Electromagnetism and earth question

[lolzwhut?]

Homework Statement

Assume that Earth’s magnetic field is everywhere perpendicular to the path of a proton
and that Earth’s magnetic field has an intensity of 4.07 × 10−8 T.

What speed would a proton need to achieve
in order to circle Earth 1660.0 km above the
magnetic equator?
Answer in units of m/s.

The Attempt at a Solution

I was told F1 = eVB, F2 = mV^2/(1.600x10^6 + R), F3 = GmM/(1.600x10^6 + R)^2. And set ±F1 + F2 = F3, where the sign is to be determined depending on the direction of the proton.

I'm completely lost, and have no idea what to do. This is the only problem left for me :/ 1 answer left or I fail the assignment :(

 

[rl.bhat]

Show your calculations.