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Angular momentum of asteroid

  • Thread starter lizzyb
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  • #1
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A planet is hit by an asteroid whose mass is M, velocity v, and it's velocity vector is 22 degrees below the Eastward horizontal. The planet has a radius R.

I used the equation [tex]l = r m v sin \phi[/tex] but the answer was wrong. Am I not doing this correctly?
 

Answers and Replies

  • #2
OlderDan
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lizzyb said:
A planet is hit by an asteroid whose mass is M, velocity v, and it's velocity vector is 22 degrees below the Eastward horizontal. The planet has a radius R.

I used the equation [tex]l = r m v sin \phi[/tex] but the answer was wrong. Am I not doing this correctly?
You have not told us the question, or anything we need to know about the planet. Please post the full statement of the problem.
 
  • #3
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ok - I didn't think the values regarding the planet were pertinent.

A planet has a mass [tex]M_p[/tex] and radius [tex]R_p[/tex] and we may approximate it as solid ball of uniform density. It rotates on its axis once every T (in hours but its easy to convert). The asteriod has a mass [tex]M_a[/tex] and a speed [tex]v_a[/tex] (relative to the planet's center); its velocity vector points [tex]\theta[/tex] below the Eastward horizontal. The impact happens at an equatorial location.

We first calculated the angular momentum of the planet before the impact - I got that right.

the next question regarding the same description is "Calculate the asteroid's angular momentum relative to the planetary axis".

So I just did [tex]l = r m v \sin \theta = R_p M_a v_a \sin \theta[/tex] but this was wrong.
 
  • #4
Doc Al
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Where did that [tex]\sin \theta[/tex] come from?
 
  • #5
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It's in my book; "The direction of the angular momentum vector in [a figure] is parallel to the z axis, in the direction of increasing z. The magnitude of this vector is given by [tex]l = r m v \sin \theta[/tex]."
 
  • #6
Doc Al
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lizzyb said:
It's in my book; "The direction of the angular momentum vector in [a figure] is parallel to the z axis, in the direction of increasing z. The magnitude of this vector is given by [tex]l = r m v \sin \theta[/tex]."
That angle theta is not the angle theta in your problem! That angle theta is the angle between the linear momentum (mv) and the radius vector (r). But in your problem, theta is the angle between linear momentum and the perpendicular to the radius (east is tangent to the earth's surface).
 
  • #7
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so I should use [tex]90 - \theta[/tex]?
 
  • #8
Doc Al
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Yep. ......
 
  • #9
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thanks!! :-)
 

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