Question

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth's mass M , for the day to become 26.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer 1

Consider the conservation of angluar momentum .
final ? * final time = initial ? * initial time

final ? * 1.26 * initialtime = initial ? * initial time

1.26 final? = initial ?

We also know that the angular momentum must be conserved. Angmomentum is just

moment of inertia * angularvelocity, or I ? so

I ? initial = I? final

Ii (1.26 ?f) = If ?f

or

I f = 1.26 Ii

We know that initial I is just that of the Earth, as asphere:

Ii = (2/5) MR² where M is mass of the Earth and Ris its radius (we'll plug in numbers later)

Also, I final is Earth plus the extra due to theasteroid mass at the edge...

If = (2/5) MR² + mR² where m is themass of the asteroid.

Now pop these expressions into the equation above:

(2/5) M R² + m R² = 1.26 * (2/5) M R²

eliminate R2 from each term

0.4 M + m = 0.504 M

m = 0.104M the asteroid mass would haveto be 0.104 times the mass of the Earth (for comparison,this is about six times the mass of the moon!!)

in kilograms, this is m = 0.104* 5.98 x10²⁴ kg = 6.2 x10²³ kg