Problem Statement
Find the angular velocity of rotation of the Earth such that the effective gravitational acceleration at the equator is zero.
Given Information
- $R_E = 6.37$ — radius
- $g = 9.8 \text{m/s}^2$ — gravitational acceleration
Physical Concepts & Formulas
These problems apply the rotational analogue of Newton’s laws about a chosen axis, using moment of inertia and torque.
- $M = I\beta$ — equation of rotational dynamics
- $I = \sum m_i r_i^2$ — moment of inertia
- $E_k = \tfrac{1}{2}I\omega^2$ — rotational kinetic energy
- $L = I\omega$ — angular momentum of a rigid body
Step-by-Step Solution
Step 1 — Identify the governing principle: We begin by recognising which physical law controls the situation and why it is the correct starting point for this problem.
$$Effective acceleration: g_eff = g_0 – \omega ^{2R}_E (centrifugal reduction at equator)$$
Step 2 — Set up the relevant equations: Next we write down the equations that follow from that principle, introducing the symbols we will carry through the algebra.
$$Set g_eff = 0: \omega = √(g/R_E)$$
Step 3 — Apply the given conditions: We now substitute the specific conditions and constraints given in the problem so the equations describe this particular situation.
$$\omega = √(9.8/6.37 \cdot 10^{6} = 1.24 \cdot 10^{-3} \text{rad/s})$$
Step 4 — Solve for the required quantity: With the equations specialised, we isolate and solve for the unknown the problem asks us to find.
$$Period: T = 2 \pi / \omega ≈ 5070\,\text{s} ≈ 1.4\,\text{h}$$
Worked Calculation
$$\omega = √(g/R_E) ≈ 1.24 \cdot 10^{-3} \text{rad/s}, T ≈ 1.4\,\text{h}$$
Answer
$$\boxed{\omega = √(g/R_E) ≈ 1}$$
This is the quantity the problem asked for, expressed in terms of the given data: $\omega = √(g/R_E) ≈ 1$.
Physical Interpretation
Earth would need to spin $17 \cdot faster$ for equatorial weightlessness. At actual rotation $( \omega)$ = $7.27$ * 10^{$-5}$ rad/s), centrifugal reduction at equator is only ~0.34%. The magnitude of the answer is consistent with everyday physical experience for this class of problem in Irodov’s Part 1 — the result shows how the answer scales with the given quantities. If we doubled the dominant input, the boxed formula tells us exactly how the output would respond, and that scaling is the key physical insight this problem trains. Comparing the answer with the appropriate limiting cases (very small or very large values of the dominant parameter) recovers the familiar Newtonian or intuitive expectation, which is a useful sanity check.
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