Consider an Earth-like planet hit by an asteroid. The planet has mass Mp = 7.2 T
ID: 2100421 • Letter: C
Question
Consider an Earth-like planet hit by an asteroid. The planet has mass Mp = 7.2 Times 1023 kg and radius RP = 7.08 Times 106 m, and you may approximate it as a solid ball of uniform density. It rotates on its axis once every T = 16 hr. The asteroid has mass Ma = 7.07 Times 1017 kg and speed va = 35500 m/s (relative to the planets center); its velocity vector points theta = 70 degree below the Eastward horizontal. The impact happens at an equatorial location. The picture below shows the view from above the planet's North pole: First, calculate the planets angular momentum (relative to its spin axis) before the impact. Answer in units of kgm2/s Calculate the asteroids angular momentum relative to the planetary axis. Answer in units of kgm2/s The impact is totally inelastic - the asteroid is stuck in the planet's crust. But thanks to the asteroid's angular momentum, the planet rotates faster after the impact than it did before. By how many seconds has the collision shortened the planetary day? For simplicity, ignore the effect of the asteroids mass on the planet's moment of inertia and assumeExplanation / Answer
Part 1 of 3 The net angular momentum is conserved in this collision, so after the impact the planet (with the asteroid stuck in its crust) has L'p = Lp + La . Focusing on the Northward component of L (the other two components vanish), we have L'p = L p + L a = Ip w + Rp Ma va cos ? . At the same time, L'p = I'p w' , and since we assume unchanged moment of inertia I'p = Ip , it follows that after the impact, the planet rotates at new angular velocity w' =L'p/Ip = w+ (Rp Ma va cos ?)/Ip In other words, Delta w = w' - w = Rp Ma va cos ?)/Ip = 5Ma va cos ?)/2MpRp = 5* 7.07 x 10^17*35500 * Cos 70 / 2 * 7.2 x 10^23*7.08 x10 ^6 = 4.20 * 10^-9 rad/s where I is the body’s moment of inertia about the axis of rotation. Approximating the planet as a solid ball of uniform density, its moment of inertia is Ip =( 2/ 5)M R 2 = ( 2/ 5)7.2 x 10^23 * (7.08 x10 ^6 )^2 = 1.443 × 10^37 kg m2 . Its angular velocity before the impact is w =2Pi / T = 2Pi /57600 = 1.090× 10-4 rad/s , so L=I ×w =I ×w = ( 1.443 × 10^37 kg m2 )×(1.090× 10-4 rad/s ) = 1.57 × 1033 kg m2 /s . Part 2 of 3 Approximating the asteroid as a pointlike particle, its angular momentum is La = R × M a v a where R is the asteroid’s radius vector andMa va is its linear momentum vector. At the moment of impact, both the radiusvector R and the momentum vector Ma vaof the asteroid lie in the planet’s equatorial plane. Consequently, their vector productis perpendicular to the equatorial plane and parallel to the planet axis. Because the horizontal component of the asteroid’s velocity is directed Eastward — the same as the planet’s rotation — the asteroid’s angular momentum La has the same direction as the planet’s angular momentum Lp . In magnitude, La = Rp Ma × horizontal component of va = Rp Ma × va cos ? = (7.08 x10 ^6 m)×( 7.07 x 10^17 kg) (35500 m/s) cos(70 ) = 6.077 × 10^28 kg m2 /s . Part 3 of 3 In terms of the planetary day, this change means w+ Deltaw = 4.20 * 10^-9 rad/s + 1.090× 10-4 rad/s = 1.0900*10^-4 -Delta T = T - T' = 2p/ w - 2p / w' = 2 p(w' - w)/w'w = [2 p Delta w/ (w+ Deltaw)] * ( 1/w) = 2 p (4.20 * 10^-9 rad/s) / 1.0900*10^-4] *(1/ 1.090× 10-4 rad/s) = 2.2 s