CHEGG A neutron star has a mass of 2.08 × 1030 kg (about the mass of our sun) and a radius of 6.73 × 103 m. Suppose an object falls from rest near the surface of such a star. How fast would it be moving after it had fallen a distance of 0.0093 m? (Assume that the gravitational force is constant over the distance of the fall, and that the star is not rotating.

Let g be the acceleration due to gravity on the surface of the star. By Newton's second law, the gravitational force felt by the object has a magnitude of

F = GMm/r ² = mg

where

• G = 6.67 × 10⁻¹¹ Nm²/kg² is the gravitational constant,

• M = 2.08 × 10³⁰ kg is the mass of the star,

• m is the unknown mass of the object, and

• r = 6.73 × 10³ m is the radius of the star

Solving for g gives

g = GM/r ²

g = (6.67 × 10⁻¹¹ Nm²/kg²) (2.08 × 10³⁰ kg) / (6.73 × 10³ m)²

g ≈ 3.06 × 10¹² m/s²

The object is in free fall with uniform acceleration and starting from rest, so its speed after falling 0.0093 m is v such that

v ² = 2g (0.0093 m)

v = √(2g (0.0093 m))

v ≈ 240,000 m/s ≈ 240 km/s

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form