In Fundamental of Physics 6th ed. by Halliday, Resnick, Walker pg. 181-183, the section is about systems with varying mass: a rocket.(adsbygoogle = window.adsbygoogle || []).push({});

It is written:

Is it okay if I change M+dM with M-dM because I know that the rocket loses its mass and -dM with dM because the exhaust product has mass in positive quantity? Assume that we are at rest relative to an inertial reference frame, watching a rocket accelerate through deep space with no gravitational or atmospheric drag forces acting on it.

For this one-dimensional motion, let M b the mass of the rocket and v its velocity at an arbitrary time t (see Fig. a)

{~~~~~~~~~~~~~}--System boundary

{|--------\........... }

{|... M .....>---> v }

{|--------/........... }

{~~~~~~~~~~~~~}

------------------------------------------- x

(a)

{~~~~~~~~~~~~~~~~~~~~}--System boundary

{ -dM ..........|--------\.......... }

{<<<<--> U|. M+dM .>---> v }

{ .................|--------/.......... }

{~~~~~~~~~~~~~~~~~~~~}

------------------------------------------- x

(b)

Figure b shows how things stand a time interval dt later.

The rocket now has velocity v+dv and mass M+dM, where the change in mass dM is a negative quantity.

The exhaust products released by the rocket during interval dt have mass -dM and velocity U relative to our inertial reference frame.

So that figure (b) becomes

{~~~~~~~~~~~~~~~~~~~~}--System boundary

{. dM ..........|--------\.......... }

{<<<<--> U|. M-dM .>---> v }

{ ................|--------/.......... }

{~~~~~~~~~~~~~~~~~~~~}

------------------------------------------- x

And dM is not a negative quantity anymore because I've taken account of it in the figure (the rocket loses its mass and the exhaust products gain mass).

Next it is written:

If I'm allowed to change M+dM with M-dM and -dM with dM then Eq. 9-38 becomes Our system consists of the rocket and the exhaust products released during interval dt.

The system is closed and isolated, so the linear momentum of the system must be conserved during dt; that is,

Pi = Pf ... (9-37)

where the subscripts i and f indicate the values at the beginning and end of time interval dt. We can rewrite Eq. 9-37 as

M * v = -dM * U + (M + dM) * (v + dv) ... (9-38)

where the first term on the right is the linear momentum of the exhaust products released during interval dt and the second term is the linear momentum of the rocket at the end of interval dt.

M * v = dM * U + (M - dM) * (v + dv) ... (Eus-1)

Then it is written:

If I do Eq. Eus-1 with little algebra it becomes We can simplify Eq. 9-38 by using the relative speed vrel between the rocket and the exhaust products, which is related to the velocities relative to the frame with

(velocity of rocket relative to frame) = (velocity of rocket relative to products) + (velocity of products relative to frame).

In symbols, this means

(v + dv) = vrel + U, or

U = v + dv - vrel ... (9-39)

Substituting this result for U into Eq. 9-38 yields, with a little algebra,

-dM * vrel = M * dv ... (9-40)

Dividing each side by dt gives us

-(dM/dt) * vrel = M * (dv/dt) ... (9-41)

We replace dM/dt (the rate at which the rocket loses mass) by -R, where R is the (positive) mass rate of fuel consumption, and we recognize that dv/dt is the acceleration of the rocket. With these changes, Eq. 9-41 becomes

R * vrel = M * a ... (first rocket equation) ... (9-42)

Equation 9-42 holds at any instant, with the mass M, the fuel consumption rate R, and the acceleration a evaluated at that instant.

The left side of Eq. 9-42 has the dimensions of a force (kg m/s^2 = N) and depends only on design characteristics of the rocket engine, namely, the rate R at which it consumes fuel mass and the speed vrel with which that mass is ejected relative to the rocket. We call this term R * vrel the thrust of the rocket engine and represent it with T. Newton's second law emerges clearly if we write Eq. 9-42 as T = M * a, in which a is the acceleration of the rocket at the time that its mass is M.

dM * vrel = M * dv ... (Eus-2)

Dividing each side by dt gives us

(dM/dt) * vrel = M * (dv/dt)

I replace dM/dt by R not -R because I've taken account of this decrement in rocket mass since the beginning.

Am I right?

Therefore, I got the first rocket equation without any problem that is R * vrel = M * a.

Now I run into a problem when finding the velocity of the rocket

It is written:

My problem is that when I use Eq. Eus-2 as below How will the velocity of a rocket change as it consumes its fuel? From Eq. 9-40 we have

dv = -vrel * (dM/M)

Integrating leads to

vi S vf dv = -vrel * [Mi S Mf (1/M) dM]

Note: I use S as integral symbol.

in which Mi is the initial mass of the rocket and Mf its final mass. Evaluating the integrals then gives

vf - vi = vrel * ln( Mi / Mf ) ... (second rocket equation) ... (9-43)

for the increase in the speed of the rocket during the change in mass from Mi to Mf.

(The symbol "ln" in Eq. 9-43 means the natural logarithm.)

dv = vrel * (dM/M)

Integrating each side leads to

vi S vf dv = vrel * [Mi S Mf (1/M) dM]

And evaluating the integrals then gives

vf - vi = vrel * ln( Mf / Mi ) ... (Eus-3)

Yup! This is the problem. I cannot replace vrel by -vrel because I don't have any reason to do that.

Because Mf is always less than Mi, Eq. Eus-3's right hand side always has a negative value and that means vf is less than vi (the rocket is slowing down)

Mmmm... do you think I've done a big mistake by changing M+dM by M-dM and -dM by dM? Why is it wrong?

If I'm not wrong then is there something that I've taken account into my calculation when I calculate the second rocket equation? What is it?

Thank you very much for you help! ^_^

I really appreciate it! ^^v

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# Homework Help: A system with varying mass - a rocket

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