Ignoring the motion of the Earth for energy vs. momentum conservation

[greypilgrim]

Hi.

If I drop an inelastic body, its potential energy first gets converted to kinetic, then to deformation energy. We use conservation of energy without taking into account the kinetic energy gain of the earth during the fall.

However, at first sight conservation of momentum seems to be violated by the body first gaining momentum and then coming to a sudden stop. For this to work out, we cannot ignore the earth.

Why can we ignore the motion of the earth for energy but not for momentum?

 

[kuruman]

Assume that the Earth moves as one with the mass after the collision. From momentum conservation, the velocity of the combined Earth+body after the collision is $$V=\frac{m}{m+M_E}v_0$$ where $v_0$ is the velocity of the body of mass $m$ just before the collision. The initial momentum of the body is $p_1=mv_0$. The momentum change of the body is $$\Delta p=m\left(\frac{mv_0}{m+M_E}\right)-mv_0$$ Assume that $m=1~$kg and take the mass of the Earth to be $M_E=6\times 10^{24}~$kg. Plug the numbers in. What do you get for $\Delta p$? Repeat the calculation for a 10-kg mass. Do you see where this is going?

 

[PeroK]

Hi.

If I drop an inelastic body, its potential energy first gets converted to kinetic, then to deformation energy. We use conservation of energy without taking into account the kinetic energy gain of the earth during the fall.

However, at first sight conservation of momentum seems to be violated by the body first gaining momentum and then coming to a sudden stop. For this to work out, we cannot ignore the earth.

Why can we ignore the motion of the earth for energy but not for momentum?

You picked the wrong scenario! If you drop an elastic object onto a hard surface, then it rebounds with (almost) no loss of KE. Hence KE appears to be conserved without considering the motion of the Earth. Momentum, on the other hand, is reversed, which represents a maximal change in momentum. This is essentially the opposite of momentum conservation, which appears to be violated in this case.

 

[kuruman]

Hi.

If I drop an inelastic body . . .

If you drop an elastic object onto a hard surface . . .

?

 

[PeroK]

?

You picked the wrong scenario!

 

[A.T.]

If I drop an inelastic body, its potential energy first gets converted to kinetic, then to deformation energy. We use conservation of energy without taking into account the kinetic energy gain of the earth during the fall.

As @PeroK points out, usually we use conservation of energy for elastic collisions because the deformation energy is hard to estimate.

However, at first sight conservation of momentum seems to be violated by the body first gaining momentum and then coming to a sudden stop. For this to work out, we cannot ignore the earth.

Why can we ignore the motion of the earth for energy but not for momentum?

Start with an Earth of the same mass as the ball and perfectly elastic collision.

What is the error on CoE and CoM if you ignore the Earth gain of KE and momentum?

What happens to those errors as you increase the Earth's mass to infinity?

 

[kuruman]

I guess I misunderstood OP's question.

The object + Earth system is assumed isolated which means that its total energy is conserved. The sum of all the energy changes must be zero. The possible energy changes are mechanical energy and deformation energy. We write the total energy conservation equation $$0=\Delta E_{\text{Total}}=\Delta K_{\text{Earth}}+ \Delta K_{\text{object}}+\Delta U_g+\Delta E_{\text{deform.}}$$Here the object and the Earth start and end at rest in the Earth-object CM frame. Thus, $\Delta K_{\text{Earth}}= \Delta K_{\text{object}}=0$. Also, near the surface of the Earth, $\Delta U_g=-mgh.$ The total energy conservation equation becomes $$0=0+0+(-mgh)+\Delta E_{\text{deform.}}\implies \Delta E_{\text{deform.}}=mgh.$$ The initial potential energy of the Earth-object system is converted to deformation energy. Nothing is ignored.

 

[Dale]

Why can we ignore the motion of the earth for energy but not for momentum?

Because KE is quadratic in velocity and momentum is linear in velocity.

(I am in the “it should be elastic” camp)

 

[Mister T]

However, at first sight conservation of momentum seems to be violated by the body first gaining momentum and then coming to a sudden stop.

Momentum is conserved in the absence of external forces. Here we have an external force exerted on the body, so we don't expect the momentum of the body to be conserved.

 

[kuruman]

I guess I still don't understand what is being asked.

To @greypilgrim : Can you write down some equation that illustrate what your question is?

 

[vanhees71]

The 10 fundamental conservation laws of Newtonian physics come from Galilei invariance, and they are valid only for closed systems, i.e., you consider a set of point particles with their mutual interactions (usually a sum of two-body interactions) but without treating some forces as "external forces" as in the approximation, where you neglect the motion of the Earth and your falling body because it can be savely neglected when describing the motion of the body relative to the Earth, because the mass of the Earth is very much larger than that of the falling body.

In (elastic) collisions, of course, kinetic energy of the involved bodies is also not conserved all the time but it's the total energy of these bodies in mutual interaction by an interaction force that can be derived from a potential. In collisions, however, you are usually not interested on the details of the entire dynamics but only in the question, how two bodies coming with certain momenta (and thus also with certain total kinetic energy) together, end up in two bodies moving away from each other after the collision. In both the initial and the final "asymptotically free state" the interaction potential can be neglected in the energy balance, because it goes (by definition) to 0 for large distances between the particles. For these asymptotically free states total momentum and energy are conserved (for an elastic collision), i.e., the total momentum and the total kinetic energy of both particles are the same for the initial and the final asymptotically free state.

 

[A.T.]

I guess I still don't understand what is being asked.

Consider this:

Start with an Earth of the same mass as the ball and perfectly elastic collision.

What is the error on CoE and CoM if you ignore the Earth gain of KE and momentum?

What happens to those errors as you increase the Earth's mass to infinity?

For m_earth = m_ball the error in both, CoE and CoM, is 100%, all of the ball's intial momentum and kinetic energy disapears.

When you let m_earth / m_ball -> inf, the CoE error goes to 0, while the CoM error goes to 200% (the ball's intial momentum is reversed).

 

[Dale]

Consider this:

For m_earth = m_ball the error in both, CoE and CoM, is 100%, all of the ball's intial momentum and kinetic energy disapears.

When you let m_earth / m_ball -> inf, the CoE error goes to 0, while the CoM error goes to 200% (the ball's intial momentum is reversed).

Yes, and that happens because momentum is first order in v and energy is quadratic in v.

 

[nasu]

It may be that the OP is not asking about the collision but about the free fall part of the motion. Or a combination of the two. The conservation of energy he mentions is the one during the fall and not during a collision.

 

[kuruman]

For m_earth = m_ball the error in both, CoE and CoM, is 100%, all of the ball's intial momentum and kinetic energy disapears.

What error? Why should there be one? In what calculation? Let me do the calculation without approximations and then perhaps it can be pointed out to me what simplifying assumption introduces an error and where. I use subscripts $o$ and $e$ for object and earth respectively.

We release the object from height $h$. The potential energy change of the object-earth system (near the surface of the Earth) is $\Delta U=-mgh$ The object is released from rest, so in the earth-object CM frame momentum conservation requires that $$0=M_eV_e+m_ov_o\implies V_e=-\frac{m_o}{M_e}v_o$$Mechanical energy conservation from point of release to point of collision requires that $$
\frac{1}{2}M_eV_e^2+\frac{1}{2}m_ov_o^2-mgh=0.$$ Mechanical energy is not conserved through the collision. When the collision is complete, the earth and object are again at rest in the CM frame. The mechanical energy lost by the system through the entire process is its initial potential energy because its kinetic energy in the CM frame is zero in the beginning and zero in the end. This is an exact result independent of the masses of the colliding objects.

We can fiddle with the equations, find the speeds of the object and the earth and then expand for $m_o/M_e<<1$ to get $$\begin{align} & v_o=\left(\frac{2gh}{1+m_o/M_e}\right)^{1/2}\approx \left(1-\frac{m_o}{2M_E}\right)\sqrt{2gh} \nonumber \\
& V_E=\frac{m_o}{M_E}\left(\frac{2gh}{1+m_o/M_e}\right)^{1/2} \approx \frac{m_o}{M_E}\left(1-\frac{m_o}{2M_E}\right)\sqrt{2gh}.\nonumber \\
\end{align}$$These expressions reinforce the idea that it's OK to pretend that the earth is at rest and remains at rest throughout the motion of the object because $m_o/M_E \sim 10^{-24}.$ However, there should be the understanding that if this were actually the case, then we are violating momentum conservation and/or Newton's third law. We cannot say that the object accelerates because the earth attracts it while asserting that the earth stays where it is because there is no unbalanced force exerted on it.

 

[A.T.]

What error? Why should there be one?

What is the error on CoE and CoM if you ignore the Earth gain of KE and momentum?

 

[sophiecentaur]

Why can we ignore the motion of the earth for energy but not for momentum?

Can we?
If you could actually measure the change in Momentum then you could infer a change in Energy. In the case of collisions involving anything but significant sized asteroids, we ignore both. Don't we?

The actual numbers count here.

 

[Ibix]

My tuppence ha'penny - let the speed of the Earth (mass $M$ be zero just before the collision and the speed of the object (mass $m$) be $v_0$. The speed of the Earth after the collision is $V=2\frac{m}{m+M}v_0\approx v_0(2m/M)$. Thus the momentum of the Earth is approximately $2mv_0$ but the kinetic energy is approximately $\frac 12m^2v_0^2\times(4m/M)$. The former is comparable to the momentum of the object; the latter is negligible.

So, in short, the rebound velocity of the Earth is $\propto m/M$, and hence the momentum doesn't depend on the mass ratio but the energy does. The Earth carries away very little energy.

This is the same concept exploited by firearms. The fast moving low mass bullet gets potentially lethal quantities of energy but the slow moving high mass gun's recoil won't bruise you.

 

[sophiecentaur]

let the speed of the Earth

Velocity of the Earth or of the CM of the two masses?

The original scenario is not idea for exploring basic collision theory. Start with a canon ball and a small ball bearing and then take the limits for more differing masses.

 

[Ibix]

Velocity of the Earth or of the CM of the two masses?

Does it matter? The energy change will be the same in every frame.

As I see it, the point is that the change in energy of the Earth is utterly negligible (in any frame, and also if you repeat the calculation for an inelastic collision). The momentum change is not. The reason, mathematically at least, is that the Earth's velocity change is approximately proportional to $1/M$, so $M$ cancels out of the momentum change but the kinetic energy change is proportional to $1/M$. Thus neglecting the velocity change of the Earth produces an error in momentum conservation comparable in size to the momentum change of the small mass, but only a negligible error in the energy available for deformation.

 

[sophiecentaur]

Does it matter? The energy change will be the same in every frame.

Yes, of course. but if you talk in terms of the "speed of the Earth", that implies declaring a frame. You can then go through the usual calculations for collisions.

 

[vanhees71]

I think, it's time to do the math. Let's consider a closed system of two particles interacting via a central-potential force. Then you have the full Galilei symmetry and thus all standard conservation laws fulfilled. Here we consider energy and momentum.

The equations of motion read
$$\begin{split}
m_1 \ddot{\vec{x}}_1 &=-\nabla_1 V(|\vec{x}_1-\vec{x}_2|) \\
m_2 \ddot{\vec{x}}_2 &=-\nabla_2 V(|\vec{x}_1-\vec{x}_2|)=+\vec{\nabla}_1 V(|\vec{x}_1-\vec{x}_2|).
\end{split}$$
From this it immediately follows that
$$\vec{p}_1+\vec{p}_2=m_1 \dot{\vec{x}}_1 + m_2 \dot{\vec{x}}_2=\text{const}.$$
Then multiplying the first equation with $\dot{\vec{x}}_1$ and the second one with $\dot{\vec{x}}_2$ you get
$$m_1 \dot{x}_1 \cdot \ddot{\vec{x}}_1+ m_2 \dot{x}_2 \cdot \ddot{\vec{x}}_2 s= -(\dot{\vec{x}}_1 \cdot \vec{\nabla}_1 + \dot{\vec{x}}_2 \cdot \vec{\nabla}_2) V(|\vec{x}_1-\vec{x}_2|).$$
Both sides are time derivatives,
$$\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t} (m_1 \dot{\vec{x}}_1^2 + m_2 \dot{\vec{x}}_2^2) = - \frac{\mathrm{d}} {\mathrm{d} t} V(|\vec{x}_1-\vec{x}_2|).$$
This is the energy-conservation law
$$E=T+V=\frac{m_1}{2} \dot{\vec{x}}_1^2 + \frac{m_2}{2} \dot{\vec{x}}_2^2 +V(|\vec{x}_1-\vec{x}_2|)=\text{const}.$$
Now you can introduce center-mass and relative coordinates
$$\vec{s}=\frac{1}{M} (m_1 \vec{x}_1 + m_2 \vec{x}_2), \quad \vec{r}=\vec{x}_1-\vec{x}_2$$
with $M=m_1+m_2$. Then you get with some simple algebra, expressing $\vec{x}_1$ and $\vec{x}_2$ in terms of $\vec{s}$ and $\vec{r}$.
$$\vec{P}=\vec{p}_1 + \vec{p}_2 = M \dot{\vec{s}}$$
and
$$E=\frac{M}{2} \dot{\vec{s}}^2 + \frac{\mu}{2} \dot{\vec{r}}^2 + V(r),$$
where $\mu=m_1 m_2/M$ is the "reduced mass".

The center of mass goes with constant velocity $\dot{\vec{s}}=\text{const}$ due to total-momentum conservation.

The equation of motion for the relative coordinate is that of a particle with mass $\mu$ in the external radial potential $V(r)$:
$$\mu \ddot{\vec{r}}=-\vec{\nabla} V(r)=-\frac{\vec{r}}{r} V'(r).$$
Now
$$\vec{x}_1=\vec{s}+\frac{m_2}{M} \vec{r}, \quad \vec{x}_2=\vec{s}-\frac{m_1}{M} \vec{r}.$$
If now $m_2 \gg m_1$ you can set $\vec{x}_2 \simeq \vec{s}$ and $\vec{x}_1\simeq \vec{s}+\vec{r}$ and also $\mu \simeq m_1$.

Then in the center-momentum frame, $\vec{s}=0=\text{const}$, you have a description, where the light particle moves in the external central potential $V(r)$.