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A neutron star has a density of $3.7 \times 10^{17} \text{ kg m}^{-3}$ - Leaving Cert Physics - Question b - 2020
Question b
A neutron star has a density of $3.7 \times 10^{17} \text{ kg m}^{-3}$. What would the radius of the Earth be if it had the same density as the neutron star? (mass... show full transcript
Worked Solution & Example Answer:A neutron star has a density of $3.7 \times 10^{17} \text{ kg m}^{-3}$ - Leaving Cert Physics - Question b - 2020
Step 1
Calculate Volume using Density
Answer
To find the radius, we first use the formula for density:
[ \rho = \frac{m}{V} ]
[ V = \frac{m}{\rho} ]
Substituting the mass of the Earth and the density of the neutron star:
[ V = \frac{6.0 \times 10^{24} \text{ kg}}{3.7 \times 10^{17} \text{ kg m}^{-3}} ]
Calculating the volume gives:
[ V \approx 1.62 \times 10^{7} \text{ m}^3 ]
Step 2
Calculate Radius from Volume
Answer
The volume of a sphere is given by:
[ V = \frac{4}{3} \pi r^3 ]
Rearranging to solve for the radius:
[ r = \sqrt[3]{\frac{3V}{4\pi}} ]
Substituting the volume we found earlier:
[ r \approx \sqrt[3]{\frac{3 \times 1.62 \times 10^{7} \text{ m}^3}{4\pi}} ]
Calculating this gives:
[ r \approx 157 \text{ m} ]