How much force would a 4 kg meteor impact earth with?

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Discussion Overview

The discussion centers around estimating the force of impact of a 4 kg meteor striking the Earth, considering various factors such as entry velocity, air resistance, and energy release. Participants explore concepts related to kinetic energy, momentum, and the effects of atmospheric entry on the meteor's mass and velocity.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether force is the correct quantity to focus on, suggesting energy, momentum, or other measures might be more relevant.
  • Another participant proposes using the kinetic energy formula to estimate the energy released upon impact, noting that mass density may also play a role.
  • Concerns are raised about accounting for air resistance, with participants debating its potential impact on the energy of the meteor upon reaching the ground.
  • Some participants discuss the complexities of calculating the final velocity of the meteor after atmospheric entry, considering factors like angle of entry and atmospheric conditions.
  • There is a discussion about the implications of the meteor's speed, with one participant explaining that 11 km/s is a minimum speed for a meteor entering the atmosphere, while others argue that actual impact speeds could be significantly higher due to additional factors.
  • One participant emphasizes that smaller meteors are likely to disintegrate in the atmosphere, complicating the estimation of impact force.
  • There is a debate about the application of F=MA in this context, with some participants suggesting that it does not adequately address the original question about energy release.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate focus of the discussion, whether on force, energy, or other factors. There is no consensus on how to accurately account for air resistance or the complexities of atmospheric entry, indicating multiple competing views remain.

Contextual Notes

Participants acknowledge the need for various assumptions regarding atmospheric conditions, entry angles, and the meteor's initial velocity, which complicate the calculations and estimations being discussed.

Pgeske
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If a meteor entered the Earth's atmosphere at 11 or so km/sec and impacted the Earth with a direct hit (by impact time it is 4 kg), what would the force of impact be? I only need a rough answer.
 
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Are you sure it is force you are interested in?
 
Energy, force, momentum, or anything... I just need something.
 
What's the formula for kinetic energy? It's a pretty simple substitution if you just want a rough answer for the amount of energy released. Convert to equivalent pounds of TNT if you want an idea of the size of the crater produced. But it also depends on the mass density, doesn't it?
 
Alright, but how would I account for air resistance? Would that reduce the energy of impact by a huge amount? I know that if there wasn't wind resistance, the energy of impact would be extremely high, but what would the energy of impact be with air resistance?
 
Pgeske said:
Alright, but how would I account for air resistance? Would that reduce the energy of impact by a huge amount? I know that if there wasn't wind resistance, the energy of impact would be extremely high, but what would the energy of impact be with air resistance?

In your OP you said by impact time it was 4kg so I assumed that was what remained after going through the atmosphere. If the 4KG is at the point of entry into the atmosphere, this is a more difficult problem. You have to take into account the angle of entry, the mass density, etc. Temperatures generated by air friction would vaporize a large amount of the meteorite and might explode it into much smaller pieces and dust, so the impact energy at ground level would have to account for the impact area, the average size of pieces, etc. It's not a problem that has an easy answer without a lot of assumptions. If this was a HW problem, maybe that's the point.
 
The 4Kg is the mass that is left over, but the velocity of impact is different than the velocity of atmospheric entry due to air resistance. Is there any way I could calculate a very rough estimate of the final velocity (impact velocity) of the mass after being slowed from air resistance?
 
At 11 km/s, and depending on the angle of entry, it would barrel through the atmosphere within about 1-10 seconds. (75% of the mass of the Earth's atmosphere is within 11 km of the surface. I wouldn't think it would lose much velocity unless the angle was pretty steep.
 
It would be unlikely to remain in one piece all the way to the ground. At those velocities, atmospheric impact generates tremendous heat and forces on the meteorite, likely exploding it into several chunks and a whole lot of dust. Those chunks would still impact with great force, but the atmospheric hypervelocity shock wave preceding the impact would be as, if not more, devastating. Also, 11 Km/s is fairly slow as far as meteorites go, that only corresponds to the Earth's gravitational pull on an object. True collision speeds would be 2, 3, or more times higher.
 
  • #10
What do you mean? How does 11 km/s corrospond to the force of gravity from earth?
 
  • #11
11 km/s is the speed an object will have if it starts far from Earth with zero speed, and is accelerated to the ground by gravity alone. If it starts with some initial speed, its impact will be faster than 11 km/s.
 
  • #12
cjl said:
11 km/s is the speed an object will have if it starts far from Earth with zero speed, and is accelerated to the ground by gravity alone. If it starts with some initial speed, its impact will be faster than 11 km/s.

That's only assuming planet without atmosphere.
 
  • #13
Yes, but it is a reasonable assumption for the speed it will have at the top of the atmosphere. Inside the atmosphere, it becomes a somewhat more challenging problem.
 
  • #14
Pgeske said:
If a meteor entered the Earth's atmosphere at 11 or so km/sec

cjl said:
Yes, but it is a reasonable assumption for the speed it will have at the top of the atmosphere.

I see what you are saying, but you don't need any assumption about the entry speed, as entry speed is given.
 
  • #15
My point about the 11 Km/s is that it's the slowest possible speed for a meteor collision because it only accounts for the Earth's attraction on a far away object without any other relative motion. When you account for the Earth's orbital speed around 30 Km/s and the meteor's speed you can see why such impacts occur at closing speeds as high as 70 Km/s. This is well above what we can achieve in lab tests aside from particle collisions. The best we can do to study meteor collisions are analytical studies of past collisions or lucky observations of asteroid impacts on other planets.
 
  • #16
Borek said:
I see what you are saying, but you don't need any assumption about the entry speed, as entry speed is given.

Yes, I was responding about why 11 km/s is just about the slowest reasonable speed for a meteor.
 
  • #17
How many asteroids in a series of asteroids hitting the moon with each both slowing the moon and deflecting it increasingly toward Earth would it take to cause the moon to hit the earth? Assuming it splashed down in the Pacific, could the resulting tsunami inundate everything below 1000 feet above sea level? How far up various major rivers would that reach?
 
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  • #18
JohnKingtamer said:
What if a series of asteroids hit the moon with each both slowing the moon and deflecting it increasingly toward earth?

What about "what if"? What question are you asking exactly?
 
  • #19
Something that small would likely burn in the atmosphere on the way down. If it did make it F=MA
Acceleration due to gravity is 9.8 m/s^2 so ~ 40 N
 
  • #20
Trevormbarker said:
Something that small would likely burn in the atmosphere on the way down. If it did make it F=MA
Acceleration due to gravity is 9.8 m/s^2 so ~ 40 N

Are you sure you're using F=MA correctly here? i.e. in the context of the question asked.
 
  • #21
ryan_m_b said:
Are you sure you're using F=MA correctly here? i.e. in the context of the question asked.

I thought so, correct me if I am wrong, the thread name was how much force would a 4 Kg meteor impact Earth with. What am I doing incorrectly?
 
  • #22
Trevormbarker said:
I thought so, correct me if I am wrong, the thread name was how much force would a 4 Kg meteor impact Earth with. What am I doing incorrectly?

This question is possibly going to reveal to PF how simplistic some of my understanding is but here goes...

F=MA refers to the energy needed to accelerate an object of given mass i.e. how much force does it take to accelerate a 4kg object at 1g = 40N per second.

However the OP isn't "what force does a 4kg asteroid undergo when falling to Earth", it's how much energy would the 4kg asteroid release. That would be worked out by figuring out the kinetic energy of the asteroid.

For another way to see why F=MA doesn't provide an answer to the OP consider this, using it the way you have done (where A is the acceleration in the gravity field of Earth) would give the same result for a 4kg asteroid traveling at 11kmps and one traveling at 111,000kmps.
 
  • #23
ryan_m_b said:
This question is possibly going to reveal to PF how simplistic some of my understanding is but here goes...

F=MA refers to the energy needed to accelerate an object of given mass i.e. how much force does it take to accelerate a 4kg object at 1g = 40N per second.

However the OP isn't "what force does a 4kg asteroid undergo when falling to Earth", it's how much energy would the 4kg asteroid release. That would be worked out by figuring out the kinetic energy of the asteroid.

For another way to see why F=MA doesn't provide an answer to the OP consider this, using it the way you have done (where A is the acceleration in the gravity field of Earth) would give the same result for a 4kg asteroid traveling at 11kmps and one traveling at 111,000kmps.

I see what your saying, sorry about this missunderstanding, So according to the OP should one not just use Ek = 1/2mv^2 and use different values for v depending on how fast the meteor is moving
 
  • #24
Trevormbarker said:
I see what your saying, sorry about this missunderstanding, So according to the OP should one not just use Ek = 1/2mv^2 and use different values for v depending on how fast the meteor is moving

The OP provided a speed for the asteroid at the surface (11kmps) which comes to 0.25GJ using http://www.csgnetwork.com/kineticenergycalc.html".
 
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