Answer to Question #157114 in Mechanics | Relativity for joel

Question #157114
Consider a planet of mass M with an orbiting satellite at a distance of R from each of their centers. Show that Kepler's third law of gravitation is true.
1
Expert's answer
2021-01-31T03:39:18-0500

Explanations & Calculations


  • The satallite is orbiting the planet at a constant distance: at a circle of radius R.
  • Apply Newton's second law towards the center of the orbit: towards center of the planet.

"\\qquad\\qquad\n\\begin{aligned}\n\\small F&= \\small ma\\\\\n \\small G\\frac{mM}{R^2}&= \\small m\\frac{v^2}{R}\\to G\\frac{\\cancel{m}M}{R^{\\cancel{2}}}=\\cancel{m}\\frac{v^2}{\\cancel{R}}\\\\\n\\small G\\frac{M}{R}&= \\small v^2\\cdots(1)\n\\end{aligned}"

  • Time spent on orbiting the planet can be written as "\\large\\frac{2\\pi R}{v}" as the orbiting speed stays constant due to the radius being constant.
  • Ths gives,

"\\qquad\\qquad\n\\begin{aligned}\n\\small T&= \\small \\frac{2\\pi R}{v}\\\\\n\\small v&= \\small \\frac{2\\pi R}{T}\n\\end{aligned}"

  • Substituting this in (1) yeilds,

"\\qquad\\qquad\n\\begin{aligned}\n\\small G\\frac{M}{R}&= \\small \\Big(\\frac{2\\pi R}{T}\\Big)^2\\\\\n\\small (GM)\\times T^2&= \\small4\\pi^2\\times R^3\\\\\n\\small T^2&\\propto \\small R^3 (\\text{as other terms are constants})\n\\end{aligned}"

  • This proves the Kepler's third which denotes "The square of the orbital period directly proportional to the cube of the length of the semi-major axis of the orbit "
  • It was originally denoted for the real elliptical orbital motions & all axes are identical in this problem as the orbit is a circle of radius R.

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