katiekat asked in Science & MathematicsPhysics · 9 years ago

What speed would a proton need to achieve in order to circle the earth 1000.0 km above the magnetic equator? Assume that Earth's magnetic field is everywhere perpendicular to the path of the proton and that the Earth's magnetic field has an intensity of 4.00 x 10^ -8 T. (Hint: the magnetic force exterted on the proton is equal to the centripetal force, and the speed needed by the proton is its tangentical speed. Remember that the radius of the circular orbit should also include the radius of the earth. Ignore relativistic effects.)

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• Logan
Lv 5
9 years ago

You should know that centripedal force is F = mv^2/r (mass times centripedal acceleration a = v^2/r). For your problem, m is the mass of the proton, v is the proton's velocity (the unknown you need to solve for), and r is 10^6 m plus the Earth's radius.

But you need the left-side of the equation. How does magnetic force relate to a magnetic field strength? Use the Right-Hand Rule: make your index finger, middle finger, and thumb all perpendicular to each other. Your index finger is the direction of the magnetic field B. Your thumb is the direction of the proton's velocity v. And your middle finger is the direction of the magnetic force F. For your situation, where the charge is moving perpendicular to B, the equation for force is F = qvB. Thus

qvB = mv^2/r

Solve for v...

v = mv^2 / qBr

v / v^2 = m / qBr

1/v = m / qBr

v = qBr / m

You have

q = charge of a proton = 1.6 x 10^-19 C

B = 4 x 10^-8 T

r = r_earth + 10^6 m = 7.37 x 10^6 m

m = mass of a proton = 1.67 x 10^-27 kg

Plugging everything in...

v = (1.6 x 10^-19 C)(4 x 10^-8 T)(7.37 x 10^6 m) / (1.67 x 10^-27 kg)

v = 2.82 x 10^7 m/s

That's almost 9% of the speed of light!