# B Rocket problem

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1. Apr 26, 2015

### Stephanus

Dear Physics Forum,
I really don't understand the concept of watt and newton.
I have two questions, perhaps someone can explain it to me.
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Case 1:
If a one newton force is applied to a one kg rocket in space with no or very little effect of gravity, for ....
A. 10 seconds, it will travel (1/2 x a x t^2 = 1/2 x 1 x10^2 = 50) 50 metres, it has consumed 50 joules, so its power is 5 watt.
B. 20 seconds, it will travel 200 metres, it consumes 200 joules, power is 10 watt.
C. 30 seconds, 450 metres, 450 joules, 15 watt
It seems that the power is increasing with each passing time(second). Its energy consume is not constant, but accelerating with each passing time.
Question1: Is that true?
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Case 2:
Before I ask, I would like to insert some equations here.
E = P.t (energy is power x time, second)
E = F.d (energy is force x distance, metre)
F = M.a (force is mass x accel)
d = 1/2 a.tt (dist is accel x time squared)
P.t = F.d
E = (F) x (d)
E = (M x a) x (1/2 a x tt)
E = 1/2 M x aa x tt
aa = 2E/M/tt
a = sqrt(2E/M)/t
F =
M x sqrt(2E/M)/t = sqrt(2EM)/t
d =
1/2 a.tt = sqrt(1/2 x E/M)/t x tt = sqrt(1/x2 x E/M) x t
Note:
F.d = sqrt(2EM)/t x sqrt(1/2 x E/M) x t = sqrt(2 x 1/2 x EE) x t / t = sqrt(EE) = E
F.d is of course (E)nergy

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A one kg rocket in space (no gravity) having a one watt power. Assuming it has a very, very efficient fuel, so the fuel mass loss is negligible.
A. After 2 seconds, its energy consume is 1 watt x 2 seconds = 2 Joule.
Its Force will be 1 newton, its distance has to be 2 metres, 1 newton x 2 metres = 2 joule
B. After 8 seconds, its energy consume is 8 joules
Its Force is 0.5 newton x its distance is 16 metres = 8 joules
C. After 32 seconds , 32 joules
Force: 0.25 Newton, distance: 128 metres.
It seems with each passing time the force is diminishing.
Is that true?

Note: It will take 5.6 billion years to push it near the speed of light
11 billion years = twice the speed of light??? But that's another question for relavity I think.

======================================================

Conclusion:
1. If the force is constant, does the power increase?
2. If the power is constant, does the force decrease?

Peace on earth

Steven

2. Apr 26, 2015

### jbriggs444

Yes. The power produced from applying a fixed force increases with the velocity of its target.

One watt power applied to what?

Rocket motors are normally rated based on their total power output -- power going to the craft plus power going to the exhaust stream. That power level turns out to be the same no matter how fast the rocket is moving.

The faster the craft is moving, the more power is going into accelerating the craft. And the less power is going into the exhaust stream. Go fast enough and the exhaust stream is actually less energetic than the unburned fuel (as viewed from the perspective of the ground that you are whizzing past).

3. Apr 26, 2015

### Stephanus

Thank you for the answer Sir, (I guess you are male judging from you avatar)

Yes. The power produced from applying a fixed force increases with the velocity of its target.

"Power increases, applying a fixed force with the velocity"
I have a little "trouble" with the word "velocity" here.
- Supposed the target is in intergalactic void, with really no or miniscule gravity. How in the hell, (hell might be right if you're in Bootes void) that the rocket knows its velocity?. Supposed there is an astronout in that rocket and he when goes to sleep, the rocket accelerates for some hours and stops before he wakes up. After he wakes up, he would feel no acceleration at all, altough its speed might be hundreds of kms per hour. Does it require more power than before he slept, to thrust or add velocity to the rocket by 1 m/s?
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One watt power applied to what?

1. Rocket motors are normally rated based on their total power output...
2. The faster the craft is moving, the more power is going into accelerating the craft...
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1. Supposed it does not use motor, but some very advanced ion thruster as in Startrek. That it ejects 1 molecule, or 1 polimer for that matter, near the speed of light to accel that 1 kg rocket by 1 m/s. Does the principle still apply?
2. The faster the craft...,
Perhaps you are refering air or ground friction or gravitation bound? Supposed the rocket is in intergalactic void where there's no friction or gravitation effect. Does it still require more power going into accelerationg the craft?

Again, thank you very much for you attention, sir.

Last edited: Apr 26, 2015
4. Apr 26, 2015

### jbriggs444

Pick a reference frame, any reference frame. The principle applies regardless.

Yes. Do the math. Take a 1.000001 kg object at rest and eject one milligram of exhaust at 1000 meters per second. Compute the total energy of the remaining 1 kg object and of the 1 milligram exhaust. Repeat the exercise but assume that the object starts with a velocity of one meter per second.

Compute the amount by which total energy increases in the first case.
Compute the amount by which total energy increases in the second case.
Compare the two.

5. Aug 7, 2015

### Stephanus

Dear PF Forum,
Can I refresh my understanding of kinetic energy?
Supposed a rocket floating in space. With relative velocity 0 wrt to a planet. Let's say that the rocket is some distance away, so that the effect of that planet gravity is very small.
The rocket is 11 kg.
Then, the rocket propels its material 1 kg to the "west".
It only produces 220 joules for that process.
In momentum conservation, we'll have...
$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
Here:
m1: 10 kg
m2: 1 kg
n: The ratio m1/m2 = 10
Because u1 and u2 is zero, the rocket is at rest first wrt planet, so
$0 = m_1v_1 + m_2v_2$
The rocket produces 220 joules for the ejecting process, so...
$\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = 220$
$m_1v_1 = -m_2v_2$
$m_2 = \frac{1}{n}m_1$
$v_2 = -nv_1$
combine those equations, we'll have
$\frac{1}{2}m_1v_1^2 + \frac{1}{2}nm_1v_1^2 = 220$
$\frac{1}{2}m_1v_1^2(1+n) = 220$
$v_1 = 2$
So, M1 will travel to the 'east' 2 m /s and m2 will travel to the 'west' 20 m/s is that so?

6. Aug 8, 2015

### Qwertywerty

Yes , I believe it's correct .

7. Aug 16, 2015

### Stephanus

Dear PF Forum,
I have a question about rocket problem. That I asked in previous thread, but I didn't get enough feed back.
Actually I want to know the relation between watt and newton/power and force. But before that I'd like to know about this answers first.
A rocket is floating in space.
The mass of the rocket: 1,000 kg.
The rocket spends energy for this action for 7,200,000 joules
Let's say n is 10
So the rocket eject 1/10th of it's mass backward to propel the rocket forward.
So $M_{rocket} = 1000 kg$
$M_{body} = M_1; M_{ejecta} = M_2$

Based on momentum equation
$M_1 * V_1 = M_2 * V_2$
Based on kinetic energy equation
$0.5 M_1 * {V_1}^2 + 0.5 M_2 * {V_2}^2 = 7,200,000 joules$

So, trough a lengthy calculation I come up with this:
$M_1 = 900 kg$
$M_2 = 100 kg$, that is easy enough.
$V_1 = 40m/s$
$V_2 = 360m/s$

I calculate the energy consumed by each objects
$Ek_{body} = 0.5 * 900 kg * 40 * 40 = 720,000 joules$
$Ek_{ejecta} = 0.5 * 100 kg * 360 * 360 = 6,480,000 joules$

So the rocket consumes energy much smaller then its ejecta?

I play around with n. Now I plug 100 into n
It means that the rocket throw 1/100th of its mass as the ejecta, to propel forward according to Newton third law:

$M_1 = 990kg$
$M_2 = 10kg$
$V_1 = 12.06m/s$
$V_2 = 1,193.98m/s$
$Ek_1 = 72,000 joules$
$Ek_2 = 7,128,000 joules$
That low?

n=1000
$Ek_1 = 7,200 joules$
$Ek_2 = 7,192,800 joules$

Is my calculation wrong? So a rocket really wastes its energy enormously?

8. Aug 16, 2015

### paisiello2

I think your calculation might be right but your assumption for n is not realistic. Most real life rockets have n=1.11 to 1.25.

Also, all rockets that have achieved orbit are multiple stage rockets so there is an additional loss of mass there.

9. Aug 16, 2015

### Staff: Mentor

In terms of energy that goes to rocket and exhaust, rockets are horribly inefficient at low speeds, that is correct. There is no more efficient way in space, however. In an atmosphere, you can accelerate the air around the rocket, which is much more effective.

The ratio is better in a frame where the rocket has some initial speed. That shows that the energy ratio is not a very useful quantity. Rockets are all about velocity changes ("delta v").

10. Aug 16, 2015

### Staff: Mentor

Now you have one thread with plenty of feedback.

As was told to you each and every time you asked, yes, most of a rocket's energy goes into the exhaust at low speeds where they are very inefficient.

Thread closed. And don't reopen another one, the answer is still the same as the first three times and there is no need to say it again.