Effect of Asteroid Impact on Earth's Rotation
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That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
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Exercise:
An asteroid of mass mO travelling at a speed of vO relative to the Earth hits the Earth at the equator tangentially and in the direction of Earth's rotation. Asing the collision is perfectly inelastic use conservation of angular momentum to estimate the percent change in the angular speed of the Earth as a result of the collision.
Solution:
Angular momentum is conserved in the Earth--asteroid system. Asing the asteroid becomes embedded in the Earth at the surface both will share the same angular velocity after the collision. Modelling the Earth as a uniform solid sphere and the asteroid as a po mass at the surface we have I_EarthIndex fracm_EarthIndex r_EarthIndex^ sscIA sscmA r_EarthIndex^ Conservation of angular momentum gives: L_ L_ I_EarthIndexomega_EarthIndex + I_aomega_a biglI_EarthIndex + I_abigromega Solving for the final angular velocity omega fracI_EarthIndexomega_EarthIndex + I_aomega_a I_EarthIndex + I_a and the change in angular velocity omega - omega_EarthIndex fracI_EarthIndexomega_EarthIndex + I_aomega_a I_EarthIndex + I_a - omega_EarthIndex fracI_aomega_a - omega_EarthIndex I_EarthIndex + I_a we get for the change as a percentage: eta fracomega - omega_EarthIndexomega_EarthIndex fracI_aomega_a - omega_EarthIndexI_EarthIndex + I_aomega_EarthIndex And since I_a ll I_EarthIndex and omega_a gg omega_EarthIndex it follows that eta &approx fracI_aomega_aI_EarthIndexomega_EarthIndex &approx fracm_a r_EarthIndex^ dfracv_ar_EarthIndexdfracm_EarthIndexr_EarthIndex^ omega_EarthIndex &approx fracm_a v_am_EarthIndexr_EarthIndexomega_EarthIndex &approx numpr. which is negligibly small.
An asteroid of mass mO travelling at a speed of vO relative to the Earth hits the Earth at the equator tangentially and in the direction of Earth's rotation. Asing the collision is perfectly inelastic use conservation of angular momentum to estimate the percent change in the angular speed of the Earth as a result of the collision.
Solution:
Angular momentum is conserved in the Earth--asteroid system. Asing the asteroid becomes embedded in the Earth at the surface both will share the same angular velocity after the collision. Modelling the Earth as a uniform solid sphere and the asteroid as a po mass at the surface we have I_EarthIndex fracm_EarthIndex r_EarthIndex^ sscIA sscmA r_EarthIndex^ Conservation of angular momentum gives: L_ L_ I_EarthIndexomega_EarthIndex + I_aomega_a biglI_EarthIndex + I_abigromega Solving for the final angular velocity omega fracI_EarthIndexomega_EarthIndex + I_aomega_a I_EarthIndex + I_a and the change in angular velocity omega - omega_EarthIndex fracI_EarthIndexomega_EarthIndex + I_aomega_a I_EarthIndex + I_a - omega_EarthIndex fracI_aomega_a - omega_EarthIndex I_EarthIndex + I_a we get for the change as a percentage: eta fracomega - omega_EarthIndexomega_EarthIndex fracI_aomega_a - omega_EarthIndexI_EarthIndex + I_aomega_EarthIndex And since I_a ll I_EarthIndex and omega_a gg omega_EarthIndex it follows that eta &approx fracI_aomega_aI_EarthIndexomega_EarthIndex &approx fracm_a r_EarthIndex^ dfracv_ar_EarthIndexdfracm_EarthIndexr_EarthIndex^ omega_EarthIndex &approx fracm_a v_am_EarthIndexr_EarthIndexomega_EarthIndex &approx numpr. which is negligibly small.
Meta Information
Exercise:
An asteroid of mass mO travelling at a speed of vO relative to the Earth hits the Earth at the equator tangentially and in the direction of Earth's rotation. Asing the collision is perfectly inelastic use conservation of angular momentum to estimate the percent change in the angular speed of the Earth as a result of the collision.
Solution:
Angular momentum is conserved in the Earth--asteroid system. Asing the asteroid becomes embedded in the Earth at the surface both will share the same angular velocity after the collision. Modelling the Earth as a uniform solid sphere and the asteroid as a po mass at the surface we have I_EarthIndex fracm_EarthIndex r_EarthIndex^ sscIA sscmA r_EarthIndex^ Conservation of angular momentum gives: L_ L_ I_EarthIndexomega_EarthIndex + I_aomega_a biglI_EarthIndex + I_abigromega Solving for the final angular velocity omega fracI_EarthIndexomega_EarthIndex + I_aomega_a I_EarthIndex + I_a and the change in angular velocity omega - omega_EarthIndex fracI_EarthIndexomega_EarthIndex + I_aomega_a I_EarthIndex + I_a - omega_EarthIndex fracI_aomega_a - omega_EarthIndex I_EarthIndex + I_a we get for the change as a percentage: eta fracomega - omega_EarthIndexomega_EarthIndex fracI_aomega_a - omega_EarthIndexI_EarthIndex + I_aomega_EarthIndex And since I_a ll I_EarthIndex and omega_a gg omega_EarthIndex it follows that eta &approx fracI_aomega_aI_EarthIndexomega_EarthIndex &approx fracm_a r_EarthIndex^ dfracv_ar_EarthIndexdfracm_EarthIndexr_EarthIndex^ omega_EarthIndex &approx fracm_a v_am_EarthIndexr_EarthIndexomega_EarthIndex &approx numpr. which is negligibly small.
An asteroid of mass mO travelling at a speed of vO relative to the Earth hits the Earth at the equator tangentially and in the direction of Earth's rotation. Asing the collision is perfectly inelastic use conservation of angular momentum to estimate the percent change in the angular speed of the Earth as a result of the collision.
Solution:
Angular momentum is conserved in the Earth--asteroid system. Asing the asteroid becomes embedded in the Earth at the surface both will share the same angular velocity after the collision. Modelling the Earth as a uniform solid sphere and the asteroid as a po mass at the surface we have I_EarthIndex fracm_EarthIndex r_EarthIndex^ sscIA sscmA r_EarthIndex^ Conservation of angular momentum gives: L_ L_ I_EarthIndexomega_EarthIndex + I_aomega_a biglI_EarthIndex + I_abigromega Solving for the final angular velocity omega fracI_EarthIndexomega_EarthIndex + I_aomega_a I_EarthIndex + I_a and the change in angular velocity omega - omega_EarthIndex fracI_EarthIndexomega_EarthIndex + I_aomega_a I_EarthIndex + I_a - omega_EarthIndex fracI_aomega_a - omega_EarthIndex I_EarthIndex + I_a we get for the change as a percentage: eta fracomega - omega_EarthIndexomega_EarthIndex fracI_aomega_a - omega_EarthIndexI_EarthIndex + I_aomega_EarthIndex And since I_a ll I_EarthIndex and omega_a gg omega_EarthIndex it follows that eta &approx fracI_aomega_aI_EarthIndexomega_EarthIndex &approx fracm_a r_EarthIndex^ dfracv_ar_EarthIndexdfracm_EarthIndexr_EarthIndex^ omega_EarthIndex &approx fracm_a v_am_EarthIndexr_EarthIndexomega_EarthIndex &approx numpr. which is negligibly small.
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Drehimpuls 2 by uz
Asked Quantity:
Verhältnis / Anteil \(\eta\)
in
Prozentsatz \(\rm \eta\)
Physical Quantity
Unit
Prozentsatz (\(\rm \eta\))
Base?
SI?
Metric?
Coherent?
Imperial?

