Question

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?

Answer 1

Complete Question

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Answer:

The amount of kinetic energy it carries is
`KE = 3.55 *10^(21) J`

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
`E_n_(mton)= 8450Mton`

Step-by-step explanation:

From the question we are told that

The diameter of the Toutatis is
`d_T = 5.4km = 5.4*10^3 m`

The mass of Toutatis is
`M_T = 0.05 * 10^(15)kg`

The radius of Toutatis orbit is
`R_T = 2.38 *10^(11) m`

The period of the of the orbit is
`T = 3.98 years = 3.98 * 12* 30 * 24* 3600 =1.23*10^(8)s`

Generally the speed of the Toutatis is mathematically represented as

`v = (2 \pi R)/(T)`

substituting values

`v = (2 * 3.142 * 2.38*10^(11))/(1.238*10^(8))`

`=11915m/s`

The kinetic energy is mathematically represented as

`KE = (1)/(2) M_Tv^2`

substituting values

`KE = (1)/(2) (0.05*10^(15)) (11915)^2`

`KE = 3.55 *10^(21) J`

The energy yield of the nuclear bomb is

`E_n=0.01 \ of \ KE`

Substituting values

`E_n =0.01 * 3.55 *10^(21)`

`= 3.55*10^(19)J`

now converting to Mton

`E_n_(mton) =(3.54 *10^(19))/(4.2*10^(15))`

`E_n_(mton)= 8450.43Mton`