Physics in troubles: the real equation of force is F = ma and not F = dp/dt

[martillo]

There's something that have passed unperceived Physics that would severely affect it in many areas because of its consequences.

Please read with attention, the problem was described with care.

 

The real equation of force is F = ma

 

Today's Physics is stating that the Equation of Force is F = dp/dt.

We will analyze the equation of motion of rockets to see that the real Equation of Force is:

F = ma

A rocket has variable mass in its trajectory and it's important to see its motion equation.

Let m be its variable mass at any instant in its movement composed by the mass of the rocket plus the mass of its contained fuel.

I have made a search in the internet about rocket motion equations and all the sites agree in the equation:

F = m(dv/dt) = –v_e(dm/dt) where v_e is the speed of the expelled fuel relative to the rocket. 

One web site:

http://www.braeunig.us/space/propuls.htm

They all agree that the force acting on the rocket is due to the expelled mass and is F = –v_e(dm/dt) and that the equation of motion is F = m(dv/dt) = ma.

I assume the equation have been completely verified experimentally with enough precision from a long time ago.

It is evident that it is used the equation:

F = ma

 for the force and not:

F = dp/dt

By definition p = mv and dp/dt = m(dv/dt) + v(dm/dt).

They derive the rocket's equation of motion based on the principle of conservation of momentum but considering the momentum of the rocket with the totality of the fuel (the contained plus the expelled fuel) at any moment and stating dp/dt = 0. After that they derive the equation of motion of the rocket as:
m(dv/dt) = –v_e(dm/dt) and specifically say that  the force on the rocket is:

F = m(dv/dt) = –v_e(dm/dt)

m(dv/dt) = ma then it is clear that what is finally applied to the rocket to determine its movement is the equation F = ma and not F = dp/dt.

 

This indicates that today's Physics is wrong stating the Equation of Force as F = dp/dt.

The right Equation of Force is F = ma even when mass varies.

 

Note that the natural derivation of the famous equation E = mc² by Relativity Theory has no sense since it is based in the wrong relation F = dp/dt.

Relativity Theory becomes a wrong theory since it is based on a wrong law.

 

By definition:

p = mv

With partial derivatives:

dp/dt = m(dv/dt) + v(dm/dt)

Now as:

F = m(dv/dt)

then the valid Equation of Momentum and Force is

dp/dt = F + v(dm/dt)

 

Then the principle of conservation of the momentum p = mv must determine that dp/dt = 0 when no forces are applied and when there's no variation on the considered mass.

It can be observed that this principle can be applied to the rocket as was applied in the cited cases giving the same result. Considering the mass m' of the composed body of the rocket and the total fuel (the contained plus the expelled fuel) which does not vary:

F’ = 0 and dm'/dt = 0

then the thrust equation can be derived (as below):

m(dv/dt) = –v_e(dm/dt)

where m is the mass of the rocket with its contained fuel.

Finally:

F = ma = m(dv/dt) = –v_e(dm/dt)  is the force exerted on the rocket.

F = –v_e(dm/dt)

 

Rocket thrust force derivation

The thrust equation of the rocket is derived here considering the approximation that the mass of the expelled fuel of the rocket is negligible compared with the mass of the mass of the rocket plus the mass of its contained fuel:

Momentum and Force equations:

p = mv, F = ma = mdv/dt

dp/dt = mdv/dt + vdm/dt = F + vdm/dt

 

Considerations:

a) masses equations:

m = mass of the rocket with its contained fuel

m_e = mass of the expelled fuel

m’ = m + me = constant = total mass of the system rocket with total fuel

dm’/dt = dm/dt + dm_e/dt = 0

Then: dm_e/dt = -dm/dt

b) velocities equations (one dimension):

v = absolute velocity of the rocket

v_e = velocity of the expelled fuel in relation to the rocket assumed constant

u = absolute velocity of the expelled fuel

v_e = v – u = constant

dv_e/dt = dv/dt – du//dt = 0

Then: du/dt = dv/dt

 

Derivation:

Total momentum of the system rocket with total fuel: p’ = 0

p’ = mv + m_eu = mdv/dt + v/dm/dt + m_edu/dt + u/dm_e/dt

Considering: du/dt = dv/dt and dm_e/dt = -dm/dt

p’ = (m + m_e)dv/dt + (v-u)dm/dt

As v – u = v_e and considering m_e << m (m + m_e ≈ m) then:

p’ ≈ mdv/dt + v_edm/dt and as p’ = 0 then:

mdv/dt ≈ -v_edm/dt

Finally F = ma = mdv/dt ≈ -v_edm/dt under the approximation m_e << m

Then the rocket thrust force is: F ≈ -v_edm/dt

 

 

[swansont]

16 minutes ago, martillo said:

specifically say that  the force on the rocket is:

F = m(dv/dt) = –v_e(dm/dt)

Where? They number the equations. What number is this equation?

I see equation 1.5, which says m(dv/dt) = –vrel(dM/dt) but it’s not equated to force.

[martillo]

2 minutes ago, swansont said:

Where? They number the equations. What number is this equation?

I see equation 1.5, which says m(dv/dt) = –vrel(dM/dt) but it’s not equated to force.

Yes, equation 1.5. Right below they say:

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

[swansont]

Just now, martillo said:

Yes, equation 1.5. Right below they say:

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

They have units of force because it’s a calculation of dp/dt. A force and its reaction force from Newton’s 3rd law. 

Newton’s second law refers to the net force, not individual forces, and is indeed given by dp/dt

[Genady]

From your link:

[martillo]

14 minutes ago, Genady said:

From your link:

Yes, but they do not apply this in the derivation of the thrust force. They just say:

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

16 minutes ago, swansont said:

Newton’s second law refers to the net force, not individual forces,

The net force on the rocket is just the thrust force. There's no other force considered.

Edited February 21, 2023 by martillo

[Genady]

9 minutes ago, martillo said:

they do not apply this in the derivation of the thrust force

Yes, they do.

They start the derivation with calculating dP/dt by expressing delta-P/delta-t and then taking delta-t to dt; see 1.3 and after it.

Then they separate two parts of the dP/dt to get the equation 1.5.

Edited February 21, 2023 by Genady

[martillo]

2 minutes ago, Lorentz Jr said:

 These are the forces the rocket and the ejected fuel exert on each other. The net force on the whole system is zero

Right, this is what is considered in the problem. No external forces. The force exerted on the rocket is the thrust force.

[zapatos]

Are you saying that rockets don't actually work?

[martillo]

8 minutes ago, Genady said:

Yes, they do.

They start the derivation with calculating dP/dt by expressing delta-P/delta-t and then taking delta-t to dt; see 1.3 and after it.

Then they separate two parts of the dP/dt to get the equation 1.5.

That is the momentum p of the total system rocket + expelled fuel.

They do not apply that on the force on the rocket. In the equation 1.5 m is the mass of the rocket alone. To relate equation 1.5 to the thrust force they just say: 

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it." That is not applying F = dp/dt anyway.

3 minutes ago, zapatos said:

Are you saying that rockets don't actually work?

I'm saying that because of rockets work very well the equation of force is F = ma and not F = dp/dt.

Edited February 21, 2023 by martillo

[Genady]

9 minutes ago, martillo said:

No external forces.

Right. And then they say, "Because there are no external forces, dP/dt=0." That is, dP/dt = external forces.

Edited February 21, 2023 by Genady

[martillo]

8 minutes ago, Genady said:

Right. And then they say, "Because there are no external forces, dP/dt=0." That is, dP/dt = extrenal forces.

They apply dp/dt to the total system rocket + expelled fuel but they do not apply it to the rocket alone, for the rocket they apply F = ma just saying:

 "The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

8 minutes ago, Genady said:

Right. And then they say, "Because there are no external forces, dP/dt=0." That is, dP/dt = external forces.

Right for the total system but what about for the rocket? They don't apply the same for the rocket.

Edited February 21, 2023 by martillo

[Genady]

2 minutes ago, martillo said:

for the rocket they apply F = ma just saying:

 "The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

No, they don't apply F=ma to say it. They take the result of previous calculation which they got from F=dP/dt, and call a component of that result thrust. It is a force, but they did not derive it from F=ma, but from F=dP/dt.

[martillo]

Just now, Genady said:

No, they don't apply F=ma to say it. They take the result of previous calculation which they got from F=dP/dt, and call a component of that result thrust. It is a force, but they did not derive it from F=ma, but from F=dP/dt.

You don't see it yet. They apply dp/dt to the total system, not to the rocket. For the rocket they apply:

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

This is to apply F = ma here.

[sethoflagos]

edit: deleted

Edited February 21, 2023 by sethoflagos

[Genady]

6 minutes ago, martillo said:

This is to apply F = ma here.

No, this is not. They don't apply anything else to get the thrust. They already got it from dP/dt=0.

Edited February 21, 2023 by Genady

[martillo]

7 minutes ago, Genady said:

No, this is not. They don't apply anything else to get the thrust. They already got it from dP/dt=0.

 

(edit: xposted with sethoflagos)

Just below equation 1.5 they just say that because of units matching the left side is the force. So, this is to apply F = ma.

Edited February 21, 2023 by martillo

[Genady]

7 minutes ago, martillo said:

Just below equation 1.5 they just say that because of units matching the left side is the force. So, this is to apply F = ma.

Saying that the units match the units of force is not the same as applying F=ma.

It is only your misinterpretation of their derivation, that they apply F=ma.

[martillo]

2 minutes ago, Genady said:

Saying that the units match the units of force is not the same as applying F=ma.

In this case it is. There's no p = mv for the rocket anywhere. They only apply dP/dt for the total system.

[Genady]

1 minute ago, martillo said:

In this case it is. There's no p = mv for the rocket anywhere. They only apply dP/dt for the total system.

That is how it should be applied.

[martillo]

2 minutes ago, Genady said:

That is how it should be applied.

Applying DP/dt they arrive at the equation: m(dv/dt) = v_rel(dm/dt) where m is the mass of the rocket at any instant. How they calculate the thrust force now? Just saying:

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

Where is the consideration of dp/dt = d(mv)/dt with the mass of the rocket? Nowhere. They just apply F = m(dv/dt).

[Genady]

Just now, martillo said:

Applying DP/dt they arrive at the equation: m(dv/dt) = v_rel(dm/dt) where m is the mass of the rocket at any instant. How they calculate the thrust force now? Just saying:

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

Where is the consideration of dp/dt = d(mv)/dt with the mass of the rocket? Nowhere. They just apply F = m(dv/dt).

OK, one more time.

They start from dP/dt=0.

Then they rewrite this equation in such a way that the characteristics of the rocket are separate from the characteristics of the ejection.

This allows them to express force on the rocket separately from the force on the ejection.

The end.

They don't need and don't apply F=ma anywhere, and they don't need dP/dt separately for the rocket.

If you don't get this, I can't help you anymore.

[martillo]

24 minutes ago, Genady said:

OK, one more time.

They start from dP/dt=0.

Then they rewrite this equation in such a way that the characteristics of the rocket are separate from the characteristics of the ejection.

This allows them to express force on the rocket separately from the force on the ejection.

The end.

They don't need and don't apply F=ma anywhere, and they don't need dP/dt separately for the rocket.

If you don't get this, I can't help you anymore.

I said at the OP that was something that passed unperceived in Physics, I don't expect it to be perceived as fast now.

Edited February 21, 2023 by martillo

[Lorentz Jr]

3 hours ago, martillo said:

F = m(dv/dt) = –v_e(dm/dt) where v_e is the speed of the expelled fuel relative to the rocket.

Using F = ma (Newton's second law) for the mass dm_gas of exhaust gasses, we have the internal force of the rocket acting on the gas:

F_int = dm_gas (-v_e/dt) = -v_e dm_gas/dt

And the action-reaction rule (Newton's third law) says the exhaust gasses exert an equal and opposite force on the rocket.
Defining F_ext as any external force that might be acting on the rocket, F = ma for the rocket is

F_tot = F_ext + F_int =  F_ext + v_e dm_gas/dt = m_rocket a = m_rocket dv/dt

Defining the final (backwards) speed of the exhaust gasses v_x = v_e - v, and remembering dm_rocket = -dm_gas,

F_ext = -v_e dm_gas/dt + m_rocket dv/dt = -(v + v_x) dm_gas/dt + m_rocket dv/dt = v dm_rocket/dt + m_rocket dv/dt - v_x dm_gas/dt

F_ext = d(m_rocketv)/dt + (-v_x) dm_gas/dt = dp_rocket/dt + dp_gas/dt = dp/dt

You can see there are two contributions to dp/dt: The change in mv of the rocket, plus however much (forward) momentum is carried away by the exhaust gasses.

 

Edited February 21, 2023 by Lorentz Jr

[martillo]

26 minutes ago, Lorentz Jr said:

Using F = ma (Newton's second law) for the mass dm of exhaust gasses, we have the internal force of the rocket acting on the gas:

F_int = dm (-v_e/dt) = -v_e dm/dt

Do you really understand what you say? You say "using F = ma...". That's the point! If you use F = ma, you are not using F = dp/dt. You know: dp/dt = mdv/dt + vdm/dt. They are different things particularly when mass varies (dm/dt not zero). What I claim is precisely that they use F = ma and not F = dp/dt for the rocket.

Edited February 21, 2023 by martillo