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?
Archived May, 2026.
Vector Fields
User 006
❤️ 1
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!