Calculating the Average Force on a Meteor Upon Entering Jupiter's Atmosphere

[brake4country]

Homework Statement

A meteor with a mass of 1 kg moving at 20 km/s collides with Jupiter's atmosphere. The meteor penetrates 100 km into the atmosphere and disintegrates. What is the average force on the meteor once it enters Jupiter's atmosphere? (Note: ignore gravity).

Homework Equations

K = 1/2 mv^2
W = F x d

The Attempt at a Solution

I initially started this problem with F = ma in mind. I then solved for time using d = vt and got 200 seconds. Then I found acceleration using V = Vi + at and got 100 m/s^2. By plugging in 100 m/x^2 into F=ma, I get (1 kg)(100 m/s^2) = 100 N.

The correct answer is 2 x 10^5. What am I doing wrong? Can't this problem be solved without using kinetics and work?

Thanks.

 

[Nathanael]

I then solved for time using d = vt and got 200 seconds.

d = 100km, right?

What did you use for "v" and why?

 

[brake4country]

Yes, d = 100 km = 1,000,000 m is what I used. I used 20,000 m/s for the initial velocity because the problem states that the meteor disintegrates in the atmosphere = 0 m/s.

 

[matineesuxxx]

What is an alternate definition for work?

 

[brake4country]

Energy via a force--kinetic energy. But I am trying to solve this problem using kinematics. Can this be done?

 

[Nathanael]

Yes, d = 100 km = 1,000,000 m is what I used. I used 20,000 m/s for the initial velocity because the problem states that the meteor disintegrates in the atmosphere = 0 m/s.

100 km is only 100,000 m

Also you can't use v = 20,000 m/s in that equation (d=vt) because the velocity is changing.




The correct answer is supposed to be 200,000 N ?
With every method I use, I keep getting an answer of 2,000 N. I probably shouldn't be helping you if I can't even solve the problem correctly..

 

[KurtWagner]

If you use a work energy approach you can use:

1/2m v^2 = f x

And solve for x. I got 2*10^5

And the velocity is in km/s so 200,000,000m/2

 

[KurtWagner]

Sorry that's m/s at the end, not m/2

 

[KurtWagner]

Wait. km/s should actually be 20,000 which gives me an answer of 2*10^3

 

[KurtWagner]

I may be wrong here, so beware. But I think the reason you cannot use a simple constant acceleration formula like v^2 = u^2 + 2ax is beacuse there is not constant acceleration inside jupiter.

So in my mind (my mind mind you) the 'x' is given beacuse there is no constant acceleration and it asks you to find the average.

 

[brake4country]

I am so sorry, yes, the correct answer is 2 s 10^3. Sorry to confuse everyone. Okay, so to clarify, I cannot use d =vt if it is assumed an object does not have constant velocity. Also, when acceleration is not constant, kinematic equations cannot be used.

Would it be safe to say that in situations like the problem, to use energy and work formulas?

Also, 100km = 100 km x 1000 m/1 km = 100,000 m. I stand corrected. I think I did this conversion correct and perhaps should keep it in scientific notation for a standardized test.

 

[KurtWagner]

100,000m

And don't take too much of what I said to heart beacuse i am not too sure myself.

However, yes. It is definitely easier to use a work-energy approach for these types of questions

 

[Nathanael]

If you use a work energy approach you can use:

1/2m v^2 = f x

And solve for x. I got 2*10^5

And the velocity is in km/s so 200,000,000m/2

Solve for x?
You mean solve for f using the fact that it traveled 100km? (x=100,000)
(P.S. What are your units? m/s? Isn't the answer supposed to be a force?)

I may be wrong here, so beware. But I think the reason you cannot use a simple constant acceleration formula like v^2 = u^2 + 2ax is beacuse there is not constant acceleration inside jupiter.

You sure can use a constant acceleration formula and get the same answer (2,000).

The possible problem you're thinking of (that acceleration is not constant,) is also apparent in your above method (the "energy method").

The "f" in your equation would not actually be a simple constant, it would be a function of some sort. But to solve for the "average force" is to solve it as if "f" were a constant.

So in the "acceleration method" you can use constant acceleration equations because you're solving the problem as if the acceleration (force) were constant (and average).

 

[brake4country]

So just to clarify; in free fall motion, kinematic equations are used because the acceleration of gravity is constant but velocity constantly changes. In other words, to use kinematics, either velocity or acceleration must be constant. Similarly, the equation d =vt can be used for problems where there is no change in velocity. Am I on the right track?

 

[KurtWagner]

Solve for x?
You mean solve for f using the fact that it traveled 100km? (x=100,000)
(P.S. What are your units? m/s? Isn't the answer supposed to be a force?)

Note that the m/s was not my answer. And yes, you are correct. I should have said solve for f

 

[Nathanael]

So just to clarify; in free fall motion, kinematic equations are used because the acceleration of gravity is constant but velocity constantly changes.

Yes, because the acceleration of gravity is approximately constant near the surface of the Earth.
(It is also approximately constant in other similar situations where the distances involved are small in comparison with the distance of the gravitation).

[edit: Also free fall questions that use those equations equation ignore air resistance (they call it negligable). In this situation, though, the forces of the atmosphere are not negligable, on the contrary, they are the basis of this problem.]

In other words, to use kinematics, either velocity or acceleration must be constant. Similarly, the equation d =vt can be used for problems where there is no change in velocity.

Yes, to use those specific equations requires constant acceleration, because those equations were derived with the assumption that acceleration is constant.

Similarly, d=vt applies to constant velocities, because that equation was "derived" with the assumption of constant velocity. (I don't know if that equation is technically "derived" because it's just a rearrangement of the definition of velocity)
BUT, those equations can be used when dealing with averages. This is because the definition of "average" is essentially "the constant that has the same result" (if that doesn't make sense don't worry about it, it's just the way I think of averages).

For example:
"Average speed" is the speed that has the same result (distance travelled) as the "actual speed" (which is non-constant)
In this example, it only makes sense to speak of average speed over a certain time period

You can also look at averages in terms of totals.
For example, average speed is the Total Distance over the Total TimeBut at any rate, an average is essentially a constant, and so you can use such equations.Side note:
If something is constant, then the average is the same thing as the "actual"

 

[Nathanael]

Also, 100km = 100 km x 1000 m/1 km = 100,000 m. I stand corrected. I think I did this conversion correct and perhaps should keep it in scientific notation for a standardized test.

Just saying it can tell you what it is. (For example, 2 times 4 would be "Two fours")

100 times 1,000 is "One hundred thousands"

But you can also use scientific notation:
$10^2*10^3=10^5=100,000$

Would it be safe to say that in situations like the problem, to use energy and work formulas?

You can also use your original method (what I call the "acceleration method"). See my above post for details.
In fact, I would personally suggest only using the "energy method" if you already understand the "acceleration method" because the "energy method" is essentially just a shortcut that is based off the "acceleration method" (No use using a shortcut if you don't understand how it works.)

 

[KurtWagner]

Note: went back and tried:

V^2 = u^2 + 2ax

Worked completely fine.
See I knew something was wrong all along. ;p

 

[Nathanael]

Okay, so to clarify, I cannot use d =vt if it is assumed an object does not have constant velocity..

Ok let's get back to the problem, starting with your first step.

You actually can use d=vt but instead of using the initial velocity, you would need to use the average velocity

So what would be the correct value for the time it takes?

 

[brake4country]

Well, the problem says to "ignore gravity" which in my interpretation, means that acceleration is constant. I will go back and review the mistakes I made. It looks like I totally messed up an easy conversion from km->m. This thread clarified so much! Thank you everyone!

 

[brake4country]

Great point Nathanael! The average velocity is (vi + vf)/2 which is just the initial velocity in this problem. Excellent point!

 

[Nathanael]

Well, the problem says to "ignore gravity" which in my interpretation, means that acceleration is constant.

The "ignore gravity" part just simplifies the problem a bit, by making the only force on the meteor the forces from the atmosphere (call it "atmospheric friction" or "air resistance")

But, just because that is the only force on the object, does not mean that the force is constant.

 

[KurtWagner]

To skip the middle man (time) you can look for an equation that does not include a time variable. Back in high school physics they had a few to memorize. One of the most useful is

V^2 = u^2 + 2ax

Where v is final velocity
U is initial velocity
A is acceleration
And x is distance

 

[haruspex]

A meteor with a mass of 1 kg moving at 20 km/s collides with Jupiter's atmosphere. The meteor penetrates 100 km into the atmosphere and disintegrates. What is the average force on the meteor once it enters Jupiter's atmosphere? (Note: ignore gravity).

Sorry, but I disagree strongly with every apparently helpful post on this thread so far.
The short answer is, you do not have enough information - the question is wrongly posed.
What is meant by "average force"?
Average acceleration is well-defined as change in velocity divided by change in time: Δv/Δt.
Assuming the mass stays constant, it is then reasonable to define average force as mass*average acceleration: mΔv/Δt = Δ(mv)/Δt = ∫F.dt/∫.dt. If you search the web you will find this is the standard definition of average force.
The information provided gives you no way to compute that. What it does allow you to compute is ΔE/Δx = ∫F.dx/∫.dx. Yes, it's an average force of a sort, namely, the "average over distance". But it is quite wrong to call that simply the "average force". If the force happens to be constant this will give the same answer, but not in general.
Note in particular that average force ought to be a vector, and Δ(mv)/Δt gives that, whereas ΔE/Δx is a bit strange - it appears to be a scalar divided by a vector.

 

[Nathanael]

I thought the question was strange, too.

The information provided gives you no way to compute that. What it does allow you to compute is ΔE/Δx = ∫F.dx/∫.dx. Yes, it's an average force of a sort, namely, the "average over distance". But it is quite wrong to call that simply the "average force". If the force happens to be constant this will give the same answer, but not in general.

You're saying there's a difference between "average over distance" and "average over time"?

I was thinking about this, but the math showed the average was the same (I guess that was because, as you said, I assumed constant force)


I've been thinking about something similar for a while, and I have a slightly irrelevant question:

If I have the force as a function of time, $F(t)$, and the distance as a function of time, $d(t)$, how would I find the Work as a function of time?

Would it be $∫F(d^{-1}(t))$? Where $d^-1(t)$ is the inverse of the distance as a function of time.
(I'm not sure if $d^{-1}(t)$ would be written as $t(d)$?)

Is that valid? This has been confusing me recently. (I know distance over time is related to force over time but I'm ignoring that aspect.)

 

[KurtWagner]

Would you not just multiply?

W = f(x) d(x)

 

[Nathanael]

Would you not just multiply?

W = f(x) d(x)

No I don't believe that works. I don't want to go into detail but essentially you want to integrate the force with respect to distance


Sorry, perhaps I shouldn't have asked that last question; I didn't intend to start a separate discussion on brake4country's thread.

 

[KurtWagner]

Unless the force/time aspect is controlled by an observer (a person), t should be able to be factored out algebraically

 

[haruspex]

Would you not just multiply?

W = f(x) d(x)

Yes, but substituting $d(x) = \dot x.dt$
For the non-constant force case, consider e.g. SHM over one quarter cycle.
F = -k²mx
$\ddot x = -k^2x$
x = A sin(kt)
$\dot x = Ak \cos(kt)$
ΔE = m(Ak)²/2
Δx = A
ΔE/Δx = mAk²/2
Δmv = mAk
Δt = π/(2k)
Δmv/Δt = mAk²(2/π) = (4/π) ΔE/Δx.

 

[Nathanael]

Thank you Harusepx! I finally did what I've been trying to do (on and off) for a couple days.

I typed a good sized private message for you (going into specific detail of what made me ask about this) but at the end the mathematics worked out correctly!
(I was trying to re-create what I did the other day, where the mathematics was incorrect. I don't know what I was doing wrong then.)

Thanks again, the following simple piece of information made it very clear :)

$d(x) = \dot x.dt$

 

[haruspex]

Unless the force/time aspect is controlled by an observer (a person), t should be able to be factored out algebraically

Not sure what you are saying. Are you maintaining that there's no difference between the time-averaged force and the distance-averaged force?