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?
To find the ratio of the distance of closest approach of Asteroid 2012 DA14 to the radius of the moon's orbit, we first need to calculate the radius of the moon's orbit.
The moon orbits the Earth in an elliptical path, with an average radius of about 384,400 kilometers (km). However, since the question states that the asteroid came within 34,200 km from the center of the Earth, we need to subtract this distance from the radius to find the actual closest approach.
Radius of moon's orbit = Average radius - Distance of closest approach
Radius of moon's orbit = 384,400 km - 34,200 km
Radius of moon's orbit = 350,200 km
Now that we have the radius of the moon's orbit, we can calculate the ratio of the distance of closest approach of DA14 to the radius of the moon's orbit.
Ratio = Distance of closest approach of DA14 / Radius of moon's orbit
Ratio = 34,200 km / 350,200 km
Dividing both numbers by 2, we get:
Ratio = 0.0978
Therefore, the ratio of the distance of closest approach of Asteroid 2012 DA14 to the radius of the moon's orbit is approximately 0.0978.