Need help with physics question about velocity!!! Any help would be greatly appreciated!!?

Question

What velocity would a proton need to circle Earth 1200 km above the magnetic equator, where Earth's magnetic field is directed horizontally north and has a magnitude of 4.00x10^-8 T?

Answer 1

Do you remember centripetal acceleration from probably a while ago?

The proton must travel in a circle, therefore it must undergo some centripetal acceleration. The interaction between the moving proton and the Earth's magnetic field will provide this force to move the proton in a circle.

Centripetal force = magnetic force

m v^2 / r = q v B

I hope this helps. If not message me.

Answer 2

m v^2 / r = q v B + mg

at an altitude of 1200 km the gravity is (earth radius is 6400 km)
g = 9.80 * 6400^2/(6400+1200)^2 = 6.94 m/s^2

The weight of the proton is
1.67x10^-27kg

So mg = 1.16x10^-26N

So equation is

m/r * v^2 - qB * v - 1.67E-27 = 0

m/r = 1.67x10^-27kg / 7600000m = 2.197E-34
qB = 1.6x10^-19C * 4x10^-8T = 6.4E-27

So

2.197E-34*V^2 - 6.4E-27*V - 1.16x10^-26 = 0

solve for V

V = 29 125 750.32 m/s

Note: This is almost 10% of the speed of light. So the mass of the proton increases by a factor of 0.4746% which does not affect the result.

Note: the gravity (mg) factor affects the velocity by 1.821103m/s so it could be neglected if you want a rough answer. But usually when computing orbits, precision is important because a small error over a long time becomes a large error.