User:Metacomet/Sandbox

Planetary orbits – Example calculations

 * The following is a work in progress... -- Metacomet 22:02, 30 December 2005 (UTC)

Using known data related to the Earth's orbit, we wish to calculate the following:


 * the distance of the Earth from the Sun at both perihelion (closest approach) and aphelion (farthest excursion),
 * the Earth's specific angular momentum h and its angular momentum L,
 * the magnitude of the force of the Sun's gravity acting on the Earth at both perihlion and aphelion,
 * the magnitude of the Earth's acceleration toward the Sun at both perihelion and aphelion,
 * the Earth's orbital speed v at both perihelion and aphelion,
 * the Earth's kinetic energy K at both perihelion and aphelion,
 * the Earth's gravitational potential U at both perihelion and aphelion, and
 * the total energy E = K + U at both perihelion and aphelion.

The known data is as follows:

Distance of the Earth from the Sun

 * at perihelion:


 * $$ r_p = a(1-e) = 147.1 \times 10^6 \ \mathrm{km}  \,$$


 * at aphelion:


 * $$ r_a = a(1+e) = 152.1 \times 10^6 \ \mathrm{km}  \,$$

Specific angular momentum

 * Equation of orbit:


 * $$r = \frac{ h^2 / GM }{1 + e \cos (\theta - \phi)}  $$


 * Calculation, using aphelion:


 * Let $$ r = r_a = a(1+e) \, $$


 * Then:


 * $$ \cos(\theta - \phi) = -1 \, $$


 * Therefore:


 * $$a(1+e) = \frac{ h^2 / GM }{ ( 1 - e )  }  $$


 * $$ a(1+e)(1-e) = { h^2 \over GM } $$


 * $$ h = \sqrt{aGM(1-e^2)} $$


 * $$ h = 4.45536 \times 10^{15} \ \mathrm{m^2/s}$$

Angular momentum
Since gravity is a central force, the displacement of the Earth from the Sun is parallel to the force of gravity between the two bodies. As a result, there is not net torque about the Sun acting on the Earth. Since torque is the time rate of change of angular momentum, the Earth's angular momentum is conserved. In other words, the angular momentum is the same at every point in the Earth's orbit. The magnitude is given as


 * $$L = |\mathbf{L}| = hm = 2.663 \times 10^{40} \ \mathrm{kg \cdot m^2/s} $$

Force of the Sun's gravity acting on the Earth

 * at perihelion


 * $$ F_p = { GMm \over r_p^2 } = 3.666 \times 10^{28}  \ \mathrm{newtons}$$


 * at aphelion


 * $$ F_a = { GMm \over r_a^2 } = 3.429 \times 10^{28}  \ \mathrm{newtons} $$

Acceleration of the Earth toward the Sun

 * $$ a_p =  { F_p \over m } = { GM \over r_p^2 } = 6.13  \times 10^{-3} \  \mathrm{m/s^2} $$


 * $$ a_a =  { F_a \over m } = { GM \over r_a^2 } =  5.74 \times 10^{-3}   \  \mathrm{m/s^2}$$

Orbital speed

 * $$ v_p = { h \over r_p }  =  { h \over a(1-e) } = 30 \, 288 \ \mathrm{m/s}$$


 * $$ v_a = { h \over r_a }  =  { h \over a(1+e) }  =  29 \, 292 \ \mathrm{m/s}$$

Kinetic energy
Since the orbital speed of the Earth is substantially smaller than the speed of light, relativistic effects are negligible. Thus we can use the non-relativistic formula to calculate the Earth's kinetic energy:


 * $$ K_p = { 1 \over 2} m v_p^2 = 2.742 \times 10^{33}   \ \mathrm{Joules} $$


 * $$ K_a = { 1 \over 2} m v_a^2 =  2.564 \times 10^{33}    \ \mathrm{Joules}  $$

Gravitational potential

 * $$ U_p  =   { -GMm \over r_p   }  =  -5.393 \times 10^{33} \ \mathrm{Joules} $$


 * $$ U_a  =   { -GMm \over r_a   }  =  -5.216 \times 10^{33} \ \mathrm{Joules} $$

Total energy

 * $$ E_p = K_p + U_p = -2.651 \times 10^{33} \ \mathrm{Joules} $$


 * $$ E_a = K_a + U_p = -2.652 \times 10^{33} \ \mathrm{Joules} $$

The difference in these two numbers, in the fourth digit, is insignificant and results entirely from rounding error in the calculations.

Planetary orbit – Table of orbital data

 * The following is a work in progress... -- Metacomet 22:02, 30 December 2005 (UTC)

Sample MatLab code


% Initialize:


 * close all;


 * clear all;


 * clc;