Calculus Without Tears

Platenary Motion and the Two-Body Problem

Introduction

In the latter half of the seventeenth century, Issac Newton solved the two-body problem and explained the motion of objects in the sky, solving a mystery that had haunted mankind from the beginning. How important was this? In the history of the intellectual development of the human race the first big step was when we came down out of the trees, this was the second, we're waiting for the third.

Newton discovered gravity, discovered the laws of motion, developed calculus, and yet the hard work still lay before him. His goal was to explain the motion of objects in the sky. In particular, Kepler had proposed three laws of planetary motion, and had recorded a great amount of data that suggested that the laws did describe the motion of the planets in particular. However, Kepler had no idea why his laws held. Newton's goal was to explain Kepler's laws. He succeeded, and his efforts mark the beginnings of modern mathematics and physics. It's the most significant event in the history of our understanding of how the world works.

Simplified Planetary Motion

We will make a good assumption, that the sun and an orbiting planet can be represented by point masses, and that the force between them is given by Newton's law of gravity. We will also make a false assumption that the sun is 'fixed' in space and does not move. This second assumption simplifies the problem; after solving the simplified problem in this section, we will drop the assumption in the next section and solve the problem without the assumption.

Polar Coordinates

The sun will be fixed at the origin of the coordinate frame. We will use polar coordinates to describe the planet's trajectory, r(t) and θ(t).

Define two unit (having length 1) vectors, e_r pointing from the origin to the planet,
e_r = (cos(θ), sin(&theta))
and the vector e_θ rotated 90 deg. counter-clockwise from e_r,
e_θ = (-sin(θ), cos(&theta))

The planet's position is given by
r(t) = r(t)e_r = r(t)(cos(θ), sin(&theta))

Note: vectors are in bold type, so r(t) is the vector from the origin to the planet, and r(t) is the length of the vector.

Note that
e_r'(t) = (cos(θ(t)), sin(θ(t))' = θ'(t)(-sin(θ(t), cos(θ(t)) = θ'(t)e_θ
and
e_θ'(t) = (-sin(θ(t)), cos(θ(t))' = θ'(t)(-cos(θ(t), -sin(θ(t)) = -θ'(t)e_r

The planet's velocity is given by
v(t) = r'(t) = r'(t)e_r(t) + r(t)e_r'(t) = r'(t)e_r + r(t)θ'(t)e_θ

The planet's acceleration is given by
a(t) = v'(t) = r'(t)e_r'(t) + r''(t)e_r + r(t)θ'(t)e_θ'(t) + (r(t)θ''(t) + r'(t)θ'(t))e_θ
thus
a(t) = r'(t)θ'(t)e_θ + r''(t)e_r - r(t)θ'(t)θ'(t)e_r + (r(t)θ''(t) + r'(t)θ'(t))e_θ
and
a(t) = (r''(t) - r(t)θ'(t)²)e_r + (r(t)θ''(t) + 2r'(t)θ'(t))e_θ

The Differential Equations of Motion

Now we'll use Newton's Law of Motion to write
F(t) = ma(t)
Where F is the force acting on the planet and m is the mass of the planet. We are assuming that the only force acting on the planet is due to gravity, and using Newton's Law of Gravity, we have
GMm/r(t)² = ma(t)
where M is the mass of the earth and G is the gravitational constant. The direction of the gravitational force is -e_r, so we have
a(t) = -GM/r(t)²e_r

From the above, we have two equations of motion,
r''(t) - r(t)θ'(t)² = -GM/r(t)²
and since the e_θ component of a is 0,
r(t)θ''(t) + 2r'(t)θ'(t) = 0

Solving the Equations of Motion

We'd like to be able to solve the DEs and write down the functions of time that describe the planet's orbit. That ain't happening. We have two coupled differential equation; one is non-linear. Non-linear equations are difficult; none of the techniques we've studied in CWT Vols 1 thru 4 apply to this problem. No closed form solution exists. Imagine Newton, after discovering gravity, the laws of motion, and calculus, finally writing down his first differential equations, and they're not solvable. Tough break!

While a closed form solution is not possible, Newton was able to show that Kepler's laws could be derived from the equations of motion. We will content ourselves with showing that the orbit is a ellipse.

Multiplying the second equation of motion by r(t) gives
r(t)²θ''(t) + 2r(t)r'(t)θ'(t) = 0
The left hand side of the equation is (r(t)²θ'(t))', so we have
(r(t)²θ'(t))' = 0
and hence
r(t)²θ'(t) = h, a constant.
r(t)θ'(t) is the tangential velocity of the planet, and mr(t)r(t)θ'(t) is the angular momentum of the planet. So this shows that the angular momentum of the planet is constant, corresponding to Kepler's second law.

At this point, things get tricky :) . We want to determine r as a function of θ (call this function r_θ), however the simpler relationship is between 1/r and θ so we begin with that and define
u(θ) = 1 / r_θ(θ)
then (since r²(t)θ'(t) = h)
r'(t) = (1 / u(θ(t)))' = - u(&theta(t))' / u(&theta(t))² = -r(t)²(du/dθ)θ'(t) = -h(du/dθ)
and
r''(t) = -h(d²u/dθ²)θ'(t) = -u²h²(d²u/dθ²)

The first equation can be re-written as (using θ'(t) = h / r²)
r''(t) - h² / r³(t) = -GM / r²(t)
Then, substituting for r''(t), we have
-u²h²(d²u/dθ²) - u³h² = -u²GM
and dividing by -u²h² gives
(d²u/dθ²) + u = GM/h²

It is easily seen that the solution to the above DE is
u(θ) = GM/h²(1 - e cos(θ - θ₀))
where e and θ₀ are arbitrary constants. We have only one initial condition to match, u(0) = 1/r_θ(0), so we can choose θ₀ = 0 giving
r_θ(θ) = (h²/GM)(1 / (1 - e cos(θ))
Et viola'. This shows that the planet's orbit is an ellipse (see last section).

The Two-Body Problem

Given that the preceeding that was the simplified version of the problem, some hesitation in proceeding to the unsimplified version is understandable. However, not to fear, all the work has been done.

In reality, both the objects move. However, if no 'outside' forces are acting on the objects, their center of mass will be non-accelerating. We can use the center mass as the origin of the coordinates for the problem. Given these coordinates, a line from one object to the other will always pass through the origin.

Let r₁ be the position vector of the 1^st object, and r₂ the position vector of the 2^nd. Then the center of mass is the point where
r₁m₁ = r₂m₂
Thus,
r = r₁ + r₂ = r₁ + m₁r₁/m₂
so
m₂r = m₂r₁ + m₁r₁
and r = (m₁ + m₂)r₁/m₂

The magnitude of the gravitational force on m₁ is
f = Gm₁m₂ / r² = Gm₁m₂ / ((m₁ + m₂)r₁/m₂)²
so
f = Gm₁m₂ / r₁² (m₂²/(m₁ + m₂)²)

Thus, since the gravitational force acts along the line between the objects, it is always directed to the origin. The magnitude of the gravitational force acting on m₁ is less than (since r > r₁)
f = Gm₁m₂/r₁²
However, by using the value m₂³ /(m₁ + m₂)² for the mass of the 2nd object, the problem is identical to the one solved in the previous section.

Note 1: this doesn't quite match the standard result so there may be a mistake in the preceding.

There is an animation of two-body orbits at the bottom of the page at Orbit Animations .

Conic Sections and Polar Coordinates

The formula for an ellipse whose major axis is along the horizontal axis, in Cartesian coordinates, is
(x - x_c²)/a² + y²/b² = 1




We will show that the formula for a ellipse in polar coordinates is


It's just a matter of going through the algebra. Multiplying:

rcos(θ) = x, so

r = x² + y², so

Rearranging:

Dividing:

Rearranging:

Completeing the square:

Rearranging and simplifying:

Et viola':

where e < 1 and:

Reference