Newton's Law of Gravitation

F = GMm/r²; gravitational field strength; orbital motion and Kepler's third law

# Newton's Law of Gravitation — IB Physics

1. Newton's Law

F=GMmr2\boxed{F = \frac{GMm}{r^2}}

G=6.67×1011G = 6.67 \times 10^{-11} N m² kg⁻²

Always attractive, acts along line joining centres.

2. Gravitational Field Strength

g=GMr2g = \frac{GM}{r^2}

At surface: g=GM/R2g = GM/R^2. gg is also free-fall acceleration.

3. Orbital Motion

Gravity provides centripetal force: GMmr2=mv2r\frac{GMm}{r^2} = \frac{mv^2}{r}

v=GMrv = \sqrt{\frac{GM}{r}}

T=2πr3GMT = 2\pi\sqrt{\frac{r^3}{GM}}

Kepler's Third Law: T2r3T^2 \propto r^3

4. Gravitational Potential Energy

Ep=GMmrE_p = -\frac{GMm}{r}

Negative: work needed to separate masses to infinity.

Total orbital energy: E=KE+PE=12mv2GMmr=GMm2rE = KE + PE = \frac{1}{2}mv^2 - \frac{GMm}{r} = -\frac{GMm}{2r}

Worked Example: Surface Gravity

Problem

Mars: M=6.42×1023M = 6.42 \times 10^{23} kg, R=3.39×106R = 3.39 \times 10^6 m. g=GM/R2=6.67×1011×6.42×1023/(3.39×106)2=3.72g = GM/R^2 = 6.67 \times 10^{-11} \times 6.42 \times 10^{23}/(3.39 \times 10^6)^2 = 3.72 m/s²

Solution

Worked Example: Orbital Period

Problem

ISS at 400 km altitude. r=6.77×106r = 6.77 \times 10^6 m. T=2πr3/(GME)=2π(6.77×106)3/(3.986×1014)=5540T = 2\pi\sqrt{r^3/(GM_E)} = 2\pi\sqrt{(6.77 \times 10^6)^3/(3.986 \times 10^{14})} = 5540 s = 92.3 min

Solution

6. Practice Questions

    1. Derive the expression T2=4π2r3/(GM)T^2 = 4\pi^2 r^3/(GM) for circular orbits. (3 marks)
    1. Calculate gg at twice Earth's radius from its centre. (2 marks)
    1. A satellite has orbital period 24 hours. Find its orbital radius. (3 marks)

    Answers

    1. GMm/r2=mv2/rGMm/r^2 = mv^2/rv2=GM/rv^2 = GM/r. v=2πr/Tv = 2\pi r/T, so 4π2r2/T2=GM/r4\pi^2 r^2/T^2 = GM/rT2=4π2r3/(GM)T^2 = 4\pi^2 r^3/(GM).
    1. g1/r2g \propto 1/r^2. At 2R2R: g=9.81/4=2.45g = 9.81/4 = 2.45 m/s².

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Summary

  • F=GMm/r2F = GMm/r^2; g=GM/r2g = GM/r^2
  • Orbits: v=GM/rv = \sqrt{GM/r}; T2r3T^2 \propto r^3
  • Ep=GMm/rE_p = -GMm/r; total orbital energy = GMm/(2r)-GMm/(2r)

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