Rockets

[skulldude]

In space there is neither friction nor gravity.Then how is it possible to stop a missile or a spacecraft in the space. We need friction to stop a moving vehicle. Is there a special mechanism behind it.

[CaptainPanic]

Regarding stopping in space (or at least slowing down)

The law of physics that you need is: Action = Reaction.

 

So, if you push in one way, your body will move back the other way (like a recoil of a gun).

 

Imagine you're moving through space in the direction of Mars. Mars is straight ahead. You fire a gun at the direction of Mars. The recoil of the gun will then actually slow you down a little. Keep firing, and eventually you will stop.

 

Rockets don't use bullets. Rockets shoot with hot gas. But the mechanism is the same. So, to slow down you have to turn the rocket around, and fire it with the back-end pointing forward.

 

Regarding gravity

In space there certainly is gravity. If you are in orbit (making circles around the earth), you have two forces on you:

1. Gravity, as always.

2. The centripetal force (hope I got the name right)... You can see this force in action if you tie a stone (or something else) to a rope, and swing it around you. The faster you swing it, the larger the force on the stone, the "heavier" it becomes. It wants to go straight, but the rope keeps it in its "orbit" around you.

 

In orbit, you also have this centripetal force. The circles you make are a lot larger, but also the speed you have is a lot larger. And you cannot go straight in orbit, because gravity keeps you close to earth (gravity is acting as the rope).

 

Funnily, the earth itself is also in orbit around the sun. And the moon is in orbit around the earth. They are all experiencing two forces: the one resulting from the speed they have around the other... and gravity. As long as those are equal, the orbit is stable.

 

Regarding friction

You're right. It is a safe assumption to say that there is zero friction. There are some particles, there is a little friction, but it is very very close to nothing.

[Dark matter]

There is gravity, and friction, but very little friction.

 

Basically, the laws of physics still apply. Space just tends to show us things in its most or near somethings most absolute forms, which causes confusion and frustration to a lot of people.

[Cap'n Refsmmat]

The Space Shuttle, if you want an example, has Reaction Control System rockets that fire to maneuver it around, and it uses the same system to stop the motion once it is in the attitude it wants. You fire the rockets once to start moving, and again to stop moving.

[SkepticLance]

Captain Panic described the basic mechanism very well. I will just make one addition, but which has many consequences.

 

The power of a rocket depends on both the mass of the gas it is firing out, and on the speed that the gas is ejected. You can make a rocket more powerful by having more mass to throw away. You can see this principle in action with the space shuttle when you look at the massive fuel tank that is strapped on. In fact, the mass of the fuel to be ejected far outstrips the mass of the shuttle.

 

However, it is actually more efficient to increase the power by ejecting the fuel at a faster speed. It is an exponential function. Increasing mass is merely linear. That means if you double the mass of the fuel you eject, you double the force. However, if you double the speed at which the fuel is ejected, you increase the force by four times. Triple the speed of ejection, and you get nine times the force.

 

Simple chemical rockets, such as used in the shuttle, have limits of speed of ejection. However, an ion drive uses an accelerator to speed up the fuel going out the back. This makes it really efficient by comparison. Ion drives are not much chop when leaving the Earth, but are great when out in space heading for Mars. A small amount of fuel gives a lot of force to accelerate, and later slow down a space probe.

[Mr Skeptic]

SkepticLance... where to begin? A square relationship is not exponential, doubling the speed of the ejected particles (at nonrelativistic speeds) will double the force, as that is based on momentum not on energy. The big savings for better fuel are that you need less fuel, so have less mass to lug around. The relation of fuel to distance is not linear due to having to lug around the extra fuel, though doubling the rate of fuel burn will double the force. The rest of your post is fairly accurate though.

[SkepticLance]

To Mr Skeptic

 

Acceleration of a rocket (assuming no friction and microgravity) is proportional to the mass of the ejected mass times its velocity squared.

[D H]

Acceleration of a rocket (assuming no friction and microgravity) is proportional to the mass of the ejected mass times its velocity squared.

That's not correct. Doubling the exhaust velocity does not quadruple the acceleration. Assuming a single engine is firing, the acceleration of a rocket is, to first order,

[math]-\,\frac {\dot m_r}{m_r}\mathbf v_e[/math]

This ignores some second-order effects such as the change in the center of the mass of the rocket and coupling between the rocket's rotation and thrusting. The above equation is the basis for the ideal rocket equation, wiki link here.

[SkepticLance]

To DH

You are correct with your equation for acceleration. I should not have used that term.

 

The kinetic energy equation is KE = m.v.v

 

That is, velocity squared. Since kinetic energy increases are equal for both rocket and ejected reaction mass (Newtons third law), the increase in that of the rocket is proportional to the velocity of the ejected mass squared. In this case, kinetic energy can be equated to thrust.

 

If you take an ion drive, and double the speed of the ejected ions, while keeping the mass ejected constant, the thrust will quadruple.

[D H]

An ion drive works the same as any other thruster; it just has a very high specific impulse. Thrust is proportional to exhaust velocity.

 

Consider the rocket and the exhaust cloud behind it to form a closed system. This means linear momentum and total energy are conserved quantities. Conservation of linear momentum dictates that the thrust is a linear function of exhaust velocity. So what is the deal with energy? Why does it seem to indicate that thrust is quadratic wrt exhaust velocity? First, it is very important to recognize that kinetic energy is not conserved. A rocket works by converting potential energy (e.g. fuel combustion) into kinetic energy. Nonetheless, it is possible to derive the thrust equation by applying conservation of energy considerations.

 

Consider a rocket moving at some velocity [math]\mathbf v_r[/math] wrt some inertial observer. Over some small interval of time [math]\Delta t[/math] the rocket ejects a quantity of exhaust [math]\Delta m_e[/math] at a velocity [math]\mathbf v_e[/math] relative to the rocket. The kinetic energy of the rocket before the mass was ejected was

 

[math]T_r(t) = \frac 1 2 m_r v_r^2[/math]

 

After ejecting the fuel, the rocket has kinetic energy

 

[math]T_r(t+\Delta t) = \frac 1 2 (m_r-\Delta m_e) (v_r^2+2\mathbf v_r\cdot \Delta \mathbf v_r + \Delta v_r^2)[/math]

 

Dropping second-order terms, the change in the rocket's kinetic energy is thus

 

[math]

\Delta T_r \approx

m_r \mathbf v_r\cdot \Delta \mathbf v_r - \Delta m_e v_r^2

[/math]

 

At the same time, the exhaust cloud gains kinetic energy:

 

[math]

\Delta T_e =

\frac 1 2 \Delta m_e ||\mathbf v_r + \mathbf v_e||^2

= \frac 1 2 \Delta m_e (v_r^2 + 2\mathbf v_r \cdot \mathbf v_e + v_e^2)

[/math]

 

The total change in kinetic energy is

 

[math]\Delta T_{tot} + \Delta T_r + \Delta T_e \approx

\mathbf v_r\cdot (

m_r \Delta \mathbf v_r + \Delta m_e \mathbf v_e)

+ \frac 1 2 \Delta m_e v_e^2

[/math]

 

Conservation of energy applies in all inertial frames. The term involving the rocket velocity is very problematic in terms of energy conservation. This term must vanish if energy is to be conserved. In other words,

 

[math]

m_r \Delta \mathbf v_r + \Delta m_e \mathbf v_e = 0

[/math]

 

is a necessary (but not sufficient) condition for conservation of energy. Dividing this condition by [math]\Delta t[/math] and taking the limit [math]\Delta t \to 0[/math] yields

 

[math]

m_r \frac{d \mathbf v_r}{dt} + \frac {d m_e}{dt} \mathbf v_e = 0

[/math]

 

or

 

[math]

\dot{\mathbf v}r = -\,\frac {\dot m_e}{m_r} \mathbf v_e

[/math]

 

Not surprisingly, this is the same result as obtained using conservation of momentum considerations.

 

Now, about that other term in the energy equation. The term [math]\frac 1 2 \dot m_e v_e^2[/math] represents the change in kinetic energy of the rocket+exhaust system. To conserve energy, this term must not exceed the change in the fuel's potential energy resulting from combustion. The term will in fact be smaller than the energy released from combustion (or some other means) because some of the released energy is always wasted in the form of heat.

[Mr Skeptic]

The kinetic energy equation is KE = m.v.v

 

Or was it [math]KE = \frac{1}{2}mv^2[/math]?

 

That is, velocity squared. Since kinetic energy increases are equal for both rocket and ejected reaction mass (Newtons third law), the increase in that of the rocket is proportional to the velocity of the ejected mass squared. In this case, kinetic energy can be equated to thrust.

 

I think you need to read up on Newton's third law. The law is about force (or change in momentum), not about energy.

 

If you take an ion drive, and double the speed of the ejected ions, while keeping the mass ejected constant, the thrust will quadruple.

 

You may consider listening to the rocket scientist on this one.