How to deduct centripetal force from the linear regression slope of a Radius vs Period^2 graph?
I have a graph representing Radius / T(period)squared, and a linear regression equation for the trendline. Question is what does the slope of the linear equation represent? How do i deduct the centripetal force from it?
- 8 months ago
Centripetal Force = 4pi^(2)mr / T^2
What you have is r / T^2. Let that be 'a'
'a' represents how fast radius changes with respect to changing period^2
To deduct Centripetal Force: Fc = 4pi^(2)ma
I actually have no idea if I'm right