Okay, imagine that you are shooting a rocket down a tunnel that is 60 miles long and the rocket is travelling at 60mph, so the rocket should reach the end of the tunnel in a hour, right? Yep, in an hour. BUT if next to that tunnel you have the same length tunnel and that tunnel is a vacuum, similar to that of space. And you shoot a rocket that can fly propel it self in the vacuum, and it's travelling at 60 mph also. Which would reach the end first?.

Also imagine that the rocket is flying straight and not zagging around, I am just wondering how the resistance of air slows down something compared to the same thing in a vacuum.


Think about what you just wrote:

rocket-1 is flying 60 mph.
rocket-2 is also flying 60 mph.

They need to travel the same distance. ... Both arrive after the same period of time.

Rocket-1 would have been bigger or more efficient to keep flying 60 mph despite air resistance, but you already compensated for that else it would not fly at the same speed.

Then there is 'a rocket that can fly propel it self in the vacuum'.
Do you know any rocket which can not do that?

Sorry if I sound harsh, but the question does not make any sense in the way you asked it.

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  • $\begingroup$ Firework rockets have a solid propellant and use the oxygen from the air, so they don't work in vacuum. $\endgroup$ – b_jonas Feb 27 '13 at 9:06
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    $\begingroup$ @b_jonas Solid propellants contain oxidizer as well as fuel, just in solid form, so air is not necessary. $\endgroup$ – deadly Feb 27 '13 at 11:19

If you specify the speed of the rocket to be 60 mph, then the rocket will take the same amount of time to traverse the tunnel regardless of whether or not it's been evacuated.

The difference comes in when instead of specifying the speed of the rocket, you specify a certain amount of thrust. If two rockets have the same amount of thrust, but if one is in vacuum while the other is in air, then the rocket in vacuum will have a higher acceleration, will move faster, and will thus traverse the tunnel in a shorter period of time.

This happens because thrust specifies the force with which the rocket is being propelled. If both rockets are being propelled by the same force, but if one encounters some air resistance, then the rocket encountering resistance will experience less net force, and it therefore (by Newton's second law) will have a lower acceleration.

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As the other two answers have said, the way you've phrased the question is a bit confusing. I'll attempt to answer what I think you're asking - if I've misunderstood you just ignore this.

Start with the rocket in a vacuum. Assuming we can ignore gravity, any object travelling in a vacuum travels at a constant speed because there is nothing to slow it down. So if your rocket enters the tube at 60 mph it doesn't need to use it's motor. It will just float along at a steady 60 mph and take an hour to reach the end of the tube.

Now take the rocket travelling through air. Calculating the drag on an object from air resistance turns out to be an extremely difficult problem, and indeed one that hasn't been solved except for simple cases. However there is a good approximation for the force due to air resistance:

$$ F = \frac{1}{2} \rho v^2 C_d A $$

where $A$ is the frontal area of the object, $C_d$ is a constant that depends on the shape of the object, $\rho$ is the density of the air and $v$ is the velocity.

So the rocket travelling through air will feel a force slowing it down. There are two options: if you leave the rocket motor off the rocket will slow down and it will take longer than an hour to reach the end of the tube. That means the rocket in vacuum will reach the end of the tube first. The second option is to keep the rocket motor running to oppose the drag force $F$, in which case the rocket still takes an hour to reach the end of the tube so both rockets reach the end of their tube at the same time. However the rocket in air will have burnt some fuel along the way while the rocket in vacuum won't have used any fuel.

It's not hard to calculate the motion of the rocket in air with the engine off, though the working involves calculus and is a bit messy. The end result is that the distance travelled by the rocket, $d$, is given by:

$$ d = k \space ln \left( \frac{v_0}{k}t + 1 \right) $$

where $v_0$ is the initial velocity (60 mph) and k is the constant:

$$ k = \frac{1}{\frac{1}{2} \rho C_d A} $$

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