Let <em>g</em> 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
<em>F</em> = <em>GMm</em>/<em>r</em> ² = <em>mg</em>
where
• <em>G</em> = 6.67 × 10⁻¹¹ Nm²/kg² is the gravitational constant,
• <em>M</em> = 2.08 × 10³⁰ kg is the mass of the star,
• <em>m</em> is the unknown mass of the object, and
• <em>r</em> = 6.73 × 10³ m is the radius of the star
Solving for <em>g</em> gives
<em>g</em> = <em>GM</em>/<em>r</em> ²
<em>g</em> = (6.67 × 10⁻¹¹ Nm²/kg²) (2.08 × 10³⁰ kg) / (6.73 × 10³ m)²
<em>g</em> ≈ 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 <em>v</em> such that
<em>v</em> ² = 2<em>g</em> (0.0093 m)
<em>v</em> = √(2<em>g</em> (0.0093 m))
<em>v</em> ≈ 240,000 m/s ≈ 240 km/s