Kinetic Energy & Megaton Calculation HELP

Click For Summary
SUMMARY

The discussion focuses on calculating the kinetic energy of a comet impacting Earth and converting that energy into megatons of TNT. The kinetic energy formula used is Kinetic energy = 1/2MV^2, where the mass of the comet is 4.2 x 10^12 kg and its velocity is 34,000 m/s. The calculated kinetic energy is 2.4276 x 10^21 joules, which translates to approximately 577 megatons of TNT when divided by 4.2 x 10^15 joules per megaton. The potential devastation from such an impact on a populated area is significant.

PREREQUISITES
  • Understanding of kinetic energy calculations
  • Familiarity with mass and velocity units
  • Knowledge of energy conversion between joules and megatons of TNT
  • Basic comprehension of impact effects on populated areas
NEXT STEPS
  • Research the implications of asteroid impacts on Earth
  • Learn about the physics of kinetic energy in different contexts
  • Study methods for calculating energy release from nuclear reactions
  • Explore historical asteroid impacts and their consequences
USEFUL FOR

Students studying physics, educators teaching energy concepts, and researchers interested in planetary defense and impact effects.

lloyd_lex
Messages
1
Reaction score
0

Homework Statement



A) A relatively small impact crater 20 kilometers in diameter could be made by a comet 2 kilometers in diameter traveling at 34.0 (3.40×104 ).

Assume that the comet has a total mass of kilograms. What is its total kinetic energy?

(B) Convert your answer from part A to an equivalent in megatons of TNT, the unit used for nuclear bombs. Comment on the degree of devastation the impact of such a comet could cause if it struck a populated region on Earth. (Hint: One megaton of TNT releases joules of energy.)


Homework Equations


Kinetic energy = 1/2MV^2


The Attempt at a Solution



my attempt:
=1/2*[4.2*10^12]*[3.4*10^4]^2
=2.4276*10^21 joules

For B
i divided my answer from A by 4.2*10^15 since One megaton of TNT releases 4.2*10^15joules of energy
 
Last edited:
Physics news on Phys.org
The formating seems to have removed some numbers - so we can't check it.
But your method is correct
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 54 ·
2
Replies
54
Views
11K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K