Gravitational Acceleration Formula


Any object placed in the field of gravitational pull of the Earth experiences gravitational force. Gravitational acceleration is defined as the acceleration gained by an object because of the force of gravity acting on it. It is represented by ‘g’ and is measured in terms of m/s2. Gravitational acceleration is a vector quantity, that is, it possesses both magnitude as well as direction.

Formula: The gravitational acceleration acting on any object can be given using the following equation,  \(g = GM/ (r + h)^{2}\)

Here, G is the universal gravitational constant whose value is fixed and is equal to 6.673×10-11 N.m2/Kg2, M is the mass of the body whose gravitational pull is acting on the object under consideration, r is the radius of the planet and h is the height of the object from the surface of the body.

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.

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


From the Newton’s Second Law of Motion, we can write


Here, F is the force acting on the object, m is its mass and ‘a’ is the acceleration.

Also, as per Newton’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 experience the same force directed towards the other.

Using the Newton’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, when the object is on or near to the surface the value of g becomes constant and does not change considerably 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 × 106 m and its mass is 7.35 × 1022 kg.


As given in the question, the radius of the moon, r = 1.74 × 106m = 1740000 m.

Also, the mass of moon = 7.35 × 1022 kg

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

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

Upon substituting the values, we get,

\(g = (6.673 * 10^{-11})(7.35* 10^{22})/1740000\)

\(g = 1.620 \frac{m}{s^{2}}\)

The acceleration due to gravity is calculated to be 1.620 m/s2.

Example 2:

The radius of the Earth is 6.38 x 106 m. The mass of the Earth is 5.98x 1024 kg. If a satellite is orbiting the Earth 250 km above the surface, what acceleration due to gravity does it experience?


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 = (6.38 x 106 m) + (250 km)

R = 6 380 000 + 250 000 m

R = 6 630 000 m

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

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

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

\(g = 9.07 \frac{m}{s^{2}}\)

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