Final answer:
The acceleration of free fall at a height equal to the radius above the surface of a planet with the given parameters is found to be 6.5 m/s².
Step-by-step explanation:
To calculate the acceleration of free fall at a height R above a planet's surface, we must consider Newton's law of gravitation, which states that the gravitational force (F) between two masses is given by F = Gm1m2/r2, where G is the gravitational constant, m1 and m2 are the masses, and r is the distance between the centers of the two masses.
The force is also equal to the mass of the object times the acceleration due to gravity (g), F = mg. If we are at distance R above the planet's surface the distance from the center of the planet becomes 2R.
The mass of the planet (M) can be found using its volume and mean density (ρ): M = (4/3)πR3ρ. Therefore, the acceleration due to gravity at height R is g = GM/(2R)2. Substituting the mass of the planet, we get:
g = (6.67 x 10-11 N·m2/kg2) · ((4/3)π(3.4 x 103 m)3 · (4.0 x 103 kg/m3))/ (2 · 3.4 x 103 m)2
Calculating this, we find that the correct answer is 6.5 m/s2.