Mars Rover Upthrust and Acceleration calculations help 🚀

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Homework Help Overview

The discussion revolves around the calculations related to the upthrust and acceleration of the Mars Rover, specifically in the context of its parachute deployment. The problem involves concepts from fluid dynamics and physics, particularly Archimedes' principle and Newton's laws of motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of upthrust using given parameters such as atmospheric density and gravitational acceleration on Mars. There are questions regarding the treatment of the parachute and the necessity of a drag coefficient for accurate calculations.

Discussion Status

Some participants express confusion about the assumptions made in the calculations, particularly regarding the parachute's characteristics. There is an ongoing exploration of whether the drag coefficient is essential for the problem, with differing opinions on the relevance of buoyancy principles versus drag effects.

Contextual Notes

Participants note the lack of information regarding the drag coefficient and question whether the problem can be solved without it. There is also mention of imposed homework rules that may limit the use of external resources or assumptions.

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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