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A stellar core of mass 3.0 x 10^30 kg and radius 4.0 x 10^8 m collapses, with no loss of mass, to become a neutron star with radius 7.5 x 10^3 m. What is the change in gravitational potential energy of 1.0 × 10^3 kg of material that was initially at the surface of the stellar core and that ends up at the surface of the neutron star?

Answer :

Answer:

[tex]\Delta U = -2.67 \times 10^{19} J[/tex]

Explanation:

Initial gravitational potential energy of the system

[tex]U_i = -\frac{GMm}{R}[/tex]

[tex]U_i = - \frac{(6.67 \times 10^{-11})(3.0 \times 10^{30})(1.0 \times 10^3)}{4.0\times 10^8}[/tex]

[tex]U_i = -5.0 \times 10^{14} J[/tex]

Now when star becomes a neutron star then it will convert into a denser star with no loss in mass

So it is given as

[tex]U_f = -\frac{GMm}{r}[/tex]

[tex]U_f = - \frac{(6.67 \times 10^{-11})(3.0 \times 10^{30})(1.0 \times 10^3)}{7.5 \times 10^3}[/tex]

[tex]U_f = -2.67 \times 10^{19} J[/tex]

Change in the potential energy of the system is given as

[tex]\Delta U = U_f - U_i[/tex]

[tex]\Delta U = (-2.67 \times 10^{19}) - (-5 \times 10^{14})[/tex]

[tex]\Delta U = -2.67 \times 10^{19} J[/tex]

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