What is the relationship between orbital altitude and orbital speed?

Question

I know that if an object has a higher orbit, it will orbit at a lower speed. However, I don't know what the exact relationship is. Is the decrease in speed linear, or quadratic, or something else?

Also, as a bonus question: higher orbits not only have the objects moving slower, but they also have a larger distance to cover. What is the relationship between orbital altitude and the time to complete a full orbit? Is it linear, quadratic, cubic...?

(For the purposes of this question, assume altitude means from the center of the Earth).

Answer 1

To orbit in a circle you need a certain amount of acceleration $$a_c=\frac{v^2}r$$ where $a_c$ stands for centripetal acceleration. Or in other words to orbit with velocity $v$ at a radius $r$ you need this specific acceleration $a_c$.

The gravitational pull between two bodies is given by $$F=G\frac{Mm}{r^2}$$ With $M$ the mass of the big body and $m$ the mass of the small body. If you plug in $F=ma$ you will get that any orbiting body accelerates with $$a=\frac{GM}{r^2}$$ If you want this orbit to be circular you can equate this acceleration to the centripetal acceleration. This leaves $v$ as the only independent variable. Which equation do you get when you do this?

For the bonus: in circular orbits you can use $v=\frac{2\pi}T r$ in your derived equation to relate $T$ with $r$. What do you get?

If you want to know where this formula for the period comes from you can use $x(t)=r\cos(\frac{2\pi}T t),\ y(t)=r\sin(\frac{2\pi}T t)$ to get $v_x(t)=-r\frac{2\pi}T\sin(\frac{2\pi}T t),\ v_y(t)=r\frac{2\pi}T\cos(\frac{2\pi}T t)$. Then use $v=\sqrt{v_x^2+v_y^2}$.