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Unformatted text preview: homework 27 BAUTISTA, ALDO Due: Apr 5 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 3. 10 points. Consider an Earthlike planet hit by an asteroid. The planet has mass M p = 6 . 35 10 23 kg and radius R p = 7060000 m, and you may ap proximate it as a solid ball of uniform density. It rotates on its axis once every T = 39 hr. The asteroid has mass M a = 3 . 09 10 17 kg and speed v a = 35700 m / s (relative to the planets center); its velocity vector points = 74 below the Eastward horizontal. The impact happens at an equatorial location. The picture below shows the view from above the planets North pole: v m R First, calculate the planets angular mo mentum (relative to its spin axis) before the impact. Correct answer: 5 . 66573 10 32 kg m 2 / s (tol erance 1 %). Explanation: Basic Concept: A rigid body rotating around a fixed axis has angular momentum L = I where I is the bodys moment of inertia about the axis of rotation. Approximating the planet as a solid ball of uniform density, its moment of inertia is I p = 2 5 M R 2 = 1 . 26603 10 37 kg m 2 . Its angular velocity before the impact is = 2 T = 4 . 4752 10 5 rad / s , so L = I = I = (1 . 26603 10 37 kg m 2 ) (4 . 4752 10 5 rad / s) = 5 . 66573 10 32 kg m 2 / s . Question 2 Part 2 of 3. 10 points. Calculate the asteroids angular momen tum relative to the planetary axis. Correct answer: 2 . 14669 10 28 kg m 2 / s (tol erance 1 %). Explanation: Approximating the asteroid as a pointlike particle, its angular momentum is ~ L a = ~ R M a ~v a where ~ R is the asteroids radius vector and M a ~v a is its linear momentum vector. At the moment of impact, both the radius vector ~ R and the momentum vector M a ~v a of the asteroid lie in the planets equatorial plane. Consequently, their vector product is perpendicular to the equatorial plane and parallel to the planet axis. Because the hori zontal component of the asteroids velocity is directed Eastward the same as the planets rotation the asteroids angular momentum ~ L a has the same direction as the planets an gular momentum ~ L p . In magnitude, L a = R p M a horizontal component of v a = R p M a v a cos = (7060000 m) (3 . 09 10 17 kg) (35700 m / s) cos(74 ) = 2 . 14669 10 28 kg m 2 / s ....
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