Throwing a baseball while standing on an Asteroid

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    Asteroid Baseball
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Homework Help Overview

The discussion revolves around a physics problem involving the dynamics of throwing a baseball on an asteroid, specifically calculating the largest radius of the asteroid that allows the baseball to achieve a circular orbit. The subject area includes gravitational physics and circular motion.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of equating gravitational force to centripetal force to find the radius. Questions arise about the calculation of mass using density and the correct formula for volume.

Discussion Status

The discussion is ongoing, with participants questioning the accuracy of the mass calculation and the formula used. There is recognition of a potential error in the volume formula, leading to further exploration of the correct approach.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for discussion. There is a focus on ensuring the correct application of physical principles and formulas.

takelgith
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Summary:: Between the orbits of Mars and Jupiter, several thousand small objects called asteroids move in nearly circular orbits around the Sun. Consider an asteroid that is spherically shaped with radius
r and density 2100 kg/m^3.

a.
You find yourself on the surface of this asteroid and throw a baseball at a speed of 28
m/s. If the baseball is to travel around the asteroid in a circular orbit, what is the largest radius asteroid on which you are capable of accomplishing this feat??

- For this I equated GMm/r^2 = mv^2/r and solved for r. I used the density to find M.
I got r = √[(28^2)/(2100π)(6.67^-11)]
it says its wrong!
 
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takelgith said:
- For this I equated GMm/r^2 = mv^2/r and solved for r. I used the density to find M.
I got r = √[(28^2)/(2100π)(6.67^-11)]
it says its wrong!
Can you elaborate on the part where you used the density to find M?
 
jbriggs444 said:
Can you elaborate on the part where you used the density to find M? You did it wrong.

So I had M = 2100π R^3.

Then I subbed that into the equation, for which the R got canceled and became R^2 and thus my result.
 
takelgith said:
So I had M = 2100π R^3
Where does that formula come from?
 
takelgith said:
So I had M = 2100π R^3.

Then I subbed that into the equation, for which the R got canceled and became R^2 and thus my result.

Oh shoot! Its supposed to be 4/3piR^3. So M= 2100 *(4/3 pi r^3) right? that should be right??
 
takelgith said:
Oh shoot! Its supposed to be 4/3piR^3. So M= 2100 *(4/3 pi r^3) right? that should be right??
I think so. I did not spot any other errors.
 
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