The correct formula for calculating the kinetic energy an asteroid hits Earth with is:

Where m is the mass of the asteroid and v is the impact speed relative Earth.

Regarding your miscalculation Newton is in kg*m/s^2 and you have used acceleration in km/s^2 so the kilo is unaccounted for in your answer.

Sorry that took so long, kinda forgot about this site. It's still possible I got this wrong like I did with the Newtons, but here it goes

951 Gaspara

Discovery

Discovered by G. N. Neujmin

Discovery date July 30, 1916

Designations

Named after Gaspra

Alternate name(s) SIGMA 45; A913 YA;

1955 MG1

Minor planet

category Main belt (Flora family)

Orbital characteristics

Epoch 6 March 2006 (JD 2453800.5)

Aphelion 2.594 AU (388.102 Gm)

Perihelion 1.825 AU (272.985 Gm)

Semi-major axis 2.210 AU (330.544 Gm)

Eccentricity 0.174

Orbital period 3.28 a (1199.647 d)

Average orbital speed 19.88 km/s

Mean anomaly 53.057°

Inclination 4.102°

Longitude of ascending node 253.2lllll18°

Argument of perihelion 129.532°

Proper orbital elements

Physical characteristics

Dimensions 18.2×10.5×8.9 km [1]

Mean radius 6.1 km[2]

Mass 2–3×1016 kg (estimate)

Mean density ~2.7 g/cm³ (estimate) [3]

Equatorial surface gravity ~0.002 m/s² (estimate)

Escape velocity ~0.006 km/s (estimate)

Rotation period 0.293 d (7.042 h) [4]

Albedo 0.22 [5]

Temperature ~181 K

max: 281 K (+8°C)

Spectral type S

Absolute magnitude (H) 11.46

Mean orbital velocity of Jupiter + mean orbital velocity of Mars

/2 = Average orbital velocity

Orbital period of Gaspara 951: 1199.647 days

and to find how many seconds that is:

1199.647*24 because 24 hours, then times 60 because 60 minutes in an hour,

then times another 60 because 60 seconds in a minute, and we get about

1.03*10^8 seconds.

average velocity to calculate the force

For acceleration, (v2-v1)/(t2-t1)

(19.88km/s-0km/s)/(1.03*10^8s-0s)= approximately 1.93^-7km/s/s

Then we multiply that by the mass to see how much force it carries if

left fully intact

2.5*10^16kg * 1.93*10^-7km/s/s = approximately 4.82*10^9N of force.

The number of meteoroids in a meteor shower various greatly, but seem to

often fall in the realm of 15-50 pieces per hour, although most of the

things in a meteor shower are pieces of dust.

http://meteorshowers...or_showers.html

Using Newtons does not make for a good conversion with dimensional analysis, so I will calculate it's

approximate kinetic energy instead.

E(sub k) = 1/2mv^2

Kinetic energy = (1/2)(2.5*10^16kg)(19.88m/s)^2=4.94*10^18J

1 joule of kinetic energy approximately equals 1 joule of thermal energy.

If all asteroid's energy was converted to thermal energy, it would release

approximately 4.94*10^18 joules of thermal energy.

2.5*10^16kg

Using the Specific heat of Earth's atmosphere to calculate the temperature change

q=McT

T = Temperature Change = ???

q = energy in joules = 4.94*10^18J = 4.94*10^15kJ

M= mass = 5.3×10^18

http://hypertextbook...LouiseLiu.shtml

http://en.wikipedia....osphere_of_Eart

c = specific heat = 1.01(kJ/kg K) = (specific heat capacity of "Air")

http://www.engineeri...ases-d_159.html

Proof:

(4.94*10^(15)kJ)=(5.3×10^(18)kg)(1.01(kJ/kg)k)(T)

Then I divide the left side by everything on the right side except by T, and dividing is the same as multiplying by 1 over that number

(4.94*10^15kJ/1)(1/5.3*10^(18)kg)(1kgK/1.01kJ) = the units cancel out and I am left with a temperature change of .0009228 Kelvin

And in case I forgot to multiply something by 1000 like I did with newtons, I will multiply the amount of joules the asteroid has by 1000 to play it safe

(4.94*10^18kJ/1)(1/5.3*10^(18)kg)(1kgK/1.01kJ) = the units cancel out and I am left with a temperature change of .9228... Kelvin

If my original theory and it's proof is correct, if the asteroid 951 Gaspara which is 6.1km large was heading for Earth and we fragmented it into pieces so small that all of it's kinetic energy would be converted to thermal energy via friction causing the heating and then vaporization of the fragments before they reached the ground, Earth would be saved and there would be a temperature change less than 1 Kelvin.

Based on this I think that it is safe to assume an asteroid 10km large would raise Earth's temperature a little over 1 K or a little over 1 hundredth of a K,

and an asteroid 20km large would definitely raise Earth's atmosphere's temperature by a little over 2 K or .002 K.

I looked about the conversion for Kelvin to Celsius and it appears 1K approximately =

**-**272.15 degrees Celsius.

**Edited by questionposter, 13 April 2012 - 11:41 PM.**