Assume that Earth’s magnetic field is every- where perpendicular to the path of a proton and that Earth’s magnetic field has an inten- sity of 4.10 × 10−8 T.

What speed would a proton need to achieve in order to circle Earth 1170.0 km above the magnetic equator?
Answer in units of m/s.

To determine the speed of the proton, we can use the equation for the centripetal force acting on a charged particle moving in a magnetic field:

F = (q * v * B) / r

where:
- F is the centripetal force
- q is the charge of the proton (1.6 × 10^-19 C)
- v is the speed of the proton
- B is the magnitude of the magnetic field (4.10 × 10^-8 T)
- r is the radius of the circular path (the distance of 1170.0 km above the magnetic equator)

The centripetal force is provided by the gravitational force:

F = (m * g)

where:
- m is the mass of the proton (1.67 × 10^-27 kg)
- g is the acceleration due to gravity on Earth (9.8 m/s^2)

Setting the two equations equal to each other, we have:

(q * v * B) / r = (m * g)

Rearranging the equation, we can solve for the speed v:

v = (m * g * r) / (q * B)

Substituting the given values, we have:

v = ((1.67 × 10^-27 kg) * (9.8 m/s^2) * (1170.0 × 10^3 m)) / ((1.6 × 10^-19 C) * (4.10 × 10^-8 T))

Calculating the above expression, we get:

v ≈ 9.47 × 10^6 m/s

Therefore, the speed of the proton would need to be approximately 9.47 × 10^6 m/s to circle Earth 1170.0 km above the magnetic equator.

To find the speed needed for a proton to circle Earth 1170.0 km above the magnetic equator, we can use the equation for the magnetic force experienced by a charged particle moving in a magnetic field.

The magnetic force (F) on a charged particle can be calculated using the equation:

F = q * v * B * sinθ,

where q is the charge of the particle, v is its velocity, B is the magnitude of the magnetic field, and θ is the angle between the velocity and the magnetic field.

In this case, the proton's charge is q = 1.6 × 10^(-19) C (coulombs), the magnitude of the Earth's magnetic field is B = 4.10 × 10^(-8) T (teslas), and the angle θ between the proton's velocity and the magnetic field is 90 degrees.

Since the proton is moving in a circular path, the magnetic force acts as the centripetal force (F_c) required to maintain the circular motion. The centripetal force is given by:

F_c = m * v^2 / r,

where m is the mass of the proton, v is its velocity, and r is the radius of its circular path above the magnetic equator.

Equating the magnetic force and the centripetal force, we have:

q * v * B * sinθ = m * v^2 / r.

Since sinθ = 1 (as the proton's velocity is perpendicular to the magnetic field), we can simplify the equation to:

q * v * B = m * v^2 / r.

Now, solving for the velocity v:

v = (q * B * r / m)^(1/2).

The mass of a proton is m = 1.67 × 10^(-27) kg, and the radius of the proton's circular path above the magnetic equator is given as 1170.0 km = 1.17 × 10^6 m.

Substituting the values, we have:

v = (1.6 × 10^(-19) C * 4.10 × 10^(-8) T * 1.17 × 10^6 m) / (1.67 × 10^(-27) kg)^(1/2).

Calculating this equation will give us the speed needed for the proton to circle Earth at the given altitude above the magnetic equator.