Landing reusable booster rockets

[paulb203]

Homework Statement

Booster rockets can be reused if they land safely. One way to slow a booster for landing is by firing its engine again. Now, thrust pushes opposite the direction the booster is moving. This technique is called propulsive landing.

Determine the thrust needed to propulsively land the booster described below.
A 25,000kg booster rocket is falling. It experiences a gravitational pull of 245,000N.
The booster's engine fires for 13 seconds before landing, slowing the booster from a velocity of
172 m/s to 0 m/s over that time.

Relevant Equations

a=v-u/(t)
a=Fnet/m

a=v-u/(t)

a=0-172/(13)

a=approx -13m/s^2

*

a=Fnet/m

-13=Fnet/25000

Fnet=-13(25000)

Fnet=-3250000N

Fg+Fth=-325000N

Fth=-325000-245000

Fth=-670000N

Any hints as to where have I gone wrong?

 

[kuruman]

What is the definition of impulse?

 

[haruspex]

a=approx -13m/s^2

I would keep one or two more digits of precision at this stage.

Fnet=-3250000N

Typo: extra 0.

Fth=-670000N

Which way are you supposed to take as positive, up or down?
Which way is the thrust?

 

[jbriggs444]

Fth=-325000-245000

Fth=-670000N

If you add 325000 and 245000, what result do you get?

 

[paulb203]

What is the definition of impulse?

Thanks.
From my Oxford Dictionary of Physics;

"The product of a force and the time for which it acts."

 

[paulb203]

Which way are you supposed to take as positive, up or down?
Which way is the thrust?

Thanks.
The thrust is up. I'm guessing that if we went back to the launch up would be positive.
But because we start at the descent I'm wondering if down is positive.
Velocity is given as 172m/s (positive) which makes me think down is positive, so up would be negative.
The phrasing of, "Determine the thrust needed..." makes me think they want the absolute value of the thrust.
Ah, the multiple choice answers are all positive; does that mean they do want the absolute value of the thrust? But how might they expect you to do the calculations; with Fg as positive and Fth as negative; or vice versa?

a=v-u(t)
a=-172/13
a=-13.2

a=Fnet/m
-13.2=Fnet/25,000
Fnet=-13.2(25,000)
Fg+Fth=-330,000
Fth=-330,000-245,000
Fth=-575,000

Abs.val. of Fth= 575,000 N

 

[paulb203]

If you add 325000 and 245000, what result do you get?

570,000
Doh!

 

[erobz]

Thanks.
The thrust is up. I'm guessing that if we went back to the launch up would be positive.
But because we start at the descent I'm wondering if down is positive.
Velocity is given as 172m/s (positive) which makes me think down is positive, so up would be negative.
The phrasing of, "Determine the thrust needed..." makes me think they want the absolute value of the thrust.
Ah, the multiple choice answers are all positive; does that mean they do want the absolute value of the thrust? But how might they expect you to do the calculations; with Fg as positive and Fth as negative; or vice versa?

a=v-u(t)
a=-172/13
a=-13.2

a=Fnet/m
-13.2=Fnet/25,000
Fnet=-13.2(25,000)
Fg+Fth=-330,000
Fth=-330,000-245,000
Fth=-575,000

Abs.val. of Fth= 575,000 N

Make a diagram if you are having trouble. Define up or down to be positive, draw vectors for the velocity, acceleration, and forces. Proceed to write equations that are consistent with the convention you chose. Then, when you are done doing that, choose the other direction as positive convention and redo the equations w.r.t. the new convention. You will get the same result if you have been consistent and there will not be a need for "absolute value".

 

[kuruman]

Thanks.
From my Oxford Dictionary of Physics;

"The product of a force and the time for which it acts."

You can go one step beyond the Oxford Dictionary definition and incorporate Newton's second law in it. Impulse $\mathbf J$ (I use boldface for vectors) is the average force $\mathbf{{\bar F}}$ times the time interval $\Delta t$ over which this force acts on mass $M$. Newton's second law then relates the impulse to the change in momentum that $$\mathbf J=\mathbf{{\bar F}}\Delta t=M\Delta \mathbf v.$$ Here, you have two forces delivering impulse to the rocket, thrust and gravity, pointing in opposite directions. You can get the answer from this directly without having to worry about the acceleration which is not necessarily constant.

 

[haruspex]

The thrust is up. I'm guessing that if we went back to the launch up would be positive.
But because we start at the descent I'm wondering if down is positive.

The positive direction isn’t inherently either, it's a matter of choice.

Velocity is given as 172m/s (positive) which makes me think down is positive, so up would be negative.

Yes, since it describes that as a velocity, not a speed, it does suggest that.

The phrasing of, "Determine the thrust needed..." makes me think they want the absolute value of the thrust.

Not really; thrust is a force and forces are vectors.

Ah, the multiple choice answers are all positive; does that mean they do want the absolute value of the thrust?

Yes, that makes it clear a positive answer is required, whether that's because the question setter only intended to ask for the magnitude or was assuming positive as down.

But how might they expect you to do the calculations; with Fg as positive and Fth as negative; or vice versa?

Entirely your choice, but to avoid confusion I would advise taking the same direction as positive for all displacements, velocities, accelerations and forces.
Of course, it gets a more complicated when some of these are oblique, unless you decompose everything to horizontal and vertical components.

 

[paulb203]

Make a diagram if you are having trouble. Define up or down to be positive, draw vectors for the velocity, acceleration, and forces. Proceed to write equations that are consistent with the convention you chose. Then, when you are done doing that, choose the other direction as positive convention and redo the equations w.r.t. the new convention. You will get the same result if you have been consistent and there will not be a need for "absolute value".

Thanks. I’ll start with up as positive direction.

I imagine at some point after launch, during the ascent, the v was 172m/s in the positive direction, or 172m/s upwards.

They, when referring to the descent, give the v as 172m/s, with downwards implied, so they’re assuming downwards as the positive direction, I think. But I’m starting with up as positive so their v becomes -172m/s.

a=v-u(t)
a=-172/13
a=-13.2

A negative a makes sense, I think, with up as the positive direction.

Now Newton’s second law.

They, when referring to Fg give 245,000N, a positive value, again assuming downwards as the positive direction. Which for me, starting with up as positive, becomes -245,000.
a=Fnet/m
-13.2=Fnet/25,000
Fnet=-13.2(25,000)

Fnet=-330,000
Fg+Fth=-330,000
-245,000+Fth=-330,000
Add 245,000 to both sides
Fth=-330,000+245,000
Fth=-85,000

Why have I ended up with a negative? If I’m assuming up as positive Fth should be positive, yeah? And I know I’m looking for a answer with magnitude 575,000.

*

I’ll now assume downwards to be the positive direction.

a=v-u/(t)

a=172/13

a=13.2

*

a=Fnet/m

13.2=Fnet/25000

Fnet=13.2(25,000)

Fnet=330,000

Fg+Fth=330,000

Fth=330,000-245,000

Fth=85,000N

Which, again, is incorrect. I'm looking for a magnitude of 575,000. And both answers should be the same in terms of positive/negative.

 

[kuruman]

When you write Fg+Fth=Fnet and think of these as magnitudes of vectors, there should be a relative negative sign on the left because the forces are in opposite directions.

 

[jbriggs444]

A negative a makes sense, I think, with up as the positive direction.

So we have a rocket at $v=-172 \text{ m/s}$ That is, 172 m/s downward.
You assume that $a$ is negative. That is, the acceleration is downward as well.
That does not sound like the scenario you are asked to assume.

A correct approach is to pick a direction that you will count as positive. Calculate what $a$ will be using that convention. If the resulting numeric value for $a$ is positive then the acceleration is in the positive direction. If the resulting numeric value for $a$ is negative then the acceleration is in the negative direction.

 

[Steve4Physics]

@paulb203, I’d guess the question should be:

A rocket (mass 25,000 kg, weight 245,000N) is descending vertically. When its speed is 172 m/s the thruster is fired for 13 s making the rocket lands at 0 m/s. Find the magnitude of the thrust.

The official answer (575,000 N) appears slightly inaccurate, If you avoid rounding errors I believe the answer is 576,000 N to 3 sig. figs.
___________

A couple of points:

It’s incorrect to write “a=v-u/(t)”. You mean a=(v-u)/t. E.g.
10 – 4/(2) = 8 but
(10 – 4)/2 = 3

Taking upwards as positive:
An object going upwards with increasing speed has positive acceleration.
An object going upwards with decreasing speed has negative acceleration.
An object going downwards with increasing speed has negative acceleration.
An object going downwards with decreasing speed has positive acceleration.

It's worth remembering that the direction of acceleration is the same as the direction of the net force which causes the acceleration.

 

[haruspex]

a=v-u(t)
a=-172/13
a=-13.2

u=-172m/s
v=0
v-u = 0-(-172) = ?

A negative a makes sense, I think, with up as the positive direction.

It doesn’t. To slow a downward velocity you need an upward acceleration.

 

[paulb203]

When you write Fg+Fth=Fnet and think of these as magnitudes of vectors, there should be a relative negative sign on the left because the forces are in opposite directions.

Thanks.
Do you mean i should write it like this:
-Fg+Fth=Fnet?

 

[paulb203]

@paulb203, I’d guess the question should be:

A rocket (mass 25,000 kg, weight 245,000N) is descending vertically. When its speed is 172 m/s the thruster is fired for 13 s making the rocket lands at 0 m/s. Find the magnitude of the thrust.

The official answer (575,000 N) appears slightly inaccurate, If you avoid rounding errors I believe the answer is 576,000 N to 3 sig. figs.
___________

A couple of points:

It’s incorrect to write “a=v-u/(t)”. You mean a=(v-u)/t. E.g.
10 – 4/(2) = 8 but
(10 – 4)/2 = 3

Taking upwards as positive:
An object going upwards with increasing speed has positive acceleration.
An object going upwards with decreasing speed has negative acceleration.
An object going downwards with increasing speed has negative acceleration.
An object going downwards with decreasing speed has positive acceleration.

It's worth remembering that the direction of acceleration is the same as the direction of the net force which causes the acceleration.

Thanks.
Yes, I prefer your statement of the problem.
And, Doh! Yes, a=(v-u)/t
And yes to the rest. Cheers.

 

[paulb203]

u=-172m/s
v=0
v-u = 0-(-172) = ?

It doesn’t. To slow a downward velocity you need an upward acceleration.

Thanks.
Doh! 0-(-172) = 172
And yes, upward acceleration.

 

[erobz]

Thanks.
Do you mean i should write it like this:
-Fg+Fth=Fnet?

If up is positive, yeah.

So given that the acceleration is positive here, we will find that the force of thrust is larger than the rockets force of weight.

 

[paulb203]

If up is positive, yeah.

View attachment 357888
So given that the acceleration is positive here, we will find that the force of thrust is larger than the rockets force of weight.

Thanks. Is the v for velocity? What about the o next to it?

 

[paulb203]

Upward = positive direction;

a=(v-u)/t

a=(0 minus -172)/t

a=172/13

a=13.2

(positive acceleration, i.e, upwards acceleration, to slow the booster down)

*

a=Fnet/m

13.2=Fnet/25,000

Fnet=13.2(25,000)

Fnet=330,000

Fg+Fth=330,000

-245,000+Fth=330,000

Fth=330,000+245,000

Fth=575,000

******

Downward = positive direction;

a=(v-u)/t

a=(0-172)/13

a=-172/13

a=-13.2

(now the acceleration is negative, but it’s still upwards acceleration, given the context, to slow the booster down)

*

a=Fnet/m

-13.2=Fnet/25,000

Fnet=-13.2(25,000)

Fnet=-330,000

Fg+Fth=-330,000

Fth=-330,000-245,000

Fth=-575,000

 

[erobz]

Thanks. Is the v for velocity? What about the o next to it?

That's initial velocity. the "o" subscript is what you will see to indicate an initial state. $v_f$ would be final velocity.

 

[erobz]

Upward = positive direction;

a=(v-u)/t

a=(0 minus -172)/t

Don't write out "minus".

$$ a = \frac{v-u}{t} = \frac{0 \frac{ \text{m}}{ \text{s}} -(-172 \frac{ \text{m}}{ \text{s}}) }{13\text{s}} = \frac{172}{13} \frac{ \text{m}}{ {\text{s}}^2} $$

Fg+Fth=330,000

It the confusion starts right here. we are dealing with signed scalars ( not vectors ) so,

$$F_{th} - F_g = 330,000 \frac{ \text{kg~m}}{ {\text{s}}^2} $$

-245,000+Fth=330,000

The value is substituted correctly, but it's not clear you understand that based on your next result.

(now the acceleration is negative, but it’s still upwards acceleration, given the context, to slow the booster down)

*

Fg+Fth=-330,000

Again, same mistake you consitently are making - you have the force of weight and thrust both positive?

One of them is negative in this convention, which one?

 

[haruspex]

It the confusion starts right here. we are dealing with signed scalars ( not vectors ) so,

$$F_{th} - F_g = 330,000 \frac{ \text{kg~m}}{ {\text{s}}^2} $$

Yes, signed scalars, and $F_g$ is negative. $F_{th} + F_g = 330,000 \frac{ \text{kg~m}}{ {\text{s}}^2}$ is correct.

 

[erobz]

Yes, signed scalars, and $F_g$ is negative. $F_{th} + F_g = 330,000 \frac{ \text{kg~m}}{ {\text{s}}^2}$ is correct.

$F_g$ is a positive scalar, and "minus" is its sign?

If something is speeding up in kinematics how often do you write $v = u - at$ where $a < 0$? If signed scalar it is not accurate choice in the name, it's surely the case in practice that $a > 0$.

If you want to teach signed scalars by definition instead of how they are used in this context and prolong this, be my guest. I'll stand aside.

 

[kuruman]

Thanks.
Do you mean i should write it like this:
-Fg+Fth=Fnet?

Yes. This is consistent with the convention up is positive and down is negative.
If you choose the convention down is positive and up is negative, you would write
Fg - Fth = -Fnet
Note that you get the second equation if you multiply both sides of the first equation by -1 but the relative sign on the left is still negative.

 

[haruspex]

$F_g$ is a positive scalar, and "minus" is its sign?

Sorry, why is $F_g$ positive? g is always taken to be a positive constant, but $F_g$ can be signed, being mg if down is positive or -mg if up is positive.
If you insist on its being positive then you are declaring it a magnitude, not a signed scalar.

If something is speeding up in kinematics how often do you write $v = u - at$ where $a < 0$? If signed scalar it is not accurate choice in the name, it's surely the case in practice that $a > 0$.

Many write $v = u - at$ in that case because they are more comfortable with magnitudes. If the intent is to work in signed scalars then the same direction should be positive for $u, v, a$, and the equation should be $v = u + at$.
The OP seems to have done this correctly in this instance.

 

[haruspex]

When you write Fg+Fth=Fnet and think of these as magnitudes of vectors, there should be a relative negative sign on the left because the forces are in opposite directions.

I see no indication that @paulb203 is thinking of them as magnitudes. Rather, the treatment appears to be consistently as scalars (preferable, to me).

Do you mean i should write it like this:
-Fg+Fth=Fnet?

No. You are quite entitled to work with scalars rather than magnitudes. Sometimes it is not at first clear which way a force will act, making scalars the natural choice.

 

[erobz]

If g is a positive constant, it was never clear to me why other variables should be treated differently because I feel like it obscures intent.

Anyhow, I think I’d just prefer they teach vectors at the point of using “signed scalers”. It seems to me I have trouble differentiating whether or not people are intending to use magnitudes or a signed scaler in equations that has already had the signs absorbed into the algebra.

But anyhow I know what you mean, and I think you know what I mean. If I didn’t see what the OP means in the 3 attempts I don’t think it hurts my case. Let’s work with vectors or let’s work with magnitudes in my opinion. I don’t think I see the utility of somewhere in the middle, and certainly have found it confusing.

 

[kuruman]

Ambiguity and confusion can be avoided if one uses unit vector notation as an intermediate step and then remove them.
Assuming down as negative and up as positive
$\mathbf F_{\text{th.}}+\mathbf F_{\text{g}}=M\dfrac{\Delta \mathbf{v}}{\Delta t}.$
$F_{\text{th.}}~(\mathbf{\hat{y}})+(245,000~\text{N})~(-\mathbf{\hat{y}})=(25,000~\text{kg})\dfrac{0-(172~\text{m/s})(-\mathbf{\hat{y}}) }{(13~\text{s})}.$
$F_{\text{th.}}-(245,000~\text{N})=(25,000~\text{kg})\dfrac{(172~\text{m/s}) }{(13~\text{s})}.$
$F_{\text{th.}}=(245,000~\text{N})+(25,000~\text{kg})\dfrac{(172~\text{m/s}) }{(13~\text{s})}.$

Assuming up as negative and down as positive
$\mathbf F_{\text{th.}}+\mathbf F_{\text{g}}=M\dfrac{\Delta \mathbf{v}}{\Delta t}.$
$F_{\text{th.}}~(-\mathbf{\hat{y}})+(245,000~\text{N})~(\mathbf{\hat{y}})=(25,000~\text{kg})\dfrac{0-(172~\text{m/s})(\mathbf{\hat{y}}) }{(13~\text{s})}.$
$-F_{\text{th.}}+(245,000~\text{N})=-(25,000~\text{kg})\dfrac{(172~\text{m/s}) }{(13~\text{s})}.$
$F_{\text{th.}}=(245,000~\text{N})+(25,000~\text{kg})\dfrac{(172~\text{m/s}) }{(13~\text{s})}.$

@haruspex and I have been over this with 1d vectors several times in the past and (I think) we have agreed to disagree. When all vector directions are known, as is the case here, I favor expressing vectors in polar form, i.e. separate the magnitude, which is always positive, from the direction which can be positive or negative. Splitting the two is easier to explain to novices who are taught that vectors have magnitude (always positive) and direction. The danger here is that some students use false logic: magnitudes are always positive, a vector is a scalar times a direction, therefore a scalar is always positive.

If both the direction and magnitude of a vector are not known, then one can assume a direction and do the algebra to find the magnitude. If the magnitude turns out to be negative, then one must conclude that the direction is in the opposite direction of the one assumed. The idea is no different from what is taught as standard practice when using Kirchhoff's laws to analyze circuits: "Assume a direction for the current and if the current turns out negative, this means that it flows in the opposite direction."