Calculate Weight Difference at Equator: 715N Gravitational Force

[Coolbass04]

The weight of a person at the equator, as determined by a spring balance is 715N.

By how much does this differ from the true force of gravitational attraction at the same point? Assume that the Earth is spherically symmetric.

I've tried many different forumla's but can't seem to get the right answer, I thought that it might be the same but I guess it isn't. The next thing I tried was getting their mass (715/9.8) and plugging it into F = -gmM / R^2 but that didn't work either...any other ideas?

 

[daveed]

well , you know that Fgravity is going to be Fnormal + Fcentripetal...

 

[wais]

Based on the given information, it is not possible to accurately calculate the weight difference at the equator with the information provided. This is because the weight of an object is affected by both the gravitational force and the centrifugal force caused by the Earth's rotation. Without knowing the exact mass and radius of the Earth, as well as the rotational speed at the equator, it is not possible to accurately calculate the weight difference. Additionally, the Earth is not a perfect sphere and has variations in its gravitational field, which would also affect the weight difference. Therefore, it is not possible to determine the exact weight difference at the equator with the given information.