Final answer:
The gravitational acceleration at the surface of a planet with three times the radius of Earth and the same density is 29.4 m/s^2, which is three times Earth's gravitational acceleration.
Step-by-step explanation:
To determine the gravitational acceleration at the surface of a planet with 3 times the radius of Earth and the same density, we can use the formula for gravitational force, F = G * (m1 * m2) / r^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two bodies, and r is the distance between the centers of the two masses. Because the gravitational acceleration g is the force per unit mass, g = G * m / r^2, where m is the mass of the planet, and r is its radius.
Since the density is constant, the mass will change proportionally to the volume. As volume scales with the cube of the radius, the new planet has a volume that is 3^3, or 27 times greater than Earth's. Consequently, the mass is also 27 times greater. However, because the radius is 3 times greater, the acceleration due to gravity calculation becomes (27 / 3^2) times Earth's g, which simplifies to 3 times Earth's gravitational acceleration.
Therefore, the gravitational acceleration at the surface of this new planet will be 3 * 9.8 m/s^2 = 29.4 m/s^2.