Homework answers / question archive / Planetary Impacts<br/><br/><br/>ENERGY CONSERVATION AND IMPACT HEATING<br/> One of the fundamental principles in physics, indeed in all of Nature is the Principle of ENERGY CONSERVATION

Planetary Impacts<br/><br/><br/>ENERGY CONSERVATION AND IMPACT HEATING<br/> One of the fundamental principles in physics, indeed in all of Nature is the Principle of ENERGY CONSERVATION

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Planetary Impacts<br/><br/><br/>ENERGY CONSERVATION AND IMPACT HEATING<br/> One

of the fundamental principles in physics, indeed in all of Nature is the Principle of ENERGY CONSERVATION. That is, although energy can be transformed from one form to another, when a 'balance sheet' is drawn up during a process when energy is moved around (like putting an ice cube in warm tap water) the total energy of the system is constant. Some simple examples: a battery powers a model car by using the chemical energy of the battery (a chemical reaction inside the battery) via the motor to be transformed into kinetic energy of motion of the model or the electric energy used to generate heat and light in a light bulb. In each of these processes one form of energy is converted into another form but the total amount of energy is conserved during the process. 
             Here we consider the dissipation of kinetic energy of motion when a projectile impacts a planetary surface to form a crater or large impact structure. In this case, the kinetic energy of the projectile impact is converted into other forms of energy: mechanical energy to launch debris on ballistic trajectories away from 'ground zero', energy in shock waves and earthquakes, sound waves and atmospheric shock waves and light, energy, heat generated via friction that causes heating, melting and possibly vaporization of portions of the target and the projectile. The details of this dynamic process requires numerical simulation using large computer "hydrocodes" such as the ones used by the military to simulate the detonation of nuclear bombs and other high energy processes. The point is, the kinetic energy of a projectile that impacts Earth (), is converted to other forms of energy. As an approximation, we usually consider that almost all of the KE is dissipated (transformed) into heat energy. That is, to a good approximation, we can consider that all of the kinetic energy of impact is converted (dissipated) into heat during high velocity impact. The heat content (or thermal energy, Q) of a body of mass m that has its temperature raised by the amount DT is given by:


                                    Q = m Cp DT                           (9)


where m is the mass of the body, Cp is the heat capacity of the body and DT is the increase in temperature of the body due to the 'deposition" of energy that is brought in by the impactor. The heat capacity is a material property that measures how much energy (Joules) must be added to 1 kilogram (kg) of a material to raise its temperature 1 degree Kelvin (K). So for example, the heat capacity of tap water is 4184 J/kg K. If one has a kilogram of water and want to increase its temperature by 1 K, you must supply 4184 J of energy.


Example:
Calculate the increase in temperature of the final body (of mass m+M) if an object of m = 1000 kg is slammed into another body of mass M = 10 000 kg. Assume that the impact velocity (i.e., the relative velocity between the impactor (or projectile) and the larger body (the target)) is 20 km/s. Assume that the heat capacity of both bodies is 1200 J/kg K. Note that the assumed impact velocity is typical of the speed at which large meteorites encounter the Earth


Solution:
Assume that all the kinetic energy of the impact is converted into heat and that all the heat goes into HEATING the combined body (mass after collision of m + M). Then the balance becomes:
1/2 mv2 = Cp (m+M) DT
where, the mass of the impactor, M is the 'before collision' mass, v is the impact velocity (that is, the relative velocity between projectile and target) of 20 km/s and Cp is the heat capacity of the objects that have collided (1200 J/kg K). Solve this for DT and then convert all quantities to the SI units and finally get:


DT = 1/2 m v2/Cp (m + M) = (0.5) (1000 kg) (20000)2  /(11000 kg) (1200 J/kg K) 
or DT = 15,152. K. That is, the entire mass heats up by ~ 15000 K! This means that the kinetic energy associated with the projectile, after dissipation (when the projectile is no longer moving relative to the target) has been converted via friction into heat. 




1.  Calculate the increase in temperature associated with each of the following scenarios.  Assume all KE of the impact goes into heat. Use an effective Cp = 2000 J/kg K.
            (a) Mercury-sized body colliding with the Earth traveling at a relative velocity equal to escape velocity from Earth. How much would this heat the Earth?  Note that you need to calculate the KE using the mass of Mercury, but calculate the heating using the mass of the Earth plus Mercury (i.e. the combined mass).


            (b) A collision between Vesta and Mars.  This time, assume Vesta approaches Mars with an approach velocity of 10 km/s.  However, the gravitational force of Mars also adds to the impact velocity, so to factor that in, we must take account of the approach velocity or speed of Vesta when it is far from Mars. The impact velocity is related to the approach velocity according to:


 (i) What is vesc,Mars? (See previous Lab handout or look this up in chart)
 (ii) What is vimpact?
(iii) If all KE from the collision is transferred to heating, what is the associated change in temperature for Mars?  Again, use Mvesta for KE and MVesta + Mmars for heating. However, since Vesta has a small mass compared to Mars the sum of the masses is essentially equal to that of Mars.