Calculating Motion of a Test Balloon on a Methane Planet

In summary: However, the balloon will rise, so it is better to consider it as a separate system.)There are two forces on the balloon: the weight of the balloon and the upward force due to the atmosphere. There is also a downwards force due to gravity, which is equal to the mass of the balloon times the gravitational constant. There is also a drag force due to the atmosphere.There are four forces on the box: the weight of the box, the upward force due to the atmosphere, the downwards force due to gravity, and the drag force due to the atmosphere.
  • #1
robax25
238
3

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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  • #2
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 ?
 
  • #3
mass of the planet
 
  • #4
You are given the density and the radius. Planets are spherical ...
 
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  • #5
i am not sure how to proceed
 
  • #6
When unsure how to proceed, make a diagram :smile:
 
  • #7
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:
  • #8
For the planet you take the density of the planet ?:) :doh:
 
  • #9
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.
 
  • #31
how can I get motion of the box?
 
  • #32
robax25 said:
how can I get motion of the box?
The starting state is that the box is 10m down in the methane and the balloon is at (on) the surface. You can suppose the balloon is not immersed at all in the methane, since it will not be for long.
What are the forces on the balloon? What are the forces on the box?
(You may be tempted to consider the balloon, rope and box as a single system and only consider the external forces on that, but that is not safe. Do you see why?)
 
  • #33
the box has force F(box)=mg and balloon is full of Helium. It goes up.The atmospheric density changes the motion of the box.I need to use Archimedes formula
 
  • #34
I can use this formula to get Drag force=Cd ρ v² *A/2 and we don't know the value of Cd, we can skip it? does It ok?
 
  • #35
robax25 said:
need to use Archimedes formula
Quite so. What is the total buoyant force?
robax25 said:
I can use this formula to get Drag force=Cd ρ v² *A/2 and we don't know the value of Cd, we can skip it? does It ok?
I wouldn't worry about drag. There is nowhere near enough information to quantify that.
 

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