The mass of this asteroid, in terms of the earth's mass m = 0.092 M.
Explanation:
The length of the day T = 2π rad / ω.
where ω = Earth's angular rotation rate
Conservation of angular momentum applied to collision between the Earth and the asteroid gives:
          [tex]\frac{2}{5} MR^{2} w_{1} = (mR^{2} + \frac{2}{5} MR^{2} )w_{2}[/tex] and
          [tex]m = \frac{2}{5} M(\frac{w_{1} -w_{2} }{w2})[/tex]
          [tex]T_{2}[/tex] [tex]= 1.23 T_{1}[/tex] which gives
          [tex]\frac{1}{w_{2} } = \frac{1.23}{w1}[/tex]
     and  [tex]w_{1} = 1.23 w_{2}[/tex]
          [tex]\frac{w_{1} - w_{2} }{w_{2} } = 0.23[/tex]
         [tex]m = \frac{2}{5} (0.23) M = 0.092 M[/tex]
 The mass of this asteroid, in terms of the earth's mass m = 0.092 M.  Â
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The mass of the asteroid of the Earth will be 0.0092M.
From the information given, the length of the day is 2Ï€rad/w. Then, the conversation of angular momentum will be:
m = 2/5M(w1 - w2)/w2
T2 = 1.23T.
Therefore,
1/w2 = 1.23/w1
Cross multiply
(1 × w1) = (w2 × 1.23)
w1 = 1.23w2
(w1 - w2) / w2 = 0.23
m = 2/5(0.23) M
= 0.092M
In conclusion, the mass is 0.092M.
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