Calculating equatorial velocity

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The discussion revolves around the confusion in calculating Earth's equatorial velocity. The initial calculation using Rω yielded approximately 465 m/s, which aligns with known values. However, a second calculation using the centripetal acceleration formula resulted in an incorrect value of around 8000 m/s. The error stems from misapplying the centripetal acceleration formula, as it assumes a scenario where weight provides the necessary centripetal force for an object in orbit just above the Earth's surface. Additionally, the gravitational acceleration 'g' is not constant and varies, further complicating the calculations.
awygle
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I'm confused about some results I've been getting. I tried calculating the equatorial velocity of Earth by the equation R\varpi=v, and I got ~465 m/s. According to various resources, this is correct. But then I tried to calculate it again, using g=a_{c}=\frac{v^{2}}{R}, and I got ~8000 m/s.

Obviously, the second is wrong. My question is why? My only guess is that maybe g doesn't work as a_{c} in this case due to the normal force from the Earth or something of the sort, but that seems a weak explanation...
 
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It's wrong because the equation is not the correct one to use.
You have calculated the speed an object would have to travel in orbit around the Earth if its path was literally just above the surface. In other words, its centripetal force was provided by its weight.
 
earth has both rectilinear and rotatory motion.but you take a=v^2/r.so your assumption is wrong.
 
hi,friend in your equation how we have take a=g.that is for surface on Earth and although 'g' is not constant at all.
 

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