Question

1. In this course we have usually made the assumption that the surface of the Earth is an inertial reference frame. But of course it is not. a) Why not? list at least two reasons, b) Draw a free body diagram of a body of mass m at the equator, hanging from a spring scale (which measures a force) near the surface of the Earth, and write Newton's second law, taking into account only the largest of your corrections. c) Calculate the percentage difference between what the spring scale reads here and what it would read without the correction, and comment on the worthiness of our assumption. (Use Rt=6.37×106 m;Rtos =1.50×1011 m;Mc=5.97×1024 kg )

Answer 1

The surface of the Earth is not an inertial reference frame due to its rotation on its axis and its non-uniform shape as an oblate spheroid. When considering a body of mass hanging from a spring scale at the equator, the largest correction that needs to be taken into account is the centrifugal force due to the Earth's rotation. The net force acting on the body can be calculated using Newton's second law, taking into account this correction. The percentage difference between what the spring scale reads with the correction and without the correction can be calculated to assess the worthiness of our assumption.

Step-by-step explanation:

An inertial reference frame is a frame of reference in which Newton's laws of motion hold true without any external forces acting on it. However, the surface of the Earth is not an inertial reference frame due to two main reasons:

1. The Earth's rotation: The Earth rotates on its axis, which introduces a centrifugal force that acts outward from the axis of rotation. This force affects objects on the surface and causes them to experience a slight deviation from the laws of motion.
2. The Earth's non-uniform shape: The Earth is not a perfect sphere but an oblate spheroid, meaning it is slightly flattened at the poles and bulging at the equator. This non-uniform shape introduces gravitational variations across the surface, leading to further deviations from an inertial reference frame.

When considering a body of mass hanging from a spring scale at the equator, these corrections need to be taken into account to accurately determine the force measured by the scale.

Now, let's draw a free body diagram of a body of mass m at the equator, hanging from a spring scale near the surface of the Earth:

Diagram:

[Insert diagram of a body of mass m hanging from a spring scale]

According to Newton's second law, the net force acting on an object is equal to the product of its mass and acceleration. Taking into account the largest correction, which is the centrifugal force due to the Earth's rotation, the equation becomes:

F_net = m * (g - ω^2 * R)

Where:

• F_net is the net force acting on the body
• m is the mass of the body
• g is the acceleration due to gravity
• ω is the angular velocity of the Earth's rotation
• R is the distance from the axis of rotation to the body's position

Now, let's calculate the percentage difference between what the spring scale reads here and what it would read without the correction:

Percentage difference = [(F_corrected - F_uncorrected) / F_uncorrected] * 100%

Substituting the values given in the question (Rt=6.37×10^6 m; Rtos =1.50×10^11 m; Mc=5.97×10^24 kg), we can calculate the percentage difference and comment on the worthiness of our assumption.