If Jupiter collapsed and became very dense like a neutron star, its size would shrink to about 7. 4% of its original size. Its radius would be approximately 5. 2×10⁴ m. This is a big reduction in size, showing how tightly packed neutron stars are.
The size of a sphere is calculated using the formula;
V = (4/3)πr³,
where V is the size and r is the distance from the center to the edge. We can show how tightly packed something is (density, ρ) by dividing its mass (m) by its volume (V): ρ = m/V.
The initial volume of Jupiter can be calculated using the given radius: (6.99×10⁷ m), Jupiter's initial volume is: V_ initial = (4/3)π * (6.99×10⁷ m)³ ≈ 1.431×10¹⁵ m³
The last density is that of a neutron star: ρ_final = 10.0¹⁸ kg/m³
.Solving for the final radius (r_final):
r_final³ = (3/4) * ρ_final * (6.99×10⁷ m)³ / π
r_final³ ≈ 3.94×10¹² m³
r_ final ≈ 5.2×10⁴ m
See complete question below
Jupiter's average density is 1310 kg/m ³ There is a type of star called a neutron star whose core density is about 10.0¹⁸ kg/m ³. If Jupiter, whose radius is about 6.99×10⁷ m, were to suddenly collapse so that its density was that of a neutron star, its radius would be about m..................?