Mars Rover Upthrust and Acceleration calculations help 🚀

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The discussion revolves around calculating the upthrust and acceleration of the Mars Rover Curiosity after parachute deployment, using principles of buoyancy and Newton's laws. The calculated upthrust is approximately 74.1 N, and the acceleration is around 3.64 m/s², but participants express confusion regarding the treatment of the parachute as an evacuated hemisphere instead of considering the drag coefficient. It is emphasized that without the drag coefficient, the calculations may not accurately reflect the situation, as the parachute interacts with the Martian atmosphere. Participants debate whether the question is meant to test buoyancy principles or if it requires additional information for a correct solution. Ultimately, it is suggested that the question may not be suitable for revising buoyancy concepts due to its complexity.
lpettigrew
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Homework Statement
Hello, I have a question which I have answered below but I do not think that I have done so correctly. Could someone please advice me how to improve my workings?
1 . Modelling the parachute as a hemisphere, use the following information to calculate the upthrust produced due to the parachute.
ρmars atmosphere = 0.02 kgm^-3
volume of a sphere = 4 / 3 𝜋 𝑟^3
gmars = 3.8 ms^-2
2. Given that Curiosity’s mass equals 900 kg, what was its acceleration once the parachute was
deployed?
Relevant Equations
ρmars atmosphere = 0.02 kgm^-3
volume of a sphere = 4 / 3 𝜋 𝑟^3
gmars = 3.8 ms^-2

F=ma
1. ρmars atmosphere = 0.02 kgm^-3
volume of a sphere = 4 / 3 𝜋 𝑟^3
gmars = 3.8 ms^-2

So: Volume of a hemisphere=2/3 𝜋 𝑟^3
r=7.75 m

Archimedes' principle states that the upthrust on an object is equal to the weight of fluid that the body displaces

To find the upthrust produced;
upthrust=weight
upthrust=mg
upthrust=p*V*g

mass=density*volume
mass=0.02*(2/3*𝜋 *7.75^3)
mass=19.498~19.5kg (3.s.f)

upthrust=19.5*3.8=74.09~74.1 N2. Given that Curiosity’s mass equals 900 kg, what was its acceleration once the parachute was
deployed?

Using Newton's Second Law; F=ma
F=mg
a=F/m
a=mg/m
a=(900*3.8)-(19.5*3.8)/900+19.5
a=3345.9N/919.5kg
a=3.6388 ~3.64 ms^-2 (3.s.f)
 
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I don't understand. It is described as a parachute, but you are treating it as an evacuated hemisphere. Isn't the parachute filled with Martian air?
Seems to me you are not given the info needed. You need the drag coefficient.
 
Last edited:
haruspex said:
I don't understand. It is described as a parachute, but you are treating it as an evacuated hemisphere. Isn't the parachute filled with Martian air?
Seems to me you are not given the info needed. You need the drag coefficient.
Thank you for your reply. Oh no, so are my answers incorrect? I have tried to utilise all of the information given, is it possible to calulate the upthrust and acceleration without the drag coefficient?
 
Thank
haruspex said:
Thank you for finding and attaching that. I am not sure if I should use the drag coefficient though as the question specifies to use the information provided, so I am not sure. Would neglecting the drag coefficient mean that my previous answers are wrong though?
 
lpettigrew said:
Thank

Thank you for finding and attaching that. I am not sure if I should use the drag coefficient though as the question specifies to use the information provided, so I am not sure. Would neglecting the drag coefficient mean that my previous answers are wrong though?
Without the drag equation, and a value for the coefficient, you have no basis for answering the question.
Treating it as an evacuated hemisphere and applying buoyancy principles is utterly wrong. If you were to detach the Rover and release the parachute, would it rise up like a helium balloon?
 
haruspex said:
Without the drag equation, and a value for the coefficient, you have no basis for answering the question.
Treating it as an evacuated hemisphere and applying buoyancy principles is utterly wrong. If you were to detach the Rover and release the parachute, would it rise up like a helium balloon?
Oh ok I seem what you mean. I was just attempting to answer the question in the format specified. Strangely, I think it is a question testing one's knowledge of buoyancy principles as you state, like Archimedes's Principle etc as opposed to drag. I found it online whilst revising but was not sure how to comprehensively answer it with the information given. So my answers do not apply then?
 
lpettigrew said:
Oh ok I seem what you mean. I was just attempting to answer the question in the format specified. Strangely, I think it is a question testing one's knowledge of buoyancy principles as you state, like Archimedes's Principle etc as opposed to drag. I found it online whilst revising but was not sure how to comprehensively answer it with the information given. So my answers do not apply then?
If you are trying to revise buoyancy, this is not a question to work on.
Please provide the link to it.
 
haruspex said:
If you are trying to revise buoyancy, this is not a question to work on.
Please provide the link to it.
My apologies but I have been looking for the question and have lost the page I found it on, which is partially why I copied it over in the first place. I had been revising some topics on fluids, denisty and upthrust which is whereabouts I found this problem classified.
 

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