Boosting Speed with Rocket Propulsion & Momentum

In summary, Shawn D. explains that a rocket has momentum (which is conserved) and that as fuel is ejected, the thrust v_{rel}\frac{dM}{dt} is increased until the desired speed is reached. He then goes on to say that the ship is a non-inertial frame and that as the rocketspeed changes, so does the mass and momentum of the fuel.
  • #1
pringless
43
0
rocket propulsion/momentum

A 1520 kg rocket has 4860 kg of fuel on board. The rocket is coasting through space at 94 m/s and needs to boost its speed to 348 m/s. It does this by firing its engines and ejecting fuel at a relative speed of 807 m/s until the desired speed is reached. How much fuel is left on board after this manuever?
 
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  • #2
Apply the Conservation of Momentum in inertial reference frame with which the rocket is moving
 
  • #3


this work is all wrong
 
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  • #4


Originally posted by pringless
A 1520 kg rocket has 4860 kg of fuel on board. The rocket is coasting through space at 94 m/s and needs to boost its speed to 348 m/s. It does this by firing its engines and ejecting fuel at a relative speed of 807 m/s until the desired speed is reached. How much fuel is left on board after this manuever?
You have to treat the rocket as a system of variable mass. As the fuel is ejected, it exerts a thrust [tex]v_{rel}\frac{dM}{dt}[/tex] on the rocket. Give it a shot (you'll need a little calculus).

ShawnD, note that 807 m/s is the relative speed: it is not the speed of the exhaust with respect to the ground.
 
  • #5
Yes Shawn D u have to apply momentum conservation w.r.t ground
 
  • #6
Ok then let's try this again. I'll change the frame of referance to the ship so that my initial is 0.

initial momentum:
p = (1520 + 4860) * 0
p = 0

relative momentum of the fuel:
p = -807x

new momentum of ship:
p = (1520 + (4860 - x )) * (348 - 94)
p = (1520 + 4860 - x) * 254
p = (6380 - x) * 254
p = 1620520 - 254x


final = initial:
1620520 - 254x - 807x = 0
1620520 - 1061x = 0
x = 1527kg of fuel spent



look better?
 
  • #7
Originally posted by ShawnD
Ok then let's try this again. I'll change the frame of referance to the ship so that my initial is 0.
No. The ship is a non-inertial frame. Its speed (and mass) are changing. See my previous post. (I'll post a solution tomorrow.)
 
  • #8
You souldn't have to factor that in. The velocity of the fuel is always the same speed relative to the ship.
 
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  • #9
Originally posted by ShawnD
The velocity of the fuel is always the same speed relative to the ship.
Well, no. The ship is speeding up. As each bit of fuel is ejected, its speed relative to the ship is the same. But the speed of the ship is different for each bit.
 
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Al can you please explain what you mean? I said the fuel is always the same speed relative to the ship. You said no, then you said it's relative speed is the same (what I said).
 
  • #11
Originally posted by ShawnD
Al can you please explain what you mean? I said the fuel is always the same speed relative to the ship. You said no, then you said it's relative speed is the same (what I said).
Sure. Each bit of fuel, as it is ejected, moves at the same speed relative to the ship. For example, at time t=1 a bit of fuel m1 is ejected. It moves at 807 m/s relative to the speed of the ship at time t=1. But the ship is accelerating. At time t=100 (say) the bit of fuel m1 is no longer moving at 807 m/s relative to the ship, since the ship is now moving faster. So... all those bits of fuel are moving at different speeds. Make sense?
 
  • #12
You are looking at fuel it has already spent, I'm looking at fuel it is currently spending.
 
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  • #13
How to solve this problem

Originally posted by ShawnD
Oh i see what you're saying. Jesus that is one hard question.
Yep. It's meant for a calculus-based college course, not high school. For your amusement, here's how I would solve this problem:

Let M stand for the mass of the ship+remaining fuel, v for the speed of the ship. Each bit dM of fuel ejected exerts a force on the ship:
[tex]v_{rel}\frac{dM}{dt}=Ma=M\frac{dv}{dt}[/tex]

Rearranging,
[tex]v_{rel}\frac{dM}{M}={dv}[/tex]

Integrating from the initial speed&mass to the final,
[tex]v_{rel}\ln{\frac{M_f}{M_i}=v_f-v_i[/tex]

Plugging in some numbers:
[tex]\ln{\frac{M_i}{M_f}=\frac{348-94}{807}[/tex]

Finally,
[tex]\Delta M= 1723[/tex]

So, of the original 4860 kg of fuel, 3137 kg is left. (Unless I made an arithmetic error.)

But the real question is: Why is this thread titled "chain question"?
 

1. How does rocket propulsion work?

Rocket propulsion works by expelling a high-speed jet of gas in one direction, which then creates an equal and opposite force in the opposite direction, propelling the rocket forward. This is based on Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction.

2. What is the difference between solid and liquid rocket propellants?

Solid rocket propellants are pre-mixed and stored in a solid form, while liquid rocket propellants are typically stored separately and then mixed and ignited during flight. Solid propellants are more stable and easier to store, but liquid propellants can be more easily controlled and have higher performance.

3. How does momentum contribute to the speed of a rocket?

Momentum refers to the quantity of motion of an object, and in the case of a rocket, it is the product of the mass and velocity. As the rocket expels gas at high speeds, it also gains an equal and opposite momentum, which helps to increase its speed.

4. What is the maximum speed that can be achieved with rocket propulsion?

The maximum speed that can be achieved with rocket propulsion depends on various factors such as the type and amount of propellant, the design and efficiency of the rocket engine, and the external forces acting on the rocket. In theory, there is no limit to the speed that can be achieved, but practical limitations and safety considerations must be taken into account.

5. How can we improve the efficiency of rocket propulsion?

Efficiency in rocket propulsion can be improved by optimizing the design of the rocket engine and its components, using more efficient propellants, and reducing the weight of the rocket through advanced materials and design techniques. Additionally, advancements in technology and research can lead to new and more efficient forms of rocket propulsion.

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