- #1
AdrianGriff
- 21
- 0
In orbits it is said that ##\vec a \cdot \vec a' = |a||a'|##
How is this possible? Two vectors multiply to get scalars, and yet we cannot do the dot product literally because we do not know either of the components of ##\vec a## or ##\vec a'##.
Nor does the Angle Between Vectors Formula work because
$$ \vec a \cdot \vec a' = |a||a'| cos (\theta)$$
And if the acceleration vector is towards the center of the orbit, and jerk, or ##a'## is orthogonal to ##a## and tangent to the circle, opposite of ##v##, then ##\theta = \pi/2##. And if that is the case, then ##cos (\pi/2) = 0##, and as such, ##\vec a \cdot \vec a' ≠ |a||a'|## but rather ##\vec a \cdot \vec a' = 0##
So, how is it possible that ##\vec a \cdot \vec a' = |a||a'|##?
Thank you for your help!
- Adrian
How is this possible? Two vectors multiply to get scalars, and yet we cannot do the dot product literally because we do not know either of the components of ##\vec a## or ##\vec a'##.
Nor does the Angle Between Vectors Formula work because
$$ \vec a \cdot \vec a' = |a||a'| cos (\theta)$$
And if the acceleration vector is towards the center of the orbit, and jerk, or ##a'## is orthogonal to ##a## and tangent to the circle, opposite of ##v##, then ##\theta = \pi/2##. And if that is the case, then ##cos (\pi/2) = 0##, and as such, ##\vec a \cdot \vec a' ≠ |a||a'|## but rather ##\vec a \cdot \vec a' = 0##
So, how is it possible that ##\vec a \cdot \vec a' = |a||a'|##?
Thank you for your help!
- Adrian