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joe_cool2

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So, I need to make an Excel spreadsheet that realistically models the motion of a toy rocket. I need to take drag, gravity, and thrust into account, keeping in mind that the latter two will be influenced by the change in mass. The equation for force I must use, then, is:

F = ma + v(dm/dt)

According to this thread: https://www.physicsforums.com/showthread.php?t=263647,

I should just use the rocket’s instantaneous rest frame to say that v=0 and therefore F = ma. D_H, a PF mentor, says that the force inside this instantaneous rest frame is constant. However, because of another post by HallsofIvy, I am convinced that the force when measured from Earth’s reference frame is not constant. This does not make sense intuitively to me because as D_H says, the acceleration is constant regardless of refrence frame. However, it does make sense mathematically.

I will post HallsofIvy’s steps here, with a slight change, because I think there was a small error towards the end. I come to a slightly different conclusion, but the basic idea is the same, that the force when measured from Earth’s reference frame is not constant:

(at+ b)dv/dt+ av= F gives dv/dt= (F- av)/(at+ b) or dv/(F-av)= dt/(at+b).

Here is where I begin to differ a little from Ivy:

Integrating both sides: (-1/a)ln|F- av|+ C = (1/a)ln|at+ b|+ C

Raise both sides to e: e^(-1/a)(F – av) = e^(1/a)|at+b|

e^(2/a)(at+b)=|F-av|

So the question is whether I’m right about this: that I have to use the force that is calculated relative to the Earth’s reference frame. I think this is true because I need to take into account drag and gravity, so I need to add all of the forces together, keeping them in the same reference frame. After all, to calculate drag, I need to know the velocity, and in order to do that, I must consider the acceleration due to the combined effects of weight and force of drag (using the assumption that these forces cause some net acceleration, which causes a velocity, which causes drag.)

Am I wrong about this? Is there some reason that I actually can just use a constant thrust in looking at the net force? And, if I could do this, does a manufacturer saying that the toy rocket's engine exerts 40.9 N mean that, as D_H seems to suggest, the manufacturer is talking about the instantaneous rest frame?

Thank you for your time.

F = ma + v(dm/dt)

According to this thread: https://www.physicsforums.com/showthread.php?t=263647,

I should just use the rocket’s instantaneous rest frame to say that v=0 and therefore F = ma. D_H, a PF mentor, says that the force inside this instantaneous rest frame is constant. However, because of another post by HallsofIvy, I am convinced that the force when measured from Earth’s reference frame is not constant. This does not make sense intuitively to me because as D_H says, the acceleration is constant regardless of refrence frame. However, it does make sense mathematically.

I will post HallsofIvy’s steps here, with a slight change, because I think there was a small error towards the end. I come to a slightly different conclusion, but the basic idea is the same, that the force when measured from Earth’s reference frame is not constant:

(at+ b)dv/dt+ av= F gives dv/dt= (F- av)/(at+ b) or dv/(F-av)= dt/(at+b).

Here is where I begin to differ a little from Ivy:

Integrating both sides: (-1/a)ln|F- av|+ C = (1/a)ln|at+ b|+ C

Raise both sides to e: e^(-1/a)(F – av) = e^(1/a)|at+b|

e^(2/a)(at+b)=|F-av|

So the question is whether I’m right about this: that I have to use the force that is calculated relative to the Earth’s reference frame. I think this is true because I need to take into account drag and gravity, so I need to add all of the forces together, keeping them in the same reference frame. After all, to calculate drag, I need to know the velocity, and in order to do that, I must consider the acceleration due to the combined effects of weight and force of drag (using the assumption that these forces cause some net acceleration, which causes a velocity, which causes drag.)

Am I wrong about this? Is there some reason that I actually can just use a constant thrust in looking at the net force? And, if I could do this, does a manufacturer saying that the toy rocket's engine exerts 40.9 N mean that, as D_H seems to suggest, the manufacturer is talking about the instantaneous rest frame?

Thank you for your time.

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