Recent content by aidos

I didn't understand how to figure out the orbital speed & distance, but I found them in the back of the book so that made it easy to plug it into the Ac formula. Thanks for your time!

OOOPS, my bad…Thanks so much! 6370 km = 6370000 m so V = (2PI(6370000m))/86400s = 463.24 m/s (463.24)^2/6370000 = .0337 .0337/9.81 = .003434 which I can label as 3.43 x (10 to the -3)g which is REAL close to the 3.44 x (10 to the -3)g in the back of the text. What formula do I use...

Ok, so V = (2PI(6370))/86400(secs in a year) = .46 m/s Ac = (.46)^2/6370 = 3.32 x 10 to the -5 m/s^2? The back of the book has 3.44 x 10 to the -3g. Help..

I’m having trouble with ch3 #73 from Tipler/Mosca Physics for scientists and engineers 5th edition. I’ve tried a few things and I keep getting stuck. Am I taking the right steps? An object resting on the equator has an acceleration toward the center of the Earth due to the earth’s...