellite in circular orbit of radius r around mass M has orbital velocity v₀ = √(GM/r) and centripetal acceleration v²/r = GM/r² = g(r). Escape velocity is derived by setting total energy to zero at infinity: ½mvₑ² − GMm/r = 0 → vₑ = √(2GM/r) = √2 · v₀.

Kepler’s third law T² = (4π²/GM)a³ follows by combining v = 2πa/T with v² = GM/a. The constant (4π²/GM) depends on the central body, so it differs for Sun–planet vs Earth–satellite systems.

Variation of g with Height and Depth

Above the surface: g(h) = g₀ [R/(R+h)]². Below the surface, only the inner sphere of radius (R − d) contributes, giving g(d) = g₀ (1 − d/R) — g falls linearly with depth and is zero at Earth’s centre.

Typical MDCAT Question Patterns

  • Numerical: find orbital period of a geostationary satellite (T = 24 h → a ≈ 42,164 km from Earth’s centre).
  • Comparison: ratio of escape velocities from two planets of known mass and radius.
  • Conceptual: why gravitational PE is negative, or why g decreases with depth but not as 1/r².

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Inertial vs Gravitational Mass

Inertial mass (from F = ma) measures resistance to acceleration; gravitational mass (from F = GMm/r²) measures how strongly a body feels gravity. Experiments show they are equivalent to better than 1 part in 10¹⁵ — this equivalence principle is the foundation of Einstein’s general relativity, which replaces Newton’s instantaneous action with spacetime curvature.

Energy in Orbit and Bound Systems

Total mechanical energy of a circular orbit: E = −GMm/(2r), exactly half the potential energy. This is why binding energy equals −E. An elliptical orbit has the same form with a replaced by the semi-major axis: E = −GMm/(2a) — a deep result used to find a comet’s orbit from a single observation of r and v.

Geostationary and Polar Orbits

A geostationary satellite orbits in the equatorial plane with T = 24 h, so it appears fixed above one point on Earth. From Kepler’s law, a ≈ 42,164 km from Earth’s centre (≈ 35,786 km altitude). MDCAT sometimes asks why the orbit radius uses R + h, not just R.

Common Mistakes in MDCAT

  • Writing U = +GmM/r — sign error; potential energy for a bound system is always negative.
  • Using g = 9.8 m/s² for a satellite at altitude h; must use g = g₀[R/(R+h)]².
  • Mixing orbital and escape velocity: remember the √2 ratio.
  • Treating G and g as the same symbol; G is universal, g is local.

Worked Micro-Example

Find the orbital speed of a satellite 500 km above Earth (R = 6.4 × 10⁶ m, M = 5.97 × 10²⁴ kg, G = 6.67 × 10⁻¹¹). r = R + h = 6.9 × 10⁶ m. v₀ = √(GM/r) = √[(6.67×10⁻¹¹)(5.97×10²⁴)/(6.9×10⁶)] ≈ 7.6 km/s.

Practice Prompts

  1. A planet has twice Earth’s mass and half Earth’s radius. Find its escape velocity relative to Earth’s (≈ 2.83 vₑ,Earth).
  2. At what depth inside Earth does g equal g₀/2? (Answer: d = R/2.)

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