Calculate Proton Speed for Earth Magnetic Equator - Magnetism Question 2

[tayi]

Homework Statement

What speed would a proton need to achieve
in order to circle Earth 1670 km above the
magnetic equator, where the Earth’s mag-
netic field is directed on a line between mag-
netic north and south and has an intensity of
4.1 × 10−8 T?
The mass of a proton is 1.673 × 10−27 kg.
Answer in units of m/s.

Homework Equations

f1=qvbsin(theta)
f2=mv^2/(d+R)
f3=Gm(proton)m(earth)/(d+R)^2 where d is 1670 km, and R is Earth distance

The Attempt at a Solution

I tried doing f1+f2=f3, b ut I am not sure if this is right and i don't know how to solve for v

 

[rock.freak667]

I think you'd want to equate f₂ to f₃ as the gravitational force of attraction will provide the centripetal force.