# Does it take longer to accelerate from high speed to higher

danielhaish
I wanted to ask this question for very long .
lets say we have rocket in mass of 1kg in space that moving in speed of 100 m/s if we want the rocket to speed up, we need to spent some energy that equal to mv^2/2.
in our case it 1*200^2/2 it 2000 Minos the energy we already have which is 100^2*1/2 =5000 .
so the total amount is 20,000-5000 it 15,000 some unit
and if we would like it to speed up from 0m/s to 100m/s we need to spent
1*100^2/ it 5000.
so is it harder to speed up from 100m/s to 200m/s then speeding up from 0m/s to 1000m/s

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Whether something is "harder" may depend on how you do it. How do you propose to accelerate a rocket in the vacuum of space?

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Edited.

Does it take longer to accelerate from:
a) 0 to 100m/s or
b) 100m/s to 200m/s?

It depends:

If the net force and mass are constant, that means the acceleration is constant (F=ma). In that case, it takes the same time for a) and b) because the velocity-changes are the same.

If the power supplied is constant, then b) takes longer because the required increase in kinetic energy is bigger so it takes longer for the extra energy to be supplied.

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hutchphd and danielhaish
Joe591
Well, yes, it is harder. You will require a larger impulse to accelerate the same mass from 0-1000m/s. Note that this can either be a small force applied over a very long period of time or it can be a very large force over a small period of time. More fuel will be required and more energy will be required.

sophiecentaur and danielhaish
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For a rocket in space, the energy from the engine accelerates the rocket forward and the spent fuel backwards. From an inertial (non-accelerating) frame of reference, for a rocket producing constant thrust, the power output to the rocket and spent fuel is constant, regardless of the rocket's speed. This is true even when the rocket velocity exceeds that of the spent fuel relative exit velocity, so that both are moving forward with respect to some inertial frame of reference.

danielhaish
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Well, yes, it is harder. You will require a larger impulse to accelerate the same mass from 0-1000m/s.
This is the opposite of what the OP suggests. But it is not clear to me if there is a type-o in the question.

Joe591
This is the opposite of what the OP suggests. But it is not clear to me if there is a type-o in the question.
I might have read the question too quickly. I see my mistake now. the explanation holds though. it will def be ''harder'' to accelerate from 0-1000 m/s, but yes, it is the opposite of what the OP suggests. it seems it won't allow me to edit the answer.

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Summary:: so is it harder to speed up from 100m/s to 200m/s then speeding up from 0m/s to 1000m/s

I wanted to ask this question for very long .
lets say we have rocket in mass of 1kg in space that moving in speed of 100 m/s if we want the rocket to speed up, we need to spent some energy that equal to mv^2/2.
in our case it 1*200^2/2 it 2000 Minos the energy we already have which is 100^2*1/2 =5000 .
so the total amount is 20,000-5000 it 15,000 some unit
and if we would like it to speed up from 0m/s to 100m/s we need to spent
1*100^2/ it 5000.
so is it harder to speed up from 100m/s to 200m/s then speeding up from 0m/s to 1000m/s
I don't agree with your logic. Because F=mA, it takes more force or more time to accelerate from 0 to 1000m/s. A rocket in space is not affected by its relative velocity to Earth (unless you consider aerodynamics or gravity). It would see acceleration from 100m/s to 200m/s as the same as acceleration from 0m/s to 100m/s.

danielhaish and Lnewqban
For a rocket in space, the energy from the engine accelerates the rocket forward and the spent fuel backwards. From an inertial (non-accelerating) frame of reference, for a rocket producing constant thrust, the power output to the rocket and spent fuel is constant, regardless of the rocket's speed. This is true even when the rocket velocity exceeds that of the spent fuel relative exit velocity, so that both are moving forward with respect to some inertial frame of reference.

I don't understand, could you clarify this part? A force has power ##\vec{F} \cdot \vec{v}##, so
$$P = \vec{v}_{\mathrm{rel}} \frac{dm}{dt} \cdot \vec{v}$$seems to depend on ##\vec{v}##, for constant ##\dot{m}## and exhaust velocity ##\vec{v}_{\mathrm{rel}}##.

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The OP seems to consider spending more energy as being 'harder', in which case the answer is yes, it requires more energy to accelerate from 100 m/s to 200 m/s compared to going from 0 m/s to 100 m/s. I assume the OP means 0 m/s to 100 m/s, even though they wrote 0 m/s to 1000 m/s more than once.

Note that the case of a rocket is more complicated, as you have changing mass and your propellant is being accelerated along with the payload.

danielhaish
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For a rocket in space, the energy from the engine accelerates the rocket forward and the spent fuel backwards. From an inertial (non-accelerating) frame of reference, for a rocket producing constant thrust, the power output to the rocket and spent fuel is constant, regardless of the rocket's speed. This is true even when the rocket velocity exceeds that of the spent fuel relative exit velocity, so that both are moving forward with respect to some inertial frame of reference.
That's very interesting (not being sarcastic!). From the rocket's frame, the engine provides a constant force and constant power. This gives constant acceleration, so each seond the speed-increase is constant and the energy used is constant.

But for an observer on the ground, the speed-increase each second is constant but the kinetic energy-increase each second gets bigger because it depends on v².

I guess it may be to do with the division of the energy between the rocket's kinetic energy and the exhaust gas's kinetic energy. Will think about it.

Science Advisor
That's very interesting (not being sarcastic!). From the rocket's frame, the engine provides a constant force and constant power. This gives constant acceleration, so each seond the speed-increase is constant and the energy used is constant.
Note that in the rocket's rest frame, all the kinetic energy goes into the exhaust, but it is not conserved over time, because the frame is not inertial.

Steve4Physics
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That's very interesting (not being sarcastic!). From the rocket's frame, the engine provides a constant force and constant power. This gives constant acceleration, so each seond the speed-increase is constant and the energy used is constant.

Remember that you're losing mass out the back, so your acceleration is actually increasing if the force remains constant.

sophiecentaur
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I guess it may be to do with the division of the energy between the rocket's kinetic energy and the exhaust gas's kinetic energy. Will think about it.

See post #2:

How do you propose to accelerate a rocket in the vacuum of space?

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The fact that everyone is getting different answers (for a car say vs. a rocket) is probably due to the problem being poorly worded. For one thing, kinetic energy is frame dependent, and thus is a poor choice for measurement of energy required to do one acceleration vs the other. That's why I think the rocket is probably the better example since it is usually thought of as proper acceleration, while a car is almost always considered in a context of coordinate acceleration, the energy of which, as I said, is dependent on the coordinate system of choice.

So what do we mean by 'harder', and what are the initial conditions for our two cases? Against what is work to be performed? The last question is probably the most important.

With a rocket, it usually takes the most energy to do the first 100 m/sec (of proper acceleration) because the rocket masses less and less as it continues on. But if we start with two rockets of equal mass, one already moving 100 m/sec relative to the other, then it will take identical energy (fuel) to accelerate either of them by that much.

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How do you propose to accelerate a rocket in the vacuum of space?

Not sure if that's rhetorical and/or sarcastic, but nevertheless...

Method 1. Expel nass from the rocket and rely on conservation of momentum. In this case, in the non-inertial frame of the rocket, we can move from 0 to 100m/s (A) and from 100m/s to 200m/s (B) using the same amount of energy. In fact it requires less energy for B because the mass of the rocket has been reduced

Method 2. Fire material at the rocket so it bounces off (solar sail principle). From the (inertial) frame of the system from whch we fire, B requires more energy than A

Method 3. Use another rocket and a tow rope. (No doubt Elon Musk will be announcing space-tugs soon.) But now we also need extra energy to accelerate the 2nd rocket.

vanhees71 and danielhaish
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Not sure if that's rhetorical and/or sarcastic, but nevertheless...

It was neither. The answer depends on the propulsion mechanism.

Method 1. Expel nass from the rocket and rely on conservation of momentum. In this case, in the non-inertial frame of the rocket, we can move from 0 to 100m/s (A) and from 100m/s to 200m/s (B) using the same amount of energy. In fact it requires less energy for B because the mass of the rocket has been reduced

Method 2. Fire material at the rocket so it bounces off (solar sail principle). From the (inertial) frame of the system from whch we fire, B requires more energy than A

These will lead to different calculations with potentially different conclusions.

danielhaish
It was neither. The answer depends on the propulsion mechanism.

These will lead to different calculations with potentially different conclusions.
sorry for not being update but I can't see the difference I used this equation for speeding up mv^2/2

danielhaish
For a rocket in space, the energy from the engine accelerates the rocket forward and the spent fuel backwards. From an inertial (non-accelerating) frame of reference, for a rocket producing constant thrust, the power output to the rocket and spent fuel is constant, regardless of the rocket's speed. This is true even when the rocket velocity exceeds that of the spent fuel relative exit velocity, so that both are moving forward with respect to some inertial frame of reference.
so more energy is spent but it speed up at the same time

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sorry for not being update but I can't see the difference I used this equation for speeding up mv^2/2

Problems arise using ##\frac {1}{2}mv^2## incorrectly. Try working through this if you want, to illustrate the problems.

You eject a 2kg object, mass m, out of your rocket engine at 10m/s relative to your rocket. The object's kinetic energy is ##\frac {1}{2}*2*10^2 = 100J##. The object has 100J kinetic energy as perceived from the rocket's frame of reference.

You then accelerate the rocket, increasing speed by 10m/s. The object is now moving at 10+10 = 20m/s away from you and has kinetic energy ##\frac {1}{2}*2*20^2 = 400J##. as perceived from the rocket's frame of reference.

The object's kinetic energy has apparently increased from 100J to 400J, but it has received no additional energy! The difference is due to the rocket's speed-change.

So ##\frac {1}{2}mv^2## has to be used with caution, or it can lead to incorrect conclusions.

sophiecentaur, vanhees71 and danielhaish