How Much Does Neutron Star Material Weigh on Earth?

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SUMMARY

The discussion focuses on calculating the weight of 1.1 cubic centimeters of neutron star material on Earth, given a neutron star mass of 1.99e+030 kg and a radius of 9 km. The user correctly applies the formula for density, resulting in a density of 6.5168e17 kg/m³. The final calculation for weight yields 7.0323e16 N, but the user reports that the answer is not being accepted, indicating a potential error in unit conversion or calculation steps.

PREREQUISITES
  • Understanding of basic physics concepts such as mass, volume, and density
  • Familiarity with gravitational acceleration (9.81 m/s²)
  • Ability to perform unit conversions between cubic centimeters and cubic meters
  • Knowledge of the formula for the volume of a sphere
NEXT STEPS
  • Review unit conversion techniques, particularly for volume measurements
  • Study the properties of neutron stars and their densities
  • Learn about gravitational force calculations in physics
  • Explore common pitfalls in physics homework problems related to density and weight
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This discussion is beneficial for physics students, educators, and anyone interested in astrophysics or gravitational calculations involving extreme materials like neutron stars.

danimal8f
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Basic mass to weigh conversion?

Consider a neutron star of mass M = 1.99e+030 kg and a radius of R = 9 km.
Assuming uniform density, how much would 1.1 cubic centimeters of neutron star material weigh on the surface of the earth?

Homework Equations


Volume of Sphere = (4/3)pi(r^3)
density = Mass/Volume
gravity = 9.81 m/s^2

First I converted 9km to 9000m, and 1.1cm^3 to .011m^3
solve for density; (M=1.99e30)/(vol=(4/3)pi(9000^3) = 6.5168e17

(.011 x 6.5168e17) x 9.81 = 7.0323e16 N

its not accepting my answer
 
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danimal8f said:
Consider a neutron star of mass M = 1.99e+030 kg and a radius of R = 9 km.
Assuming uniform density, how much would 1.1 cubic centimeters of neutron star material weigh on the surface of the earth?

Homework Equations


Volume of Sphere = (4/3)pi(r^3)
density = Mass/Volume
gravity = 9.81 m/s^2

First I converted 9km to 9000m, and 1.1cm^3 to .011m^3
solve for density; (M=1.99e30)/(vol=(4/3)pi(9000^3) = 6.5168e17

(.011 x 6.5168e17) x 9.81 = 7.0323e16 N

its not accepting my answer


Welcome to PF.

Maybe you want to check that conversion?
 


why thank you... I'm an idiot
 
Last edited:

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