Question

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

Answer:

Explanation:

Mass of earth = M

initial Time period, T = 24 hr

final time period, T' = T + 26% of T = 24 + 24 x 26 / 100 = 30.24 hr

The angular momentum of the system is conserved as no external torque is applied on the system

$\left ( \frac{2}{5}MR^{2} \right )\frac{2\pi }{T}=\left ( \frac{2}{5}MR^{2} + mR^{2}\right )\frac{2\pi }{T'}$

$\frac{2M}{5T}=\left ( \frac{2}{5}M + m\right )\frac{1}{T'}$

$\frac{2M}{5\times 24}=\left ( \frac{2}{5}M + m\right )\frac{1}{302.4}$

0.504 M = 0.4 M + m

m = 0.104 M

Answer 2

The mass of the asteroid, in terms of the earth's mass M should be 0.104 M.

Calculation of the mass of asteroid:

Since

Mass of earth = M

initial Time period, T = 24 hr

So,

final time period, T' = T + 26% of T

= 24 + 24 x 26 / 100

= 30.24 hr

Now

(2/5MR^2)2π/T = (2/5MR^2 + mR^2) 2π/T

2M/5T = (2/5M + m) 1/T'

2M/5*24 = (2/5M + m) 1/302.4

0.504 M = 0.4 + m

So, m = 0.104 m

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