Simple Rocket Equation in Gravity

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The discussion focuses on deriving the velocity equation for a rocket under constant gravitational acceleration with constant exhaust velocity. The user initially struggles with the integration process, leading to a negative logarithmic result for velocity. They realize that the discrepancy arises from the sign convention of exhaust velocity, which should be treated as negative in the momentum conservation equation. The correct formulation yields a positive logarithmic expression for velocity, aligning with established answers. Ultimately, the user corrects their approach by recognizing the importance of sign in the exhaust velocity.
Dextrine
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Homework Statement


I'm having trouble deriving the equation for the velocity of a rocket in constant G given that it has constant exhaust velocity

Homework Equations



I know that a=dv/dt=U/M(dM/dt)-g

The Attempt at a Solution


so from here dv=U/M dM-gdt

v=U ∫1/M dM - g ∫dt v from 0 to v, M from Minitial to Mfinal, and t from 0 to t

which leads to v=-ln[Minitial/Mfinal]-gtHOWEVER, everywhere I look, the answer is positive ln[Minitial/Mfinal] -gt and I can't seem to get why this is so
 
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Dextrine said:

Homework Statement


I'm having trouble deriving the equation for the velocity of a rocket in constant G given that it has constant exhaust velocity

Homework Equations



I know that a=dv/dt=U/M(dM/dt)-g

The Attempt at a Solution


so from here dv=U/M dM-gdt

v=U ∫1/M dM - g ∫dt v from 0 to v, M from Minitial to Mfinal, and t from 0 to t

which leads to v=-ln[Minitial/Mfinal]-gt

HOWEVER, everywhere I look, the answer is positive ln[Minitial/Mfinal] -gt and I can't seem to get why this is so

From conservation of momentum you get the equation MΔv-UΔm=-Mg, where Δm is the exhausted mass during Δt time. But it is negative of the change of mass of the rocket. So the differential equation for the rocket mass is M dv/dt+UdM/dt=-Mg ---->## v= -\int _{Minitial}^{Mfinal}(\frac{dM}{M})##
 
Ah, i see what I was doing wrong, my answer is correct if U is taken to to be negative
 
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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