Mathematics of the Planets

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Mathematics of the Planets — Radii, Distances & Key Calculations

This post gives a compact mathematical tour of the eight major planets: their mean radii, average distance from the Sun (semi-major axis) in AU and km, masses, and orbital periods — followed by general formulas (circumference, surface area, volume, density) and two worked examples (Earth and Jupiter).

Quick reference table

Constants used: 1 AU = 149,597,870.7 km

Planet Mean radius (km) Distance from Sun (AU) Distance (km) Mass (kg) Orbital period
Mercury 2,439.7 0.387 57,894,376.0 3.3011 × 1023 88 days (≈ 0.241 yr)
Venus 6,051.8 0.723 108,159,260.5 4.8675 × 1024 225 days (≈ 0.615 yr)
Earth 6,371.0 1.000 149,597,870.7 5.97237 × 1024 365.256 days (1.000 yr)
Mars 3,389.5 1.524 227,987,154.9 6.4171 × 1023 687 days (≈ 1.88 yr)
Jupiter 69,911 5.204 778,507,319.1 1.8982 × 1027 4,331 days (≈ 11.86 yr)
Saturn 58,232 9.582 1,433,446,797.0 5.6834 × 1026 10,747 days (≈ 29.46 yr)
Uranus 25,362 19.201 2,872,428,715.3 8.6810 × 1025 30,589 days (≈ 84.01 yr)
Neptune 24,622 30.047 4,494,967,220.9 1.02413 × 1026 ~60,190 days (≈ 164.8 yr)

Key geometric formulas (use radius r)

  • Circumference: \(C = 2\pi r\) (same units as r)
  • Surface area: \(A = 4\pi r^2\)
  • Volume: \(V = \dfrac{4}{3}\pi r^3\)
  • Mean density: \(\rho = \dfrac{M}{V}\) (mass divided by volume; ensure M and V use consistent units)

Worked example 1 — Earth (step-by-step)

Given: Earth mean radius \(r = 6,371.0\) km and mass \(M = 5.97237\times10^{24}\) kg.

  1. Circumference:
    \(C = 2\pi r = 2 \times \pi \times 6{,}371.0\ \text{km}.\)
    Compute digits: \(2 \times 6{,}371.0 = 12{,}742.0\).
    Then \(12{,}742.0 \times \pi \approx 12{,}742.0 \times 3.141592653589793 = 40{,}030.173592041145\ \text{km}.\)
    So Earth's circumference \(C \approx 40{,}030.17\ \text{km}.\)
  2. Surface area:
    \(A = 4\pi r^2 = 4 \times \pi \times (6{,}371.0)^2\ \text{km}^2.\)
    First square radius: \(6{,}371.0^2 = 40{,}589{,}641.0\).
    Multiply: \(4 \times 40{,}589{,}641.0 = 162{,}358{,}564.0\).
    Then \(162{,}358{,}564.0 \times \pi \approx 162{,}358{,}564.0 \times 3.141592653589793 = 510{,}064{,}471.90978825\ \text{km}^2.\)
    So \(A \approx 5.1006\times10^{8}\ \text{km}^2.\)
  3. Volume:
    \(V = \dfrac{4}{3}\pi r^3.\)
    Compute cube: \(6{,}371.0^3 = 258{,}518{,}952{,}400.999\) (≈ \(2.585189524\times10^{11}\)).
    Multiply by \(4/3\): \(\dfrac{4}{3}\times 258{,}518{,}952{,}401 \approx 344{,}691{,}936{,}534.0\).
    Then times \(\pi\): \(344{,}691{,}936{,}534.0 \times 3.141592653589793 \approx 1{,}083{,}206{,}916{,}845.7535\ \text{km}^3.\)
    So \(V \approx 1.0832\times10^{12}\ \text{km}^3.\)
  4. Mean density (ρ):
    Convert volume to m³ for use with mass in kg:
    \(1\ \text{km}^3 = (1000\ \text{m})^3 = 10^9\ \text{m}^3.\)
    So \(V = 1.0832069168457535\times10^{12}\ \text{km}^3 = 1.0832069168457535\times10^{21}\ \text{m}^3.\)
    Then \(\rho = \dfrac{M}{V} = \dfrac{5.97237\times10^{24}\ \text{kg}}{1.0832069168457535\times10^{21}\ \text{m}^3}\).
    Divide: \(5.97237\times10^{24} / 1.0832069168457535\times10^{21} \approx 5{,}513.600316910137\ \text{kg/m}^3.\)
    So Earth's mean density ≈ 5513.6 kg·m⁻³.

Worked example 2 — Jupiter (step-by-step)

Given: Jupiter radius \(r=69{,}911\) km, mass \(M=1.8982\times10^{27}\) kg.

  1. Circumference:
    \(C = 2\pi r = 2 \times \pi \times 69{,}911\).
    \(2 \times 69{,}911 = 139{,}822.\)
    \(139{,}822 \times \pi \approx 139{,}822 \times 3.141592653589793 = 439{,}263.7680102321\ \text{km}.\)
  2. Surface area:
    \(A = 4\pi r^2\). First \(r^2 = 69{,}911^2 = 4{,}887{,}885{,}921.\)
    \(4 \times r^2 = 19{,}551{,}543{,}684.\)
    Multiply by \(\pi\): \(19{,}551{,}543{,}684 \times \pi \approx 61{,}418{,}738{,}570.72667\ \text{km}^2.\)
  3. Volume:
    \(V = \dfrac{4}{3}\pi r^3\). Cube: \(69{,}911^3 \approx 341{,}184{,}722{,}444{,}000\) (≈ 3.41184722444×10^14).
    Multiply by \(4/3\) and \(\pi\): \(V \approx 1.4312818107393572\times10^{15}\ \text{km}^3.\)
  4. Mean density:
    Convert \(V\) to m³: \(1.4312818107393572\times10^{24}\ \text{m}^3.\)
    \(\rho = \dfrac{1.8982\times10^{27}}{1.4312818107393572\times10^{24}} \approx 1{,}326.2237986657897\ \text{kg/m}^3.\)
    So Jupiter's mean density ≈ 1326.2 kg·m⁻³.

How to use this mathematically

• To compare sizes: compute radius ratios, e.g. Jupiter/Earth radius = \(69{,}911 / 6{,}371 \approx 10.97\) (Jupiter ≈ 10.97 × Earth's radius).
• To compare volumes: volume scales as \(r^3\); Jupiter's volume / Earth's volume ≈ \((69{,}911 / 6{,}371)^3 \approx 10.97^3 \approx 1{,}320\) — Jupiter holds about 1,320 Earths by volume.
• To compare gravitational acceleration roughly at surface: \(g \propto \dfrac{M}{r^2}\). You can compute \(g\) ratio Earth vs Planet by using the mass and radius in consistent units.

Final notes

• All radii are mean (planets are not perfect spheres). Distances are average (semi-major axes). Masses and orbital periods are standard accepted values.
• If you'd like this post with inline LaTeX rendered, or a downloadable DOCX/Google Doc / printable poster version, tell me which format and I’ll produce it next.

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