How big is the comet based on a 735x731px surface picture taken from 3km away?

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Homework Help Overview

The discussion revolves around determining the width of a comet based on a photograph taken from a distance of 3 kilometers. The image has a resolution that is 3.5 times lower than a specified reference resolution, and participants are exploring how to relate the pixel dimensions of the image to the actual size of the comet.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to use the relationship between angular size and linear dimensions to find the comet's width. Questions are raised about how to calculate the size represented by each pixel and the implications of the camera's resolution.

Discussion Status

Some participants are providing hints and guidance on how to approach the problem, particularly regarding the conversion of angular measurements and the use of relevant equations. There is an ongoing exploration of different interpretations of the problem setup, but no consensus has been reached.

Contextual Notes

Participants note that the original calculations seem inconsistent with their expectations of the comet's size, leading to further questioning of assumptions and methods used. The discussion is framed within the constraints of homework rules that prohibit direct solutions.

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Homework Statement


There is a picture, size 735x731px, that show the surface of comet that was taken from a probe distanced 3 kilometers from the comet. If the resolution of the camera that recorded the surface is 3.5 times lower then the resolution of Tycho Brache (that is 1 angular minute, 1/60 of a degree), what is the width of the comet? Use expresion for α (angle), arch (l) and distance (r)

Homework Equations


α=l/r
l=α*r

The Attempt at a Solution


l=(3.5/60)*3000meters
l=175meters
To me this seems to small, the comet is like 2,2km wide or more. Also I have no idea in what way to use the pixels to calculate the right size.
Any help would be nice
 
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Hi Boxblax, Welcome to Physics Forums.

Boxblax said:

Homework Statement


There is a picture, size 735x731px, that show the surface of comet that was taken from a probe distanced 3 kilometers from the comet. If the resolution of the camera that recorded the surface is 3.5 times lower then the resolution of Tycho Brache (that is 1 angular minute, 1/60 of a degree), what is the width of the comet? Use expresion for α (angle), arch (l) and distance (r)

Homework Equations


α=l/r
l=α*r

The Attempt at a Solution


l=(3.5/60)*3000meters
l=175meters
To me this seems to small, the comet is like 2,2km wide or more. Also I have no idea in what way to use the pixels to calculate the right size.
Any help would be nice
Wouldn't the resolution information imply that each image pixel represents angular size of 3.5 arc minutes? How many pixels across is the picture?
 
It is 731px across. Could you please explain more how to solve it
 
If each pixel represents an angular size of 3.5 arc minutes, the what linear size does each pixel represent at 3 km distance?
 
Yes, that is what i need to find out. I need to find out the real size of comet. How to find out the size of 1 pixel in meters?
 
Boxblax said:
Yes, that is what i need to find out. I need to find out the real size of comet. How to find out the size of 1 pixel in meters?
Draw it out. You have an angle of 3.5 arc minutes and a range of 3 km. What's the length of the arc on the end of that radius? Look at your Relevant equations.
 
175meters? I don't quite understand what you want to do. Could you solve it?
 
Helpers are not allowed to solve homework problems for you. We can only provide hints and suggestions, or point out errors. It's in the forum rules (which you read right? :smile:).

175 meters is not correct. One pixel spans an angle of 3.5 arc minutes. That's 3.5 x (1/60) degrees. With a radius of 3 km, what is the length of the arc that is swept out by that angle? Show your calculations.
 
my calculations are in a photo. still doesn't seem right to me, maybe i am wrong
 

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  • #10
Remember that geometrical calculations involving angles should use radians. Otherwise, you can approximate the figure with a right angle triangle since the radius is so large, and use the tangent function.

I'd convert the angle in arcminutes to radians in your calculation.
 
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  • #11
thanks mate