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August 29, 2014

August 29, 2014

Posted by **janie** on Monday, May 31, 2010 at 1:06pm.

(a) Find the speed of the asteroid at closest approach, assuming its speed at infinite distance to be zero and considering only its interaction with the Earth.

(b) Observations indicate the asteroid to have a diameter of about 2.0 km. Estimate the kinetic energy of the asteroid at closest approach, assuming it has an average density of 3.33 g/cm3. (For comparison, a 1-megaton nuclear weapon releases about 5.6 1015 J of energy.)

- physics -
**Damon**, Monday, May 31, 2010 at 1:29pmF = G m Me/r^2

potential energy relative to zero at infinity = -G m Me/r

here r = 84600 miles

convert that to meters

then

potential energy lost = Kinetic energy gained

G m Me/r = (1/2)m v^2

so

v^2 = 2 G Me/r

for part b you need m

diameter = 2*10^3

so

diameter/2 = radius = 1,000 meters

convert density to kg/m^3

3.33 g/cm^3 * (1kg/10^3g) * (10^6 cm^3/m^3) = 3330 kg/m^3

m = density * (4/3)pi (diameter/2)^3

then do Ke = (1/2) m v^2

- physics - reply is here for asteroid -
**Damon**, Monday, May 31, 2010 at 1:38pmThis is where I posted the reply to the multiple postings.

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