From One North Pole to the Other

Classical Mechanics Level 5

There are two identical uniform spherical planets of radius $R$ . The first has its center at the origin of the $xyz$ coordinate system. The second has its center at $(x,y,z) = (2 R,0,0)$ . The planets are touching.

A projectile is launched from the "North Pole" of the first planet at $(x,y,z) = (0,0,R)$ with its initial velocity pointed in the direction of the vector $(1,0,1)$ .

Let the escape speed relative to the planet's surface be $v_e$ . Note that here, the escape speed is for a single planet in isolation (following the typical convention).

With the given launch vector, let $v_0$ be the minimum launch speed for the projectile to reach the "North Pole" of the second planet at $(x,y,z) = (2R,0,R)$ . The two speeds are related as follows:

$\large{v_0 = \alpha \, v_e}$

Determine the value of $\alpha$

The answer is 0.813.

1 solution

Mark Hennings
Mar 15, 2019

We are being asked to solve the paired differential equations $\begin{aligned} \ddot{x} & = \; -gR^2\left[\frac{x}{(x^2 + y^2)^{\frac32}} + \frac{x-2R}{((x-2R)^2 + y^2)^{\frac32}}\right] \\ \ddot{y} & = \; -gR^2\left[\frac{y}{(x^2 + y^2)^{\frac32}} + \frac{y}{((x-2R)^2 + y^2)^{\frac32}}\right] \end{aligned}$ together with the initial conditions $x(0) = 0$ , $y(0) = R$ , $\dot{x}(0) = \dot{y}(0) = \frac{\alpha}{\sqrt{2}}v_e$ , where, of course, $v_e = \sqrt{2gR}$ .

We can render these equations dimensionless by changing variables so that $x = RX$ , $y = RY$ , $t = \sqrt{\tfrac{R}{g}}T$ , and the equations become $\begin{aligned} x'' & = \; -\left[\frac{X}{(X^2 + Y^2)^{\frac32}} + \frac{X-2}{((X-2)^2 + Y^2)^{\frac32}}\right] \\ y'' & = \; -\left[\frac{Y}{(X^2 + Y^2)^{\frac32}} + \frac{Y}{((X-2)^2 + Y^2)^{\frac32}}\right] \end{aligned}$ with the initial conditions $X(0) = 0$ , $Y(0) = 1$ , $X'(0) = Y'(0) = \alpha$ . Here a dash represents differentiation with respect to $T$ .

Given a particular value of $\alpha$ , we want to find the first value of $T$ for which $X(T) = 2$ , and determine the corresponding value of $Y(T)$ . We need to find the value of $\alpha$ such that $Y(T) = 1$ when first $X(T) = 2$ . Searching numerically in the range $0.7 < \alpha < 1$ , we find that the required value of $\alpha$ is $\boxed{0.8131305}$

Is the path traversed by the projectile an ellipse , parabola , circle or neither ? I can only figure out that it is symmetric about the plane passing through the contact point of the planets. Thanks in advance.

Jacob Sony - 2 years, 2 months ago

I would guess that it is none of these. I doubt that these equations have an exact solution...

Mark Hennings - 2 years, 2 months ago

it will move in ellipse with foci the centre of two planets

Karthik Sai - 2 years ago

This is not true. The tangents at the ends of the latus rectum of an ellipse have gradient $\pm e$ , where $0 < e < 1$ is the eccentricity. The path in this question has gradient $1$ , so the trajectory is not an ellipse as you suggest.

Mark Hennings - 2 years ago

@Mark Hennings How do you reach this differential equation?. And how solved it??
Thanks in advance

A Former Brilliant Member - 1 year, 1 month ago

From One North Pole to the Other

The answer is 0.813.

1 solution

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