Earth orbit
Follow in the steps of Newton and find the path that the earth
follows around the sun.
Age
16 to 18
Challenge level
Problem
The force which drives the motion of the planets around the sun is gravity and Newton showed that this force is inversely proportional to the square of the distance of the planet from the sun. The constant of proportionality is called Newton's constant $G$.
If the mass of the planet is $m$ and the mass of the sun is $M$ then the force between the two is $$F = -\,\frac{GmM}{r^2}\;,\quad\quad G = 6.674 \times 10^{-11}\mathrm{m}^3\mathrm{kg}^{-1}\mathrm{s}^{-2}\;.$$ In polar coordinates in the plane of motion, the equation of motion is$$ \frac{d^2 r}{dt^2}-\frac{h^2}{r^3} = -\frac{GM}{r^2} $$ where the angular momentum $h$ of the planet is given by $$
h = r^2\frac{d\theta }{dt}\;. $$ This equation is very tricky to solve directly, but making the substitution$$ u = \frac{1}{r} $$ leads to an inhomogenous linear second order differential equation.
Find this equation and show that its solution is$$ u=GM/h^2 +A\cos \theta + B \sin \theta\;. $$ Show that numbers $e$ and $f$ can be found so that$$ h^2 = GMr(1+e\cos(\theta+f))\;. $$
For a fun and very much simpler practical activity, why not try the problem Making Maths: Planet paths.
Extension: Investigate the elliptic paths observed in the solar system using the real data below. Why not try to draw a scale model of the solar system.
NOTES AND BACKGROUND
Solving the problem of the motion of the planets around the sun required the invention of differentiation and integration, credited to Newton about 350 years ago. The fact that orbits follow such beautiful, pure paths shows how elegantly the universe is put together. The results are incredibly accurate and were only challenged by Einstein whose theory of General Relativity provides very small corrections to the orbits.
Essentially, this mathematics is sufficient to send space probes all the way from earth to the far reaches of the solar system with sufficient accuracy to meet up with various planets and moons along the way.
The following table comprises real astronomical data (compiled from Wikipedia) which describe the elliptical paths taken by some key objects in our solar system.
| Name | Diameter relative to Earth | Mass relative to Earth | Orbital radius | Orbital period | Inclination to sun's equator |
Orbital Eccentricity
e
|
Rotation period (days) |
| The Sun | 109 | 332946 | -- | -- | -- | -- | 26.38 |
| The Moon | 0.273 | 0.0123 | -- | 29.5 days | -- | 0.0549 | -- |
| Halleys Comet | -- | -- | -- | 73.3 | 162.3 | 0.967 | -- |
| Mercury | 0.382 | 0.06 | 0.39 | 0.24 | 3.38 | 0.206 | 58.64 |
| Venus | 0.949 | 0.82 | 0.72 | 0.62 | 3.86 | 0.007 | -243.02 |
| Earth | 1.00 | 1.00 | 1.00 | 1.00 | 7.25 | 0.017 | 1.00 |
| Mars | 0.532 | 0.11 | 1.52 | 1.88 | 5.65 | 0.093 | 1.03 |
| Jupiter | 11.209 | 317.8 | 5.20 | 11.86 | 6.09 | 0.048 | 0.41 |
| Saturn | 9.449 | 95.2 | 9.54 | 29.46 | 5.51 | 0.054 | 0.43 |
| Uranus | 4.007 | 14.6 | 19.22 | 84.01 | 6.48 | 0.047 | -0.72 |
| Neptune | 3.883 | 17.2 | 30.06 | 164.8 | 6.43 | 0.009 | 0.67 |
The actual numbers for the earth are
| Diameter | Mass kg | Distance from sun | Orbital period | Rotation time |
| 12756 km | 5.9736 x 10^24 | 147.1-152.1 million km | 365.256366 days | 23 hours 56 minutes |
Getting Started
For the first part you will need to change variables in an equation. You will need to use this idea:$$\frac{df}{dr} = \frac{df}{du}\frac{du}{dr}\;.$$ For the second part, you need to redefine your angle by shifting it by a certain amount.
Student Solutions
We have to solve $$\frac{d^2r}{dt^2} - \frac{h^2}{r^3} = -\frac{GM}{r^2}\;.$$ $h = r^2\frac{d\theta}{dt}$ is constant. Putting $u = \frac{1}{r}$ we have $\frac{d\theta}{dt} = hu^2$ and $$\frac{dr}{dt} = \frac{dr}{du}\frac{du}{d\theta}\frac{d\theta}{dt} = -\frac{1}{u^2}\frac{du}{d\theta}hu^2 = -h\frac{du}{d\theta}$$ Therefore, $$\frac{d^2r}{dt^2} = -h\frac{d}{dt}\left(\frac{du}{d\theta}\right) = -h\frac{d\theta}{dt}\frac{d^2u}{d\theta^2} = -h^2u^2\frac{d^2u}{d\theta^2}\;.$$ Our differential equation now becomes $$-h^2u^2\frac{d^2u}{d\theta^2} - h^2u^3 = -GMu^2\quad \Rightarrow\quad \frac{d^2u}{d\theta^2} + u = \frac{GM}{h^2}$$ which has a general solution $u(\theta) = \frac{GM}{h^2} + A\cos{\theta} + B\sin{\theta}$ for constants $A$ and $B$.
Now we substitute back to $r$, and get $$r = \frac{1}{\frac{GM}{h^2} + A\cos{\theta} + B\sin{\theta}} = \frac{\frac{h^2}{GM}}{1 + \frac{Ah^2}{GM}\sin{\theta} + \frac{Bh^2}{GM}\cos{\theta}} = \frac{\frac{h^2}{GM}}{1 + e\cos{(\theta + f)}}$$ for some $e$ and $f$. Therefore $h^2 = GMr(1 + e\cos{(\theta + f)})$.
Teachers' Resources
Why do this problem?
This problem is a difficult exercise in algebra,
differentiation and trigonometry which draws together strands from
polar coordinates and mechanics, although these are not necessary
for the solution of the problem. The rich ideas covered lead to a
genuinely beautiful result which is well within the reach of the
more technically skilled 6th former.
Possible approach
The technical aspect of the transformation is well suited to
an individual activity. The ideas concerning the derivation of the
equation would lend themselves to class discussion, although this
requires the knowledge of the acceleration of a particle moving in
polar coordinates. Is the class working together able to derive the
equation used in the process?
Key questions
Students should be enouraged to understand the variables in
the problem before attempting to make the transformation
- How do we make a change of variables in an equation?
- What is the meaning of the variables used when rewriting the solution?
- What are the shapes of the solutions?
Possible extension
Those who are keen to make an extension should be encouraged
to derive the equation. Alternatively, they can use physical data
for the earth and the sun to investigate how closely the solution
obtained here matches reality (it is actually exceedingly
accurate).