Archived May, 2026.

Vector Fields

User 006

Hi Mr. Mark,
Today we discussed launching an object horizontally on path \mathbf{r}(t) = \langle 10t, 64 - 16t^2 \rangle, where 64 represents the height in feet above the ground. In this problem, we discussed gravity as being a constant. However, we know that due to the curvature of the Earth, this isn't exactly the case. Though the effect will be incredibly small, I am wondering how we can solve this equation when accounting for the curvature of the Earth?

mark

You can use a set of two or three differential equations together with initial conditions. In 2D, the equations look like so:

\begin{align} x''(t) &= -G \frac{x(t)}{(x(t)^2 + y(t)^2)^{3/2}} \\ y''(t) &= -G \frac{y(t)}{(x(t)^2 + y(t)^2)^{3/2}}. \end{align}

I used equations like that to generate this model of planetary motion:

Of course, that's really counter productive when you're modelling motion right here on earth. It's critically important, though, for studying objects moving around in the solar system!