# Earth-Asteroid collision problem

Nathew

## Homework Statement

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass, based on its size, of 5 × 10^9 kg. It is approaching the Earth on a head-on course with a velocity of 595 m/s relative to the Earth and is now 5.0 × 10^6 km away. With what speed will it hit the Earth's surface, neglecting friction with the atmosphere?

## Homework Equations

V = 2$\pi$R/T

a = (v^2)/R

## The Attempt at a Solution

I am very confused because wouldn't the speed with which the asteroid hits the earth depend where the asteroid is in relation to the Earth? or does head on imply something I'm missing?

Either way what I thought I should do is calculate the velocity of the Earth around the sun plus the velocity of the asteroid, but that's not right.

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Dick
Homework Helper

## Homework Statement

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass, based on its size, of 5 × 10^9 kg. It is approaching the Earth on a head-on course with a velocity of 595 m/s relative to the Earth and is now 5.0 × 10^6 km away. With what speed will it hit the Earth's surface, neglecting friction with the atmosphere?

## Homework Equations

V = 2$\pi$R/T

a = (v^2)/R

## The Attempt at a Solution

I am very confused because wouldn't the speed with which the asteroid hits the earth depend where the asteroid is in relation to the Earth? or does head on imply something I'm missing?

Either way what I thought I should do is calculate the velocity of the Earth around the sun plus the velocity of the asteroid, but that's not right.

Ignore the motion around the sun. You don't have enough information to account for that. You just have to account for the effect of the Earth's gravity on the asteroid. Use the notion of gravitational potential energy.

Nathew
Ignore the motion around the sun. You don't have enough information to account for that. You just have to account for the effect of the Earth's gravity on the asteroid. Use the notion of gravitational potential energy.

Thanks! That worked!

The asteroid would strike the earth at a much higher speed than stated. Since the force of gravity is inversely proportioned to the distance from the center of gravity squared, the gravitational effects on the asteroid would raise it's impact velocity exponentially. In fact, the true velocity at impact would far outweigh the combined tangential velocity from rotation and the orbital velocity around the sun. Since the mass is considerable, the total KE of the asteroid at impact would be a world ender.

gneill
Mentor
The asteroid would strike the earth at a much higher speed than stated. Since the force of gravity is inversely proportioned to the distance from the center of gravity squared, the gravitational effects on the asteroid would raise it's impact velocity exponentially. In fact, the true velocity at impact would far outweigh the combined tangential velocity from rotation and the orbital velocity around the sun. Since the mass is considerable, the total KE of the asteroid at impact would be a world ender.
No speed was stated as the Original Poster did not share his final working. The rest of your statement is just handwaving without a calculation to support it. Can you support your argument with some math?

lsume
Yes! As you know KE = 1/2 m v^2. So let's make an assumption about the mass of this hypothetical asteroid. Let us assume that the density of the asteroid is 490 pounds per cubic ft. which is an average density of steel. One common misconception is that an astronaut appearing to float in space is weightless. Since generally, the astronaut is in a geosynchronous orbit and his centripetal force cancels the gravitational force exerted by the earth. The earth is still exerting a force on the astronaut. Now for our asteroid, assume that the asteroid is 300' in diameter and it's shape mirrors a sphere. The volume of this spherical asteroid is (4/3) pi r^3 = 14,130,000 cubic ft. with a mass of
((14,130,000)(490)/32.174) = 215,195,499 slugs

If this asteroid is going to hit earth then logically it's under the gravitational pull of the earth. The force exerted is inversely proportional to the distance from the center of gravity squared but the acceleration component is always present albeit very weak at great distance but present none the less. Make your own assumption about the speed at impact but rest assured it will be a very high speed. I am not going to give a velocity which implies direction I am assuming a direct hit normal to the earth. As I recall, the escape velocity from earth is 7 miles per second, and for no reason I'll use that scaler for the kinetic energy calc. ( 215,195,499 ) (0.5 ) (360,960 )^2 = ONE GREAT BIG NUMBER which would blow away planet earth. Now, as to the tangential velocity of earth and the orbital velocity of earth around the sun, let's assume they are working together for a worst case scenario. The circumference of earth is 24,901 miles and at 1 revolution every 24 hours the tangential velocity at the equator would be 1,521 ft./sec. Assume the average distance that the earth is from the sun is 93 million miles and rather than calculating based on the elliptical orbit just use a circular orbit. So, based on these assumptions, the earth travels ( pi ) D every 365 days. The speed of orbit would then be 9.25 miles per second. Your point was well taken. If we couple the tangential and orbital scalers or we don't couple these scalers, the orbital speed is more than enough to blow away planet earth. Thanks for the heads up Gneill. I never did this simple calc before this am.

No speed was stated as the Original Poster did not share his final working. The rest of your statement is just handwaving without a calculation to support it. Can you support your argument with some math?
Check my reasoning. You were absolutely correct as I stated at the bottom of my calc. I'd never done this simple calc before thus am and now I'm glad I did. Good catch!

Jim

gneill
Mentor
Yes! As you know KE = 1/2 m v^2. So let's make an assumption about the mass of this hypothetical asteroid. Let us assume that the density of the asteroid is 490 pounds per cubic ft. which is an average density of steel. One common misconception is that an astronaut appearing to float in space is weightless. Since generally, the astronaut is in a geosynchronous orbit and his centripetal force cancels the gravitational force exerted by the earth. The earth is still exerting a force on the astronaut. Now for our asteroid, assume that the asteroid is 300' in diameter and it's shape mirrors a sphere. The volume of this spherical asteroid is (4/3) pi r^3 = 14,130,000 cubic ft. with a mass of
((14,130,000)(490)/32.174) = 215,195,499 slugs
So about 3 x 109 kg. Note that SI units are preferred internationally.
If this asteroid is going to hit earth then logically it's under the gravitational pull of the earth. The force exerted is inversely proportional to the distance from the center of gravity squared but the acceleration component is always present albeit very weak at great distance but present none the less. Make your own assumption about the speed at impact but rest assured it will be a very high speed.
Assuming your desired conclusion is not a good approach. Make no assumptions about the final speed or what constitutes "very high speed". Very high compared to what?
I am not going to give a velocity which implies direction I am assuming a direct hit normal to the earth. As I recall, the escape velocity from earth is 7 miles per second, and for no reason I'll use that scaler for the kinetic energy calc. ( 215,195,499 ) (0.5 ) (360,960 )^2 = ONE GREAT BIG NUMBER which would blow away planet earth.
About 2 x 1017 Joules. You cannot conclude that because the number looks big to you that that it will "blow away planet earth". You haven't presented an argument to support an actual required amount of energy to accomplish this feat.

You might start by determining the gravitational binding energy of the Earth itself. Ultimately it's gravity that's holding the planet together. Assuming approximately uniform density for the Earth with mass Me and radius Re,

##U = \frac{3 G {M_e}^2}{5 R_e}##

Which is on the order of 2 x 1032 Joules, which is about 15 orders of magnitude larger than your estimated impact energy.

This is not to say that the results of such an impact would be insignificant to us surface dwellers, but it would not be of concern to the Earth's overall integrity as a planet.
Now, as to the tangential velocity of earth and the orbital velocity of earth around the sun, let's assume they are working together for a worst case scenario. The circumference of earth is 24,901 miles and at 1 revolution every 24 hours the tangential velocity at the equator would be 1,521 ft./sec. Assume the average distance that the earth is from the sun is 93 million miles and rather than calculating based on the elliptical orbit just use a circular orbit. So, based on these assumptions, the earth travels ( pi ) D every 365 days. The speed of orbit would then be 9.25 miles per second. Your point was well taken. If we couple the tangential and orbital scalers or we don't couple these scalers, the orbital speed is more than enough to blow away planet earth. Thanks for the heads up Gneill. I never did this simple calc before this am.
Again, you can't leap to the conclusion that the Earth will be "blown away" without running the actual numbers. The speed of the Earth's surface due to its rotation is dwarfed by the orbital speed of the Earth (about half a km per second versus about 30 km per second), so ignoring the rotational contribution the worst case scenario would add 30 km per second to the impact velocity used above. Let's call it 41 km/sec in total. That yields a kinetic energy of about 2.5 x 1018 Joules. Still a drop in the bucket compared to Earth's gravitational binding energy.

Running these sorts of calculations for "what if" scenarios is excellent practice. I applaud your enthusiasm and encourage you to keep at it. They help to temper our tendency to leap to unwarranted conclusions when numbers seem large (but turn out not to be when placed in context) and help to hone our intuitions and instincts about likely outcomes.

Edit: I made a couple of calculation errors (futzed the conversion of the diameter of the proposed iron sphere). The mass of the 300 foot diameter sphere would be about 2.1 x 107 kg, not the 3 x 109 kg that I wrote. So the impact KE at escape speed would be even smaller than I wrote, about 1.3 x 1015 Joules rather than 2 x 1017 Joules. For the 41 km/s case it would be about 2 x 1016 Joules. Sorry for any confusion this might have caused.

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I didn't keep the result but I could have probably compared it to how many megatons of TNT without to much trouble. I learned years ago to always have as many peers as I could get check my calculations for the real world. I know better than to make statements without backup and I truly do appreciate it when some one points out an error. I still like to do calculations in general and, honestly, I'm glad you caught the error in my assumption. Whenever I hear the asteroid scenario on some science documentary, I seem to recall that, they say a nuke would not mitigate the problem. My thought has been that if you shatter the asteroid into tinier pieces, the energy would be dissipated through our atmosphere. I suppose that the people who have analyzed the problem have surly figured that postulate into account.

Thanks again for the correction
Jim

ellis1954
I didn't keep the result but I could have probably compared it to how many megatons of TNT without to much trouble. I learned years ago to always have as many peers as I could get check my calculations for the real world. I know better than to make statements without backup and I truly do appreciate it when some one points out an error. I still like to do calculations in general and, honestly, I'm glad you caught the error in my assumption. Whenever I hear the asteroid scenario on some science documentary, I seem to recall that, they say a nuke would not mitigate the problem. My thought has been that if you shatter the asteroid into tinier pieces, the energy would be dissipated through our atmosphere. I suppose that the people who have analyzed the problem have surly figured that postulate into account.

Thanks again for the correction
Jim
Based on the assumed numbers above, the KE would be 3.82 X 10^16 ft-Lbs. Using the conversion calculator for ft-lbs to megatons of explosives

http://www.convertunits.com/from/foot+pounds/to/megaton+[explosive]

The result is 1.238193612253279e-15 megatons of explosives. I don't know how many megatons of explosives it would take to annihilate earth but the number of megatons above seems like it would do the trick.

gneill
Mentor
Based on the assumed numbers above, the KE would be 3.82 X 10^16 ft-Lbs. Using the conversion calculator for ft-lbs to megatons of explosives

http://www.convertunits.com/from/foot+pounds/to/megaton+[explosive]

The result is 1.238193612253279e-15 megatons of explosives. I don't know how many megatons of explosives it would take to annihilate earth but the number of megatons above seems like it would do the trick.
That doesn't look right. The negative value on the exponential is a giveaway that something went awry in the conversion.

A megaton explosive is the equivalent of about 4.18 x 1015 Joules. Using my revised numbers for the energy released (see EDIT at bottom of post #8), the 300 foot diameter steel sphere hitting at 41 km/second would yield less than 5 megatons equivalent energy.

ellis1954
I didn't go any further than the online calculator. Your logic seems sound so I will have to test out with some given constants and see how it works. That won't be the first time I've found an error on the net. I'm glad you checked it out by hand. When I get around to it I'll post any errors I find.

Thanks again
Jim

ellis1954
I didn't go any further than the online calculator. Your logic seems sound so I will have to test out with some given constants and see how it works. That won't be the first time I've found an error on the net. I'm glad you checked it out by hand. When I get around to it I'll post any errors I find.

Thanks again
Jim
After checking the online calculator one thing I am sure of and that is the result you gave is significantly higher than the result the calculator gives when I enter the numbers without scientific notation. The write up below the calculator helps to put things in perspective. I May toy with the calculator some more but I am finished posting on the subject for now.