Final answer:
The student's question pertains to the necessary rotational velocity of Earth for weights at the equator to be 3/5 of their original value, which is solved using the centripetal acceleration and force of gravity equations to find the velocity that provides the required centrifugal force.
Step-by-step explanation:
The student is asking about the necessary rotational velocity of Earth for the weight of objects at the equator to become 3/5 of their original value due to the centrifugal effect of Earth's rotation. To solve for this, we can use the formula for centripetal acceleration, which at the original speed is provided by both Earth's gravity and its rotation.
The original centripetal acceleration due to Earth's rotation at the equator is v^2/r, where v is the linear velocity due to Earth's rotation and r is the radius of the Earth. According to the question, the weight (which is the force of gravity on a mass) at the equator should be 3/5 of the original, meaning the centrifugal force due to the Earth's rotation must provide the remaining 2/5.
The formula for gravitational force (weight) is F_g = mg, where m is the mass and g is the acceleration due to gravity. To drop to 3/5 of this value, the new force due to the Earth's gravity is (3/5)mg. The centrifugal force, F_c, provided by Earth's rotation is mv^2/r. Therefore, at the new velocity, we would want F_c to be such that: (3/5)mg + F_c = mg, simplifying to F_c = (2/5)mg.
From the above equation, substituting F_c as mv^2/r and solving for v, we can determine the necessary velocity for the Earth's rotation at the equator to achieve the desired weight effect.