Physics Formulas

Acceleration due to Gravity Formula

We have no doubt seen gravity work in our life. After all, it is the force that is helping us to keep our feet on the ground. If we throw a ball in the upward direction in the air. Then it will come down on its own. Why? When the ball is going in upwards direction, its speed will be less as compared to when it comes down. This is because of the acceleration, which is produced due to the force of gravity. In this topic, we will discuss acceleration due to Gravity formula.  Let us learn about acceleration due to gravity in detail.

Acceleration due to Gravity Formula


Acceleration due to Gravity Formula

What is Acceleration due to Gravity?

Acceleration due to gravity is the acceleration that is gained by an object due to the gravitational force. Its SI unit is ms². It has a magnitude as well as direction. Thus it is a vector quantity.

We represent acceleration due to gravity by the symbol g. Its standard value on the surface of the earth at sea level is 9.8 ms². Its computation formula is based on Newton’s Second Law of Motion and Newton’s Law of Universal Gravitation.

Acceleration Due to Gravity Formula

Near the surface of Earth, the acceleration due to gravity is approximately constant. But, at large distances from the Earth, or around other planets or moons, it is varying. The acceleration due to gravity depends on the terms as the following:

Mass of the body,

Distance from the center of mass,

Constant G i.e. Universal gravitational constant.

g = \( \frac {G M}{r^2} \)


g Acceleration due to gravity \((units ms^ {-1} ) \)
G The universal gravitational constant,\(  = 6.673 \times 10^{-11} N m^2 Kg^2 \)
m Mass of a very large body like Earth.
r The distance from the center of mass of the large body

Variation of g with Height:

Acceleration due to gravity varies with the height from the surface of the earth. Its computation can be done as follows:

\( g_h = g \; (1+\frac {h}{R})^{-2} \)


g Acceleration due to gravity at the surface.
\(g_h\) Acceleration due to gravity at the height h.
R The radius of the earth.
h Height from the earth’s surface.

It is clear that the value of g decreases with an increase in height of an object. Hence the value of g becomes zero at infinite distance from the earth.

Solved Examples

Example-1: The radius of the moon is \( 1.74 \times 10^6 m\). The mass of the moon is taken as \(7.35 \times 10^{22}\) kg. Find out the acceleration due to gravity on the surface of the moon.

Solution: On the surface of the moon, the distance to the center of mass will be the same as the radius.

Thus,  r = \(1.74 \times 10^6 m\).

Mass of the object i.e. moon,

m = \(7.35 \times 10^{22}\) kg

As we know that, universal gravitational constant G = \( 6.673 \times 10^{-11} \)

The acceleration due to gravity on the surface of the moon can be computed by using the formula as below:

g = \( \frac {G M}{r^2} \)

Substituting the values,

g = \( \frac {6.673 \times 10^{-11} \times 7.35 \times 10^{22}}{(1.74 \times 10^6)^2}\)

g = \( 1.620 \; ms^{-2} \)

Hence, value of the acceleration due to gravity is \( 1.620 \; ms^{-2} \)

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5 responses to “Spring Potential Energy Formula”

  1. Typo Error>
    Speed of Light, C = 299,792,458 m/s in vacuum
    So U s/b C = 3 x 10^8 m/s
    Not that C = 3 x 108 m/s
    to imply C = 324 m/s
    A bullet is faster than 324m/s

  2. Malek safrin says:

    I have realy intrested to to this topic

  3. umer says:

    m=f/a correct this

  4. Kwame David says:

    Interesting studies

  5. Yashdeep tiwari says:

    It is already correct f= ma by second newton formula…

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