Explanation Given: Radius of the asteroid, r = 123 m Mass of the asteroid, M = 2.25 × 10 10 kg Speed of rotation of the asteroid, ω = 4 rev day = 4 × 2 π 24 × 60 × 60 rad s = 2.9 × 10 − 4 rad s Force applied by the spaceship, F = 285 N Angle rotated by the spaceship, θ = 5 ° Formula used: The torque acting on the asteroid because of the force from the spaceship’s engine is equal to the rate of change of angular momentum of the asteroid i.e. τ = Δ L Δ t --- (i), where ΔL is the change in angular momentum of the asteroid in time Δt If we know the torque and the change in angular momentum between the initial position and when the spaceship has rotated the axis of rotation by 5°, we can find out the time taken for this process. The torque because of the force from the spaceship is given by, τ = F × r --- (ii) The change in the angular momentum of the asteroid is given by Δ L = Δ L x 2 + Δ L y 2 , where L x is the change in angular momentum along the x-axis and L y is the change in angular momentum across the y-axis. If the initial direction of the angular momentum is taken as the y-axis, the change in angular momentum across this direction is, Δ L y = L – L c o s θ , where L is the initial angular momentum of the asteroid L = I ω , where I is the moment of inertia of the asteroid about its axis of rotation or L = ( 2 5 ) M r 2 ω Calculation: Therefore, Δ L y = L ( 1 – c o s θ ) = ( 2 5 ) M r 2 ω ( 1 – c o s 5 ° ) = ( 2 5 ) × 2.25 × 10 10 kg × ( 123 m ) 2 × 2.9 × 10 − 4 rad s × ( 1 – 0

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ISBN: 9780137453276

Author: GIANCOLI

Publisher: SAVVAS L

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Chapter 8, Problem 93GP

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The time taken by the spaceship to rotate the asteroid’s axis of rotation through an angle of 5°

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Chapter 8, Problem 92GP

Chapter 8, Problem 94GP

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