Final answer:
The magnitude of the linear momentum (Po) of a 13.4 kg object on the Equator, as measured from the center of the Earth, is approximately 6210.70 kg·m/s.
Step-by-step explanation:
The linear momentum, Po, of an object on the equator can be calculated using the formula Po = m * v, where m is the mass of the object and v is its velocity. In this case, the object's mass is given as 13.4 kg. To calculate the velocity, we need to consider the object's motion due to the rotation of the Earth. The velocity can be calculated as the product of the object's radius from the center of the Earth, R, and the angular velocity of the Earth, ω. The angular velocity of the Earth can be calculated using the formula ω = 2π/T, where T is the period of rotation of the Earth.
The period of rotation of the Earth is approximately 24 hours, which is equivalent to 86400 seconds. Plugging in the values, we get ω ≈ 2π/86400 rad/s. The radius of the Earth, R, is given as 6371 km, which is equivalent to 6371000 m. Plugging in the values, we get v = R * ω ≈ 6371000 * (2π/86400) ≈ 464.24 m/s.
Now, we can calculate the linear momentum, Po, using the formula Po = m * v ≈ 13.4 * 464.24 ≈ 6210.70 kg·m/s. Therefore, the magnitude of the linear momentum, Po, of the object is approximately 6210.70 kg·m/s.