A near miss

Classical Mechanics Level 3

Recently, Asteroid 2012 DA14 came within 34,200 km from the center of the earth at its point of closest approach. If the moon goes around the earth once every 27.5 days, what is the ratio of the distance of closest approach of DA14 to the radius of the orbit of the moon?

Details and assumptions

The mass of the earth is $6 \times 10^{24}~\mbox{kg}$ .
One day is 24 hours.
Assume the moon orbits the earth in a circle.
Newton's constant is $G=6.67 \times 10^{-11}~\mbox{N}~\mbox{m}^2/\mbox{kg}^2$ .

The answer is 0.0887.

1 solution

David Mattingly Staff
May 13, 2014

The angular frequency of the moon is $\omega=2\pi/T=2\pi/(27.5 \times 24 \times 3600)=2.64\times 10^{-6}~\mbox{rad/sec}$ . Since the moon is in a circular orbit, we can equate the centripetal acceleration to the acceleration provided by gravity,

$a_c=\omega^2 r=\frac {G m_E}{r^2}$ .

We therefore can solve this for $r$ , yielding $r=3.85 \times 10^{8}~\mbox{m}$ , and so the ratio of the distance of closest approach to the orbit of the moon is $3.42 \times 10^7/3.85 \times 10^8=0.0887$ .

A near miss

The answer is 0.0887.

1 solution

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