- #1
papazulu
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Neglecting equatorial bulge and using 'g' (9.81m/s^2) as standard before calculations, I wanted to see if I could calculate the change in 'g' before looking it up. Obviously this not a difficult sort of problem but I am finding myself a tad confused about something. It seemed intuitive to me that based on Newton's 3rd law I could find radial accel. at equator A-rad=(465.1^2 m/s) / (6378137 m) = (.033912 m/s^2) then subtract it from 9.81m/s^2.
'g' at equator = (approx.) 9.776 m/s^2
My question is exactly which direction is the acceleration vector responsible for this effect pointing? I've read that there is no 'true' force acting outward in rotational motion. How can something weigh less at the equator if there is no component of force that is in the opposite direction of the radial acceleration?
I'm not doing this for a class so please (if you care to) just go right to the basic explanation of it.
thank you
'g' at equator = (approx.) 9.776 m/s^2
My question is exactly which direction is the acceleration vector responsible for this effect pointing? I've read that there is no 'true' force acting outward in rotational motion. How can something weigh less at the equator if there is no component of force that is in the opposite direction of the radial acceleration?
I'm not doing this for a class so please (if you care to) just go right to the basic explanation of it.
thank you