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To calculate the acceleration due to gravity on the surface of the planet, we use the formula:

\[
g = \frac{G \cdot M}{R^2}
\]

where:
- \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \),
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

Given:
- \( M = 2.50 \times 10^{22} \, \text{kg} \),
- \( R = 1.20 \times 10^6 \, \text{m} \).

### Step-by-Step Solution

1. Substitute the values into the formula:

   \[
   g = \frac{(6.674 \times 10^{-11}) \times (2.50 \times 10^{22})}{(1.20 \times 10^6)^2}
   \]

2. Calculate the square of the radius:

   \[
   (1.20 \times 10^6)^2 = 1.44 \times 10^{12} \, \text{m}^2
   \]

3. Calculate the numerator:

   \[
   (6.674 \times 10^{-11}) \times (2.50 \times 10^{22}) = 1.6685 \times 10^{12} \, \text{Nm/kg}
   \]

4. Divide the results:

   \[
   g = \frac{1.6685 \times 10^{12}}{1.44 \times 10^{12}} \approx 1.158 \, \text{m/s}^2
   \]

### Answer
The acceleration due to gravity on the surface of the planet is approximately:

\[
g \approx 1.16 \, \text{m/s}^2
\]

Would you like more details or have any questions?

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### Related Questions
1. How does the mass of a planet affect the surface gravity?
2. What would happen to the acceleration due to gravity if the radius of the planet were doubled?
3. How does surface gravity compare between planets with different densities?
4. What is the relationship between gravitational acceleration and distance from the planet's center?
5. How would the acceleration change if this planet had an atmosphere?

### Tip
Remember that gravitational acceleration decreases with the square of the distance from the planet's center.

A planet has a radius of 1.20 × 10^6 m and a mass of 2.50 × 10^22 kg. What is the acceleration of gravity on the surface of the planet?

Learn how to calculate the acceleration of gravity on a planet's surface given a radius of 1.20 x 10^6 m and a mass of 2.50 x 10^22 kg. This physics problem uses Newton's Law of Universal Gravitation and includes a step-by-step solution for students in Grades 11-12.

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