The average gravitational force of the Earth is 980 cm/s².

Explanation:

Gravitational force is the force of attraction between two objects with mass. In the case of the Earth, it exerts a force on any object near its surface due to its mass. This force is commonly referred to as the acceleration due to gravity and is denoted by 'g'.

The average gravitational force of the Earth can be calculated using Newton's law of universal gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

F = G * (m1 * m2) / r^2

Where:
- F is the gravitational force
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
- m1 and m2 are the masses of the two objects
- r is the distance between the centers of the two objects

In the case of an object near the surface of the Earth, the other object is the Earth itself, and the distance is the radius of the Earth.

Now, let's calculate the average gravitational force of the Earth.

Step 1: Convert the given value of 980 cm/s^2 to m/s^2.
- 1 cm = 0.01 m
- Therefore, 980 cm/s^2 = 980 * 0.01 m/s^2 = 9.8 m/s^2

Step 2: Use the formula to calculate the mass of the Earth (m1) in kilograms.
- Rearrange the formula to solve for m1: m1 = (F * r^2) / (G * m2)
- The mass of an object near the surface of the Earth (m2) can be considered negligible compared to the mass of the Earth itself.
- The radius of the Earth (r) is approximately 6,371 kilometers (or 6,371,000 meters).

Step 3: Substitute the known values into the formula.
- m1 = (9.8 * (6,371,000)^2) / (6.67430 × 10^-11 * m2)
- Since m2 is negligible compared to the mass of the Earth, we can ignore it in this calculation.

Step 4: Calculate the mass of the Earth.
- m1 = (9.8 * (6,371,000)^2) / (6.67430 × 10^-11)
- m1 ≈ 5.972 × 10^24 kg

Conclusion:
The average gravitational force of the Earth is approximately 9.8 m/s^2, which is equivalent to 980 cm/s^2.