Rocket Equation and Orbit questions?

picklepie159
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Homework Statement


#1.) A rocket exhausts fuel with a velocity of 1500 m/s, relative to the rocket. It starts from outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted the speed is...

(Answer was 2400 m/s)


Homework Equations



Vrel (dm/dt) = m (dv/dt)

(Does conservation of momentum apply here? I know it applied to an earlier problem, but the velocity of the ejected gasses weren't relative to the rocket)

The Attempt at a Solution



1500 (0.8) = 0.2 v
v = 600 m/s

1500 (0.8) = 1 v
v = 1200... still wrong.

Can anyone help me out here??


Also, if a man in a circular orbit around Earth fires his forward thrusters to drop his KE, would he fall into a larger elliptical orbit with a greater period? It was a question in the book I was confused about.
I know K = -E and E = - GMm/2r , or - GMm/2a
 
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picklepie159 said:
Vrel (dm/dt) = m (dv/dt)

(Does conservation of momentum apply here?
Yes, you need to use conservation of momentum, but the mass of the rocket is changing, so you need it in the form of a differential equation. The equation you quoted is fine, but you don't really care about time here. Can you rewrite it without using dt?
 
I understand it's not okay to treat the dt's as a part of a fraction, but I'm learning calc alongside with physics and this is the only thing I could think of. It's not correct, but I would love to learn how to rewrite it as you suggested.

Vrel * dm = m * dv
1500 * 0.8 = dv
 
picklepie159 said:
I understand it's not okay to treat the dt's as a part of a fraction, but I'm learning calc alongside with physics and this is the only thing I could think of. It's not correct, but I would love to learn how to rewrite it as you suggested.

Vrel * dm = m * dv
Yes
1500 * 0.8 = dv
No, dm is a small change in mass. You need to rearrange and integrate the eqn.
 
dv/dm = Vrel/m

v = Vrel * ln(m)

2414 = -1500 * ln(0.2).

Awesome, thanks a lot!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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