Calculating Motion of a Test Balloon on a Methane Planet

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SUMMARY

The discussion focuses on calculating the motion of a test balloon on a methane planet with a gaseous atmosphere density of 0.1 g/cm³ and a liquid methane density of 0.4 g/cm³. The balloon, with a volume of 1 m³ and a mass of 15 kg, is attached to a 4 kg box by a 10 m rope. The gravitational acceleration on the planet is calculated to be approximately 0.425 m/s², and the buoyant force acting on the box and balloon is derived using Archimedes' principle. The final calculations yield a time of approximately 3.33 seconds for the box to reach the surface of the methane.

PREREQUISITES
  • Understanding of Archimedes' principle and buoyancy
  • Knowledge of gravitational force calculations using the formula g = mG/r²
  • Familiarity with kinematic equations, specifically y = -0.5at²
  • Basic concepts of fluid dynamics, including density and drag force
NEXT STEPS
  • Study the application of Archimedes' principle in various fluid scenarios
  • Learn about gravitational force calculations on different celestial bodies
  • Explore kinematic equations and their applications in motion analysis
  • Investigate the effects of drag force in fluid dynamics and its calculations
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Students in physics or engineering, educators teaching fluid dynamics, and anyone interested in gravitational effects and buoyancy in extraterrestrial environments.

robax25
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Homework Statement


A planet has a gaseous atmosphere with a density of 0.1 g/cm³. The planet itself
completely consists of liquid methane with a density of 0.4 g/cm³ and has a radius of
3800 km. A test balloon is dropped onto this planet, which consists of a helium spherical
balloon (1 m³, mass 15 kg) and a box of test equipment (10 cm x 10 cm x 10 cm, mass 4 kg)
attached by a 10 m long rope.
Consider the case when the box is in the liquid methane and the balloon is at the surface.
Calculate the motion of the box and draw x(t), v(t) and a(t) diagrams until the boxes reaches
the surface, quantitatively

Homework Equations


y=-0.5at²

G=6.67 *10^(-11) m³/kgs²
g(for the planet)=mG/r²

The Attempt at a Solution


g=9,81 kgm/s²

I need value of g for the planet.
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robax25 said:
I need value of g for the planet.
Well, in your own words:
robax25 said:
g(for the planet)=mG/r²
So what are you missing ?
 
mass of the planet
 
You are given the density and the radius. Planets are spherical ...
 
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i am not sure how to proceed
 
When unsure how to proceed, make a diagram :smile:
 
I get it how to do it. can you tell me please which density should I consider?Should I sum up both density.?.
 
Last edited:
For the planet you take the density of the planet ?:) :doh:
 
you mean 0,1g/cm³
 
  • #10
robax25 said:
The planet itself completely consists of liquid methane with a density of 0.4 g/cm³
 
  • #11
Then I get the value of g=0.414 m/s²
 
  • #12
I get 0.425
 
  • #13
yes.0 425
 
  • #14
i just draw Displacement vs time graph
 

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  • #15
can you tell me please whether it is right or wrong
 
  • #16
What am I supposed to see ? Perhaps you want to work it out a little more ?
And: are you sure you have the right scenario in mind ?
robax25 said:
Consider the case when the box is in the liquid methane and the balloon is at the surface.
Calculate the motion of the box and draw x(t), v(t) and a(t) diagrams until the boxes reaches the surface, quantitatively
 
  • #17
yes, he just throws it from 10m. Total mass=19kg.
 
  • #18
Balloons have a tendency to go up. Not this one ?
 
  • #19
but if you attach Ballons and box together and throw them from 10m on a Liquid surface. You will see the same thing that he explains.
 
  • #20
If I miss something, please explain it correctly.
 
  • #21
I would prefer it if you check your own assumptions -- do you really think the balloon is going down ?
 
  • #22
balloon will not go down. However, If you put heavy thing attached with balloon, they will go down.
 
  • #23
I think i should only consider the mass of box
 
  • #24
Box is 4 kg, balloon 15.

What did you use the density of the atmosphere of that planet for ? Or was it given just to distract you ?
 
  • #25
if you consider air resistance, then, atmosphere of the planet is necessary.If you neglect air resistance, then, it is not necessary.I would like to consider. If i consider, then, the value of acceleration would change.
 
  • #26
What is the mass of 1 m3 of this planets atmosphere at the surface ?
 
  • #27
100kg as ρ=0.1 g/cm³ if i multiply with 1000 then the value is 100kg/m³
 
  • #28
Good. So according to Archimedes, the planet pulls harder on 1m3 of atmosphere than on 1 m3 of helium balloon with a mass of 4 kg hanging from it, isn't it ? Make a free body diagram and see what the resulting force is.
 
  • #29
resulting force =-40.8 N as g=0.425 m/s² so should the balloon weight be negligible? Yes balloon force acts up, it always blows up.
 
  • #30
'Blowing up' is not how I would describe it :wink: .

I wonder if the exercise composer really intended to make things so complicated with this scenario:
robax25 said:
Consider the case when the box is in the liquid methane and the balloon is at the surface
We are supposed to look at the box, which initially is at -10 m.

I don't see how you can find an upward net force of 40.8 N. Can you show the details ?

BvU said:
Make a free body diagram and see what the resulting force is
Then you can decide if the mass of the balloon can be ignored (not). Idem mass of box (not). And the upward force on the submerged box (probably).

Nothing is said about the viscosity of the liquid or gaseous methane, so I propose to ignore that. The exercise is already tough enough.
 

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