Calculating Speed at Equator: 91.3kg Person

[AdnamaLeigh]

I thought I set up the equations correctly, but the speed I got seemed too fast.

An early objection to the idea that the Earth is spinning on its axis was that Earth would turn so fast at the equator that people would be thrown into space. Show the error in this logic by calculating the speed of a 91.3kg person at the equator.

I set the equations for centripetal force and gravitational force equal to each other:

(m2v^2)/r = (Gm1m2)/(r^2)

I set m1 = Earth's mass and m2 = person's mass

r was given: 6.37e6m
mass of earth: 5.98e24kg
G = 6.67259e-11

I solved for velocity and got 7914.584m/s and that seems too fast. I'm worried because they also provided the moon's radius in the givens, I hope I wasn't supposed to use that because I can't find any use for that.

 

[Janus]

Your answer is correct for the speed you would need to exceed in order to be thrown into space(or to be more precise, into Earth orbit). Now compare it to the actual speed at the equator.

 

[Tide]

I'm not sure what you're trying to calculate but it appears you calculated how fast a person near the equator would have to travel in order to "be in orbit" or to be weightless. The speed you calculated is correct. Did you notice how the mass of the person cancels from both sides of the equation?

The question is a little unclear and it appears to ask how fast a person standing at the equator moves -- due to the rotation of the Earth. In relation to the context given it would further seem you would have to compare that number with the one you calculated above.