Orbital Energy

Estimate the total energy of Earth’s orbit in J \si{\joule} .

Details and Assumptions:

  • G = 6.6 × 1 0 11 N m 2 / k g 2 G = \SI[per-mode=symbol]{6.6e-11}{\newton\per\kilo\gram\squared\meter\squared}
  • M Earth = 6 × 1 0 24 kg M_\textrm{Earth} = \SI{6e24}{\kilogram}
  • M Sun = 2 × 1 0 30 kg M_\textrm{Sun} = \SI{2e30}{\kilogram}
  • 1 au = 1.5 × 1 0 11 m \SI{1}{\astronomicalunit} = \SI{1.5e11}{\meter} (This is the mean distance from the Earth to the Sun.)

I created this problem for the 2013 UBC Physics Olympics (Problem 2).
Answers to a 2 % 2 \% margin of error will be accepted.


The answer is -2.64E+33.

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2 solutions

Pranshu Gaba
Nov 23, 2016

Relevant wiki: Characteristics of Circular Orbits

The Sun and the Earth move in circular paths about their center of mass. However, since the mass of the Sun is so much greater than the mass of the Earth, the center of mass of Earth and Sun lies very close to the Sun, and the radius of the circle that the Sun moves in is very small. Thus, we can approximately consider the Sun to be fixed, and the Earth to be revolving around the Sun.

The potential energy of Earth is given by U = G M Earth M Sun r U = - \dfrac{GM_{\text{Earth}} M_{\text{Sun}}}{r} .

The kinetic energy of Earth is K = 1 2 M Earth v 2 K = \dfrac{1}{2} M_{\text{Earth}}v^2 . The Earth moves with speed v = G M Sun R v = \sqrt{\dfrac{GM_{\text{Sun}}}{R}} , therefore the kinetic energy of Earth is K = G M Earth M Sun 2 r K = \dfrac{GM_{\text{Earth}} M_{\text{Sun}}}{2r} .

The total energy is given by E = K + U = G M Earth M Sun 2 r E = K + U = - \dfrac{GM_{\text{Earth}} M_{\text{Sun}}}{2r} .

When we substitute the given values, we get E 2.64 × 1 0 33 E \approx - 2.64 \times 10^{33} Joules

William G.
Feb 16, 2017

bold text CHINESE PHYSICS OLYMPIAD MUST BE EASY

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