On February 15, 2013, Asteroid 2012 DA14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8 kilometers per second [km/s ]. This is considerably closer than the orbit of geosynchronous satellites (26,200 miles). This is the closest recorded approach of an object this large. The asteroid 2012 DA14 was estimated to have a diameter of 30 meters [m] and a specific gravity of 3. If 2012 DA14 had hit the Earth, what is the total amount of energy that would have been released (i. e., what was the kinetic energy of the asteroid)

Question

Jaylin Lozano

Physics

27 May, 10:16

On February 15, 2013, Asteroid 2012 DA14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8 kilometers per second [km/s ]. This is considerably closer than the orbit of geosynchronous satellites (26,200 miles). This is the closest recorded approach of an object this large. The asteroid 2012 DA14 was estimated to have a diameter of 30 meters [m] and a specific gravity of 3. If 2012 DA14 had hit the Earth, what is the total amount of energy that would have been released (i. e., what was the kinetic energy of the asteroid)

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Diamond · Answer 1

The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 * 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ

1 PJ = 10¹⁵ J

Explanation:

Kinetic energy = mv²/2

velocity of the asteroid is given as 7.8 km/s = 7800 m/s

To obtain the mass, we get it from the specific gravity and diameter information given.

Density = specific gravity * 1000 = 3 * 1000 = 3000 kg/m³

But density = mass/volume

So, mass = density * volume.

Taking the informed assumption that the asteroid is a sphere,

Volume = 4πr³/3

Diameter = 30 m, r = D/2 = 15 m

Volume = 4π (15) ³/3 = 14137.2 m³

Mass of the asteroid = density * volume = 3000 * 14137.2 = 42411501 kg = 4.24 * 10⁷ kg

Kinetic energy of the asteroid = mv²/2 = (4.24 * 10⁷) (7800²) / 2 = 1.29 * 10¹⁵ J