Suppose a 5.0×10 10 kg meteorite struckthe Earth at the equator with a speed2.8×

Question

Suppose a 5.0×1010 kg meteorite struckthe Earth at the equator with a speed2.8×104 m/s, as shown in the figure andremained stuck. The velocity vector of the meteorite makes a 45degree angle with the radius of the earth. By what factor would this affect the rotational frequency ofthe Earth (1rev/ day)? Delta omega/ omega = ? Thanks for your help! Suppose a 5.0×1010 kg meteorite struckthe Earth at the equator with a speed2.8×104 m/s, as shown in the figure andremained stuck. The velocity vector of the meteorite makes a 45degree angle with the radius of the earth. By what factor would this affect the rotational frequency ofthe Earth (1rev/ day)? Delta omega/ omega = ? Thanks for your help!

Explanation / Answer

if we assume the earth is a uniform sphere (its not... butwe'll approximate it as such) then the angular momentum of theearth is... .        L for Earth = I = (2/5) M R2 * = 0.4* 6 x 1024 * (6.38 x 106)2 * (2 radians / 86400 sec ) = .             =    71.043 x 1032 . The angular momentum of the meteoriteis     L = m v r sin = 5.0 x1010 * 2.8 x 104 * 6.38 x 106 *sin45 = .             =   63.159 x 1020 . This angular momentum is imparted to the Earth, and the ratioof the added angular momentum to the Earth's original is .         63.159 x1020 / 71.043 x 1032   =     0.889 x10-12        . Since the angular speed is directly proportional to theangular momentum of the earth, this is also the ratio of /     roughly one part in abillion. . This angular momentum is imparted to the Earth, and the ratioof the added angular momentum to the Earth's original is .         63.159 x1020 / 71.043 x 1032   =     0.889 x10-12        . Since the angular speed is directly proportional to theangular momentum of the earth, this is also the ratio of /     roughly one part in abillion.

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