An asteroid, named Toutatis, is in a near Earth orbit. The asteroid is 5.4 km in diameter, more precisely it is 4.6 km x 2.4 km x 1.9 km. The asteroid has a mass of 0.05 x 1015 kg. If the path of the asteroid is assumed circular around the Sun, the radius of the orbit is 2.38 x 10 m and it takes 3.98 years to make one orbit, how much kinetic energy does it carry? If it collided with Earth, where would this energy go? If we could divert the asteroid by exploding a nuclear bomb on its surface that had an energy yield of 0.01% of its kinetic energy, how big would the nuke have to be in Mtons, where 1 Mton = 4.2 x 1015 J?

Complete Question

The complete is shown on the first uploaded image

The amount of kinetic energy it carries is

There two condition that would determine where the energy would go

 First if it is moving in the same direction as the earth then both earth and  Toutatis would maintain their orbit and the energy from  collision  would be lost as heat , sound and vibrations

Second if the collision is head on then both Toutatis and earth would loss their orbit and the energy from the collision will be lost as heat, sound and vibrations

The energy of the nuke is  

Explanation:

   From the question we are told that

      The diameter of the  Toutatis is

       The mass of Toutatis is

        The radius of Toutatis orbit is

         The period of the of the orbit is

 Generally the speed of the Toutatis is mathematically represented as

substituting values

The kinetic energy is mathematically represented  as

       substituting values

The energy yield of the nuclear bomb is

Substituting values

now converting to Mton

yup its correct its: c) b b

0.19 m/s and 0.45 m/s