radius of the earth = 6.37 × 10^6

what is the speed of a 90.6 kg person at the equator?
I know that F=m(v^2/r) but I don't know the force. Is there another equation that I'm missing or I'm using the wrong one?
Thanks

They are not asking for the force. They want the speed due to the rotation of the earth. That equals R*w.

R is the earth radius and w is its rotation rate in radians per second.

How would I know W if that wasn't given?

So I figured this out myself a=v^2/r

which is also equal to (4pi^2*r^2)/T^2
and T is 86400 secs which is how long it takes for a day to pass

To find the speed of a person at the equator, you can consider the centripetal force acting on the person due to the rotation of the Earth. The equation you mentioned, F = m(v^2/r), is indeed applicable in this situation.

However, instead of finding the force directly, you can use the fact that the force of gravity acting on the person provides the necessary centripetal force. This is because the person remains on the Earth's surface due to the gravitational force holding them in place as they rotate with the Earth.

The equation for gravitational force is given by F = mg, where m is the mass of the person and g is the acceleration due to gravity. In this case, g can be approximated as 9.8 m/s^2.

At the equator, the person is moving in a circular path with a radius equal to the radius of the Earth, which is 6.37 × 10^6 meters. Therefore, r in this equation is the radius of the Earth.

Setting these forces equal to each other, we have:

mg = m(v^2/r)

Now, we can solve for v:

v^2 = rg

v = √(rg)

Substituting the given values:

v = √((6.37 × 10^6)(9.8))

Calculating this expression would give you the speed of a 90.6 kg person at the equator.