Complete Question
The complete is shown on the first uploaded image
Answer:
The amount of kinetic energy it carries is [tex]KE = 3.55 *10^{21} J[/tex]
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 [tex]E_n_{mton}= 8450Mton[/tex]
Explanation:
From the question we are told that
The diameter of the Toutatis is [tex]d_T = 5.4km = 5.4*10^3 m[/tex]
The mass of Toutatis is [tex]M_T = 0.05 * 10^{15}kg[/tex]
The radius of Toutatis orbit is [tex]R_T = 2.38 *10^{11} m[/tex]
The period of the of the orbit is [tex]T = 3.98 years = 3.98 * 12* 30 * 24* 3600 =1.23*10^{8}s[/tex]
Generally the speed of the Toutatis is mathematically represented as
[tex]v = \frac{2 \pi R}{T}[/tex]
substituting values
[tex]v = \frac{2 * 3.142 * 2.38*10^{11}}{1.238*10^{8}}[/tex]
[tex]=11915m/s[/tex]
The kinetic energy is mathematically represented as
[tex]KE = \frac{1}{2} M_Tv^2[/tex]
substituting values
[tex]KE = \frac{1}{2} (0.05*10^{15}) (11915)^2[/tex]
[tex]KE = 3.55 *10^{21} J[/tex]
The energy yield of the nuclear bomb is
[tex]E_n=0.01 \ of \ KE[/tex]
Substituting values
[tex]E_n =0.01 * 3.55 *10^{21}[/tex]
[tex]= 3.55*10^{19}J[/tex]
now converting to Mton
[tex]E_n_{mton} =\frac{3.54 *10^{19}}{4.2*10^{15}}[/tex]
[tex]E_n_{mton}= 8450.43Mton[/tex]
