The weight of a star is usually balanced by two forces: the gravitational force, acting inward, and the force created by nuclear reaction, acting outward. Over a long period of time, the force due to nuclear reactions gets weaker, causing the gravitational collapse of the star and crushing atoms out of existence. Under such extreme conditions, protons and electrons are squeezed to form neutrons, giving birth to a neutron star. Neutron stars are massively heavy-a teaspoon of the substance of a neutron star would weigh 100 million metric tons on the Earth.  If a satellite is to circle 12.1 km above the surface of such a neutron star whose radius is 12.1 km, how many revolutions per minute will it make? Do not enter unit. (You may enter your calculation using scientific notation.)

To solve this problem, we need to use the formula for the orbital period of a satellite around a celestial body, which is given by:

T = 2π√(r³/GM)

where:

• T is the orbital period (the time it takes for the satellite to make one complete revolution),
• r is the distance from the center of the celestial body to the satellite,
• G is the gravitational constant (6.674 x 10^-11 m³ kg⁻¹ s⁻²), and
• M is the mass of the celestial body.

Given that the radius of the neutron star is 12.1 km and the satellite is to circle 12.1 km above the surface of the star, the distance r is 2 * 12.1 km = 24.2 km = 24.2 x 10^3 m.

The mass of the neutron star can be estimated from its density. The density of a neutron star is approximately 5 x 10^17 kg/m³. Therefore, the mass M of the neutron star is its density times its volume, or M = ρV = ρ(4/3πr³) = 5 x 10^17 kg/m³ * 4/3π(12.1 x 10^3 m)³ ≈ 2.3 x 10^30 kg.

Substituting these values into the formula for T, we get:

T = 2π√((24.2 x 10^3 m)³/(6.674 x 10^-11 m³ kg⁻¹ s⁻² * 2.3 x 10^30 kg)) ≈ 0.0002 s.

To convert this to revolutions per minute, we divide 60 seconds by T and get:

60 s / 0.0002 s ≈ 300,000 revolutions per minute.