- #36
mfb
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The asteroid is a point mass: "Assume that the asteroid is very small compared to the earth" - sure it is completely unrealistic, but we can ignore that issue.
Such a massive asteroid (Mars just has 10% of the mass of earth, our "asteroid" is larger!) would change the center of rotation of earth. This makes the new moment of inertia a bit smaller than calculated before.
Also, a day 28% longer than before does not correspond to a new angular velocity 28% smaller than before.
Putting all together, I get about 12.6% of the mass of Earth for the asteroid.
18.4% with the 28%-mistake, 11.2% with the wrong center of rotation and 15.56% with both at the same time.By the way, you do not need any data about Earth (apart from the constant density) - the result is true independent of its size, mass or length of a day.
Such a massive asteroid (Mars just has 10% of the mass of earth, our "asteroid" is larger!) would change the center of rotation of earth. This makes the new moment of inertia a bit smaller than calculated before.
Also, a day 28% longer than before does not correspond to a new angular velocity 28% smaller than before.
Putting all together, I get about 12.6% of the mass of Earth for the asteroid.
18.4% with the 28%-mistake, 11.2% with the wrong center of rotation and 15.56% with both at the same time.By the way, you do not need any data about Earth (apart from the constant density) - the result is true independent of its size, mass or length of a day.