The Earth has an angular speed of 7.272·10 -5 rad/s in its rotation. Find the ne

Question

The Earth has an angular speed of 7.272·10-5 rad/s in its rotation. Find the new angular speed if an asteroid (m = 2.07·1022 kg) hits the Earth while traveling at a speed of 4.79·103 m/s (assume the asteroid is a point mass compared to the radius of the Earth) in each of the following cases:

a) The asteroid hits the Earth dead center.

b) The asteroid hits the Earth nearly tangentially in the direction of Earth's rotation.

c) The asteroid hits the Earth nearly tangentially in the direction opposite of Earth's rotation.

Explanation / Answer

conservation of angular momentum ;

inintial momentum = final momentum ;

( 2 / 5 ) MR^2 * ( 7.272 e -5 ) + m * (4.79 e 3 ) R = ( I1 + I2 )

2 / 5 * ( 6 e 24*6 e 6 * 6e6 * 7.272 e -5 ) + (2.07 e 22* 4.79 e 3* 6 e 6 ) = [ 2 /5 MR^2 + mR^2 ] ;

= ( 6.88 e 33 ) / [ 2/5 MR^2 + mR^2 ]

= 6.88 e 33/( (2*5.98 e 24 * 6e6*6e6*/5 + 2.07 e 22 * 6e6*6e6)

= ( 6.383 e 33 )/(8.68 e 37)

W = 7.35 e -5 rad /s

-------------------------------------------

( b )

( 2 / 5 ) MR^2 * ( 7.272 e -5 ) - m * (4.79 e 3 ) R = ( I1 + I2 )

(2/5 * 5.98 e 24 * 6e6 * 6e6) - ( 2.07 e22 * 4.79 e 3 *6e6 ) = [ 2/5 MR^2 + mR^2 ] W

= [ 8.61 e 37 /8.68e 37

or = 0.99 rad/s

-----------------------------------
C>

( 2 / 5 ) MR^2 * ( 7.272 e -5 ) _ m * (4.79 e 3 ) R = ( I1 + I2 )

(2/5 * 5.98 e 24 * 6e6 * 6e6) + (2.07 e22 * 4.79 e 3 *6e6) = [ 2/5 MR^2 + mR^2 ] W

= 8.61 e 37 /8.68e 37

or = 0.99 rad/s

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