How Do You Calculate the Kinetic Energy of an Asteroid?

[MJC8719]

In the early morning hours of June 14, 2002, the Earth had a remarkably close encounter with an asteroid the size of a small city. The previously unknown asteroid, now designated 2002 MN, remained undetected until three days after it had passed the Earth. Suppose that at its closest approach, the asteroid was 71000 miles from the center of the Earth -- about a third of the distance to the Moon. (Astronomical data needed for this problem can be found on the inside back cover of the text.)


(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.
2.64011 km/s

(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.)

It is part B i am getting stuck with...

I know that density = mv and once i get the mass of the asteriod I can get the KE = .5mv^2...

So D = m/v
3.33 g/cm^3 = x/(4/3)pi r^3

My question is what units do i use here...for instance, should the 3.33 g/cm^3 be converted to meters and the 2.0km converted to meters...if so, how to i convert cm^3 to m^3...

If this is not correct, could someone please guide me on the correct units to use.

thanks so much

 

[berkeman]

I know that density = mv and once i get the mass of the asteriod I can get the KE = .5mv^2...

So D = m/v

Your second equation is correct, D = m/v (not the first one "density = mv", which is a typo).

For this problem, you should convert all units into the mks system of units (meters, kilograms, seconds). The easiest way to convert units is to multiply by 1. That is, say you want to convert 2g/cm^3 into mks units. You do it like this:

2 \frac{g}{{cm}^3} * \frac{1 kg}{1000g} * \frac{10^6 cm^3}{1 m^3} = ? \frac{kg}{m^3}

Can you fill in the ? number? You treat the units as if they were numbers, and cancel out cm/cm = 1, etc. BTW, it is helpful in most problems to carry units along in your calculation, to be sure that you have not made any mistakes in the formulation and solution of your equations. The units of the left hand side (LHS) must always match the units of the RHS, and the units of each term in a sum must always match.

Make sense?


EDIT -- It took me a couple passes to get the latex above correct. Sorry if I confused you with an intermediate version as I re-edited.

 

[MJC8719]

So, if i understand you correctly,
then my 3.33 g/cm^3 = (3.33 x 10^6)/1000 (i can't do the fancy stuff like you did lol) which gives me a value of 3330 kg/m^3

So then the equation would read: m = (3330)(4/3pi(1000^3))
so m = 1.3949 x 10^13 which is them multipled by 0.5v^2

And in part A, i solved for the velocity as being 2.64011 km/s which i then need to convert to m so the final equation to solve for the KE would be:
KE = (0.5)(1.3949 x 10^13)(2640.11 m^2)

Thanks again

 

[berkeman]

I didn't check the math, but it looks like you are on the right track.

Remember what I said about carrying along units in your equations, though. Even if you don't show the units while you type here on the forum, hopefully you are now showing units for each of the quantites in the equations that have them, and cancelling them out when they appear on top and bottom like cm/cm = 1, so cross them both out. That will save you many, many errors over the years, and also help you figure out what goes on the top and bottom of a fraction when you aren't sure.

Like, what is the relationship between wavelength, speed and frequency? Well, if wavelength's units are m, speed is m/s, and frequency is 1/s (=Hz), then you can figure out the equation just based on keeping the units consistent, right?

 

[MJC8719]

yes..i do write out the units on my paper...its just quicker to not type them out on the computer...

thanks for all the help, greatly appreciated.