Calculating Motion of a Test Balloon on a Methane Planet

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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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on Phys.org
mass of the planet
 
i am not sure how to proceed
 
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:
you mean 0,1g/cm³
 
Then I get the value of g=0.414 m/s²
 
i just draw Displacement vs time graph
 

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can you tell me please whether it is right or wrong
 
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
 
yes, he just throws it from 10m. Total mass=19kg.
 
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.
 
If I miss something, please explain it correctly.
 
balloon will not go down. However, If you put heavy thing attached with balloon, they will go down.
 
I think i should only consider the mass of box
 
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.
 
100kg as ρ=0.1 g/cm³ if i multiply with 1000 then the value is 100kg/m³
 
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.
 
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.
 
'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.