Variable mass and rockets

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username2436711
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#1
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#1
When solving variable mass problems with rockets ejecting fuel/ gas, can I calculate the thrust by considering the force exerted on the exhaust since it is equal and opposite?

i.e. dp/dt=(m)dv/dt + (v)dm/dt

where dm/dt = mass of exhaust ejected at speed v (cancelling first term since speed of ejection constant) ?

many thanks, sorry I don't have my specific example atm
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Eimmanuel
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#2
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(Original post by marinacalder)
When solving variable mass problems with rockets ejecting fuel/ gas, can I calculate the thrust by considering the force exerted on the exhaust since it is equal and opposite?

i.e. dp/dt=(m)dv/dt + (v)dm/dt

where dm/dt = mass of exhaust ejected at speed v (cancelling first term since speed of ejection constant) ?

many thanks, sorry I don't have my specific example atm

I hope the physics would be so simple. Unfortunately, it is NOT.

The thrust can be calculated as  ={{v}_{gas}}\frac{dm}{dt}

where vgas is the velocity of the exhaust gas dm relative to the rocket m.

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

This is abused of calculus.

The v in (m)dv/dt and (v)dm/dt is actually “different”. The dv/dt should be the acceleration of the rocket + exhaust gas while the v is the relative velocity of the exhaust.


You can look at the actual derivation in this Wikipedia.

https://en.wikipedia.org/wiki/Variable-mass_system


You can try solving the following
Spoiler:
Show


Name:  rocket01.jpg
Views: 65
Size:  71.9 KB


Ans : (a) 3.90 × 107 N (b) 3.20 m/s2

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Forest 4ever
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#3
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#3
sounds like a receipe for a nuclear war
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steve44
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#4
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#4
You Kim Jong-un?
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username2436711
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#5
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#5
(Original post by Eimmanuel)
I hope the physics would be so simple. Unfortunately, it is NOT.

The thrust can be calculated as  ={{v}_{gas}}\frac{dm}{dt}

where vgas is the velocity of the exhaust gas dm relative to the rocket m.

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

This is abused of calculus.

The v in (m)dv/dt and (v)dm/dt is actually “different”. The dv/dt should be the acceleration of the rocket + exhaust gas while the v is the relative velocity of the exhaust.


You can look at the actual derivation in this Wikipedia.

https://en.wikipedia.org/wiki/Variable-mass_system


You can try solving the following
Spoiler:
Show





Name:  rocket01.jpg
Views: 65
Size:  71.9 KB


Ans : (a) 3.90 × 107 N (b) 3.20 m/s2

Thanks for clearing that up
When you say it’s abused of calculus, do you mean I’ve differentiated wrong? Or that the terms just have the different meanings? Like I used implicit/chain rules, but was that expression wrong?




Thanks for clearing that up When you say it’s abused of calculus, do you mean I’ve differentiated wrong? Or that the terms just have the different meanings? Like I used implicit/chain rules, but was that expression wrong?
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Eimmanuel
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#6
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#6
(Original post by marinacalder)
Thanks for clearing that up When you say it’s abused of calculus, do you mean ….….


I mean that you are NOT supposed to use product rule from calculus.



A lot of physics teachers or tutors had applied product rule to differentiate the product of mass and velocity, thinking that it is ok but it is wrong.

https://www.youtube.com/watch?v=oGe_dvjz9kg

https://www.youtube.com/watch?v=BB_B7Vkj2XY&t=151s

The result does not give the correct answer for all the situation in variable mass problems.



If you go to the Wikipedia link, you would see there was a paper that wrote about the “On the use and abuse of Newton's second law for variable mass problems
You don't need to read the whole paper, just the first two pages and it will reveal why it is wrong.
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