**Definition: **

Any object located in the field of the earth experiences a gravitational pull. Gravitational acceleration is described as the object receiving an acceleration due to the force of gravity acting on it.Â It is represented by â€˜gâ€™ and its unit is m/s^{2}. Gravitational acceleration is a quantity of vector, that is it has both magnitude and direction.

**Formula: **

\(g = GM/ (r + h)^{2}\)

Here, G is the universal gravitational constant (G = 6.673Ã—10-11 N.m2/Kg2.)

r is the planet radius.

h is the height of the object from the body surface.

When the object is on or near the surface of the body, the force of gravity acting on the object is almost constant and the following equation can be used.

\(g = GM/ r^{2}\)

**Derivation:**

From Newtonâ€™s Second Law of Motion, we can write

F=ma

Here, F is the force acting on the object.

m is its mass and

â€˜aâ€™ is the acceleration.

Also, as per Newtonâ€™s Law of Gravity, we can write,

\(F_{g} = GMm / (r + h)^{2}\)

It is the gravitational force acting between two bodies lying in the gravitational field of each other. This force acts inwards and is attractive in nature. Each of the two bodies experiences the same force directed towards the other.

Using Newtonâ€™s second law of motion, in order to find the acceleration of the body under this condition,

\(a = F_{g}/m\)

Here, m is the mass of the object for which the gravitational acceleration is to be calculated.

\(a = g = GMm / (r + h)^{2}m\)

\(g = GM/ (r + h)^{2}\)

Also, the value of g is constant when the object is on or near the surface and there is no considerable change with the height. Hence we can write,

\(g = gM/r^{2}\)

**Real Life examples: **

Let us consider a satellite that has to revolve in the upper part of the atmosphere surrounding the Earth. In order to calculate the velocity with which it has to move so as to remain in its path, we must know the gravitational acceleration acting on the object.

**Solved examples **

**Example 1:**

**Calculate the acceleration due to gravity for an object placed at the surface of the Earth, given that, the radius of the Moon is 1.74 Ã— 10 ^{6} m and its mass is 7.35 Ã— 10^{22} kg. **

**Solution: **

The radius of the moon, r = 1.74 Ã— 10^{6}m = 1740000 m

r^{2} = 3.0276Â Ã— 10^{12}m

The mass of moon = 7.35 Ã— 10^{22} kg

Using the formula for the acceleration due to gravity, we write,

\(g = GM/ r^{2}\)

Upon substituting the values, we get,

\(g = (6.673 Ã—10^{-11})(7.35Ã— 10^{22})/3.0276 Ã— 10^{12}\)g = (4.905Â Ã—10^{12})/(3.0276Â Ã— 10^{12})

**g = 1.620 m/s ^{2}**

The acceleration due to gravity is 1.620 m/s^{2}.

**Example 2:**

**The radius of the Earth is 6.38 x 106Â m. The mass of the Earth is 5.98x 10 ^{24}Â kg. If a satellite is orbiting the Earth 250 km above the surface, what acceleration due to gravity does it experience?**

**Solution:**

It can be seen that the satellite is present at a considerable height from the surface of the Earth, hence the height cannot be neglected. Using the first formula, we can write,

*R=r+h*Â = (6.38 x 10^{6}Â m) + (250 km)

*R*Â = 6 380 000 + 250 000 m

*R*Â = 6 630 000 m

The acceleration due to the gravity of the satellite can be found from the formula:

\(g = GM/ (r + h)^{2}\)

\(g = (6.673 Ã— 10^{-11})(5.98Ã— 10^{24})/(6630000)^{2}\)

g = (3.9704Ã—10^{14})/(4.396Ã—10^{13})

**g= 9.031 m/s ^{2}**