Earth orbit
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.
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  147.1152.1 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?
Possible approach
Key questions
 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?