How Massive Must an Asteroid Be to Extend Earth's Day by 28%?

[BOAS]

Homework Statement

[/B]
Suppose that an asteroid traveling straight toward the centre of the Earth were to collide
with our planet at the equator and bury itself just below the surface. What would have
to be the mass of this asteroid, in terms of the earth’s mass, for the day to become 28.0%
longer than it presently is as a result of the collision? Assume that the asteroid is very
small compared to the Earth and that the Earth is uniform throughout.

Homework Equations

The Attempt at a Solution

[/B]
I have completed this question, but I think my reasoning is rather questionable. It hasn't convinced me, which doesn't give me much hope...

Moment on Inertia, $I = \frac{2}{5}MR^{2}$
Angular Momentum, $L = I \omega$

$\omega = 7.27$x$10^{-5}$rads$^{-1}$

$M_{e} = 5.97$x$10^{24}$kg

$R = 6371000m$

$L = 7.05$x$10^{33}$

So that is the angular momentum of the Earth before the impact, and I thought perhaps I can keep this constant, and using an angular velocity 28% slower, calculate the extra mass that must be added. The difference between this mass and the initial mass being the mass of the asteroid.

Through some algebra I arrive at an asteroid that has a mass $M = 0.39M_{E}$ or $2.33$x$10^{24}$kg.

This sounds very big, but it's hard to judge if that's a sensible number.

Anyway, my assumption of constant angular momentum is where I think a problem lies,

what do you think?

Thanks!

 

[SteamKing]

Consider that the mass of the moon is only about 1.2% of the Earth's mass, and I would say that your calculation of the mass of the 'asteroid' lies somewhere between the masses of Mars and Venus.

I think that if angular momentum is used, assuming I = (2/5) MR², then both M and R would need some adjustment after the collision, given the quantity of material added to the earth.

Perhaps you should look at this problem from an energy standpoint. As the Earth turns on its axis each day, a certain amount of rotational energy is generated. If the Earth slows down its rotation, for whatever reason, the amount of energy generated is reduced. This approach may produce an asteroid of more reasonable size.

 

[BOAS]

Consider that the mass of the moon is only about 1.2% of the Earth's mass, and I would say that your calculation of the mass of the 'asteroid' lies somewhere between the masses of Mars and Venus.

I think that if angular momentum is used, assuming I = (2/5) MR², then both M and R would need some adjustment after the collision, given the quantity of material added to the earth.

Perhaps you should look at this problem from an energy standpoint. As the Earth turns on its axis each day, a certain amount of rotational energy is generated. If the Earth slows down its rotation, for whatever reason, the amount of energy generated is reduced. This approach may produce an asteroid of more reasonable size.

I actually tried to solve this question using rotational kinetic energy before trying angular momentum, and I still get an absurd answer. I will carefully check all of my algebra, but both methods of working rely on this idea that the angular momentum and the rotational kinetic energy remain the same after the collision.

Is this not problematic?

 

[BOAS]

I checked my algebra for the kinetic energy solution, and got the same result that I first got; $0.93M_{E}$ :/

 

[SteamKing]

I checked my algebra for the kinetic energy solution, and got the same result that I first got; $0.93M_{E}$ :/

I think you mean 0.39 M_e.

Still, I think that with so much material being added to the earth, the mass moment of inertia of the planet undergoes a significant change in its value, and since this quantity is used to calculate both angular momentum and rotational kinetic energy, it must be adjusted somehow to account for the mass added by the asteroid.

 

[BOAS]

I think you mean 0.39 M_e.

Still, I think that with so much material being added to the earth, the mass moment of inertia of the planet undergoes a significant change in its value, and since this quantity is used to calculate both angular momentum and rotational kinetic energy, it must be adjusted somehow to account for the mass added by the asteroid.

Actually, I don't - I found the answer to be 0.93M_E, but 0.39M_E in my angular momentum calculation.

The only relevant shape that I can think of for this modified moment of inertia is treating the Earth as a hollow sphere, since so much mass will be added 'just below the surface', but that's simply not true.

 

[SteamKing]

Actually, I don't - I found the answer to be 0.93M_E, but 0.39M_E in my angular momentum calculation.

The only relevant shape that I can think of for this modified moment of inertia is treating the Earth as a hollow sphere, since so much mass will be added 'just below the surface', but that's simply not true.

I don't think a hollow sphere is the correct shape to use for the MOI calculation after impact. After all, you still have the original mass of the Earth spinning away, albeit slower, and its distribution has not changed. You do, however, have a lot of new mass which is either lumped onto the original surface of the earth, so that the new planet has a shape resembling a dumbbell, or the new mass has been spread evenly over the entire surface, increasing R, which in turn increases the MOI.

 

[NTW]

By other way, that is, assuming a linear inelastic collision of a body of the Earth's mass at a velocity of 293,7 m/s [at that velocity, its linear KE is the same as the Earth's rotational KE] with a immobile body of a certain mass, and calculating that mass for a reduction of velocity of 28%, I get practically the same results as you: 2,312 * 10²⁴ kg or 38,72% of the Earth's mass... The difference may be in rounding here and there...

 

[haruspex]

Perhaps you should look at this problem from an energy standpoint.

That will get you nowhere. There is a huge and unknown loss of mechanical energy as heat. The only way to find it is by first applying conservation of angular momentum.

Through some algebra I arrive at

Please post that algebra.
I get less than half your number.

 

[BOAS]

That will get you nowhere. There is a huge and unknown loss of mechanical energy as heat. The only way to find it is by first applying conservation of angular momentum.

Please post that algebra.
I get less than half your number.

Ok,

$I = \frac{2}{5}M_{E}R_{E}^{2}$

$L = I \omega$

$\omega = \frac{d \theta}{dt} = \frac{2 \pi}{86400}$

The angular momentum of the Earth prior to collision;

$L = \frac{2}{5}M_{E}R_{E}^{2} \omega = 7.05$x$10^{33} Kgm^{2}s^{-1}$

A 28% slower day corresponds to $0.72 \omega$

<Insert questionable logic of assuming angular momentum remains unchanged after the impact>

$M_{E + A} =$ mass of the Earth + mass of asteroid

$L = \frac{2}{5}M_{E+A}R_{E}^{2} (0.72 \omega)$

$M_{E+A} = \frac{5L}{2 R_{E}^{2} (0.72 \omega)} = 8.29$x$10^{24}Kg$

$M_{A} = M_{E+A} - M_{E} = 2.33$x$10^{24}$kg or $0.39 M_{E}$

 

[gneill]

Can you explain why everyone is assuming that the asteroid's mass is going to be spread throughout the volume of the Earth? After all, it seems that you're applying the same moment of inertia formula before and after collision, just assuming a change in mass.

The problem states that the asteroid buries itself just below the surface. If you assume that it remains in one concentrated location at one Earth radius from the center of rotation, what's the mass moment of inertia associated with it?

 

[BOAS]

Can you explain why everyone is assuming that the asteroid's mass is going to be spread throughout the volume of the Earth? After all, it seems that you're applying the same moment of inertia formula before and after collision, just assuming a change in mass.

The problem states that the asteroid buries itself just below the surface. If you assume that it remains in one concentrated location at one Earth radius from the center of rotation, what's the mass moment of inertia associated with it?

Could I use the moment of inertia for a dumbbell and use the parallel axis theorem to make it rotate about one of the spheres?

The only other shapes we've really looked at are disks, cylinders, and rods.

 

[gneill]

Could I use the moment of inertia for a dumbbell and use the parallel axis theorem to make it rotate about one of the spheres?

The only other shapes we've really looked at are disks, cylinders, and rods.

You've got a sphere (the Earth) and what is essentially a point mass (the asteroid) located at the equator. What's the moment of inertia of a point mass rotating at a distance R from the center of rotation?

 

[BOAS]

You've got a sphere (the Earth) and what is essentially a point mass (the asteroid) located at the equator. What's the moment of inertia of a point mass rotating at a distance R from the center of rotation?

$I = MR^{2}$

But I also need to take into consideration the moment of inertia of the earth, and I don't know how I combine these quantities.

 

[gneill]

$I = MR^{2}$

But I also need to take into consideration the moment of inertia of the earth, and I don't know how I combine these quantities.

Moment of inertia is the rotational analog of mass. They are combined in the same way: just sum them.

 

[BOAS]

Moment of inertia is the rotational analog of mass. They are combined in the same way: just sum them.

Great, thanks!

I'm still unsure about my method of keeping angular momentum constant. I can't think of a good reason why this would be the case.

 

[gneill]

Great, thanks!

I'm still unsure about my method of keeping angular momentum constant. I can't think of a good reason why this would be the case.

Great, thanks!

I'm still unsure about my method of keeping angular momentum constant. I can't think of a good reason why this would be the case.

Think about what comprises the system under consideration. What's the total angular momentum prior to impact? (The nature of the trajectory of the asteroid is important!). Are there any external forces or torques involved or is the system isolated?

 

[BOAS]

Think about what comprises the system under consideration. What's the total angular momentum prior to impact? (The nature of the trajectory of the asteroid is important!). Are there any external forces or torques involved or is the system isolated?

Ah ha!

The question states that the asteroids trajectory is at the equator and straight towards the center - Therefore the torque is zero and the angular momentum does remain constant.

Thanks a lot!

 

[SteamKing]

Moment of inertia is the rotational analog of mass. They are combined in the same way: just sum them.

I don't think it's that simple.

For a sizable mass located a distance away from the axis of rotation, the moment of inertia of the Earth combined with the asteroid should be calculated by applying the parallel axis theorem.

For

M_e - mass of the earth
M_a - mass of the asteroid

J_e - moment of inertia of the Earth about its axis of rotation
J_a - moment of inertia of the asteroid about its centroid
J_e+a - resultant moment of inertia of the Earth and asteroid

D_a - distance of the center of the asteroid from the rotation axis of the earth

J_e+a = J_e + J_a + M_a * D_a²

The complete analysis would entail calculating the distance of the centroid of the combined earth-asteroid body from the original axis of rotation, and then using that centroid to calculate the moment of inertia.

Of course, if M_a is small compared to M_e, then you can neglect the additional moment due to transfer. However, it appears that M_a is a sizable fraction of M_e, so the additional term should be evaluated.

 

[gneill]

Work it out the simple way first and see if it's a significant mass. The level of the problem and understanding demonstrated by the OP suggests that a simple approach may be warranted.

 

[BOAS]

Work it out the simple way first and see if it's a significant mass. The level of the problem and understanding demonstrated by the OP suggests that a simple approach may be warranted.

It comes out to be 0.16 Earth masses, which sounds like what haruspex got.

This is a year 1 mechanics problem, and I have studied the parallel axis theorem.

 

[NTW]

$I = MR^{2}$

But I also need to take into consideration the moment of inertia of the earth, and I don't know how I combine these quantities.

I have tried to combine them in the following way: First, you calculate the rotational KE of the Earth, before the arrival of the asteroid of mass M_a. You get a number J.

You now the moment of inertia of the Earth, MI_e. After the arrival of the asteroid, you have to count with its contribution, its own moment of inertia, say MI_a. Of course, you know the 'reduced' (72% of the original) rotation rate of the planet, ω_e.

You can now set up the equation:

((MI_e + MI_a)*ω_e²)/2 = J

And solve for MI_a. Once you get it, remember that it's M_a*R², and solve for M_a

 

[haruspex]

First, you calculate the rotational KE of the Earth

KE is not going to help. Huge amounts of mechanical energy will be converted to heat. Angular momentum is the only way.

 

[haruspex]

It comes out to be 0.16 Earth masses, which sounds like what haruspex got.
.

Yes, that's what I got.

M_e - mass of the earth
M_a - mass of the asteroid

J_e - moment of inertia of the Earth about its axis of rotation
J_a - moment of inertia of the asteroid about its centroid
J_e+a - resultant moment of inertia of the Earth and asteroid

D_a - distance of the center of the asteroid from the rotation axis of the earth

J_e+a = J_e + J_a + M_a * D_a²

That's exactly what gneill and I did, but with J_a = 0 (since we know nothing about that, and anyway it would be tiny compared with J_e). And D_a = Earth radius, yes? Or are you suggesting something else there?

The complete analysis would entail calculating the distance of the centroid of the combined earth-asteroid body from the original axis of rotation, and then using that centroid to calculate the moment of inertia.

That's true, but surely that's the point of being told in the OP to assume the asteroid's mass is much smaller than the Earth's. Have you done the full calculation to see how much difference it makes?

 

[SteamKing]

Yes, that's what I got.

That's exactly what gneill and I did, but with J_a = 0 (since we know nothing about that, and anyway it would be tiny compared with J_e). And D_a = Earth radius, yes? Or are you suggesting something else there?

D_a can be whatever it needs to be.

That's true, but surely that's the point of being told in the OP to assume the asteroid's mass is much smaller than the Earth's. Have you done the full calculation to see how much difference it makes?

No, you guys seemed to be having too much fun with this problem. I just wanted the OP to know that the transfer moment for the MOI can be significant on occasion.

 

[NTW]

KE is not going to help. Huge amounts of mechanical energy will be converted to heat. Angular momentum is the only way.

I don't think so... In the problem's statement: (...)'an asteroid traveling straight toward the centre of the Earth were to collide
with our planet at the equator'(...); 'straight toward the center' would mean following any of the Earth's radii... Thus, for an observer close to the impact point, the asteroid would be seen as coming down from the zenith, and -on contact- no angular moment transfer would take place at all, as with any object falling from the zenith... If you add that -always according to the original problem statement, (...) 'and bury itself just below the surface' (...), then very little dissipation of energy could be expected...

 

[haruspex]

I don't think so... In the problem's statement: (...)'an asteroid traveling straight toward the centre of the Earth were to collide
with our planet at the equator'(...); 'straight toward the center' would mean following any of the Earth's radii... Thus, for an observer close to the impact point, the asteroid would be seen as coming down from the zenith, and -on contact- no angular moment transfer would take place at all,

The impact is at the equator. To an observer there, the asteroid would have a horizontal speed of 1670 kilometers/hour relative to the surface of the Earth. The angular momentum of the Earth+asteroid system does not change, but the moment of inertia of the Earth increases by acquisition of the extra mass, so the angular velocity decreases. That is the whole point of the question.

If you add that -always according to the original problem statement, (...) 'and bury itself just below the surface' (...), then very little dissipation of energy could be expected...

It's the rotational KE of the Earth that is relevant, agreed? Forget the vertical collision, that's not relevant here - there is a massive tangential collision. You could lower the asteroid gently onto the surface and get the same result.
Try your energy-conserving approach. You will get quite a different answer from the (completely reliable) angular momentum approach, and it is therefore invalid.

 

[BiGyElLoWhAt]

Homework Statement

[/B]
Suppose that an asteroid ...
Assume that the asteroid is very small compared to the Earth and that the Earth is uniform throughout.

I don't think it's that simple.

For a sizable mass located a distance away from the axis of rotation...

The mass is most definitely not sizeable, however, it is massive (perhaps this is what you meant?). It's meant to be treated as a point mass. $M_a R_{earth}^2 + I_e = I_{combined}$

 

[SteamKing]

The mass is most definitely not sizeable, however, it is massive (perhaps this is what you meant?). It's meant to be treated as a point mass. $M_a R_{earth}^2 + I_e = I_{combined}$

An asteroid 16% of Earth's mass sounds both sizable and massive. After all, the moon is only 1.2% of Earth's mass, and no one neglects it.

By comparison, the biggest asteroid (dwarf planet?) known is Ceres, and it has a mass of only 0.015% of earth.

http://en.wikipedia.org/wiki/Ceres_(dwarf_planet)

If you throw Pluto into the mix, its mass is 0.22% of earth.

http://en.wikipedia.org/wiki/Pluto

 

[BiGyElLoWhAt]

That's all fine and dandy, but the problem statement says that the size of the asteroid is neglegable. This is so you can treat it as a point mass. Since when do Mechanics 1 problems need to be realistic?

 

[SteamKing]

That's all fine and dandy, but the problem statement says that the size of the asteroid is neglegable. This is so you can treat it as a point mass. Since when do Mechanics 1 problems need to be realistic?

Turns out whoever wrote this problem either lied or didn't know what the answer was. Having something 10 times as massive as the moon hitting Earth is not 'negligible' in any sense of the word.

 

[BiGyElLoWhAt]

Ok, so then op needs to integrate r^2 dm over the density function of the asteroid (which we don't have) and sum that with the earth.
[edit] [earth]'s moment.

 

[SteamKing]

Ok, so then op needs to integrate r^2 dm over the density function of the asteroid (which we don't have) and sum that with the earth.

You can make a list of things we don't know about this asteroid, which is why this problem is poorly constructed.

 

[BiGyElLoWhAt]

This was just a problem that someone pulled out of their rear end. Think about it. An asteroid hits the Earth aimed at towards the center of the earth, and causes an increase in our day by 28% (!). That's huge for an asteroid that exerts 0 torque. It's all about recognizing you need to use angular momentum and applying it properly.

 

[haruspex]

28.0% longer than it presently is

BOAS, are you sure it's 28%, not perhaps 0.28%?