Acceleration due to gravity can be defined as the acceleration produced in the motion of the body under the effect of gravity. Gravity is the force by which a heavier object like a planet attracts a smaller object towards its center.

- It is denoted by a small â€˜gâ€™ and its
**SI unit is meter-per-square-second (m/s**^{2}**)**. - It may also be referred to as Gravitational acceleration.
- The
**dimensional formula of acceleration due to gravity is [ M**^{0}L^{1}T^{-2}]

## Acceleration Due to Gravity on the Surface of the Earth

Consider the earth as a sphere that is made up of a large number of concentric shells with its center as the center of the earth.

Let a particle lie inside the earth at a distance r from the center of the earth. Let R_{e} be the radius of the earth and M_{e} be the mass of the earth.

**Acceleration due to gravity of the surface of the earth**

The density of the earth, $$ \rho = \frac{M_e}{\frac{4}{3}\pi R^3_e}~~~~~ …..(i)\\ \Rightarrow M_e = \frac {4}{3}\pi R^3_e \rho $$

Also, the mass of the sphere of radius r will be $$ M_r =\frac{4}{3}\pi r^3 \rho $$

Now the gravitational force on mass m due to M_{r} is given by $$ F= \frac {GmM_r}{r^2}=\frac {Gm \frac {4}{3}\pi r^2 \rho}{r^2}=\frac {4}{3}Gm\pi r \rho $$

Using equation (i), $$ F= \frac {4}{3}Gm\pi r\frac{3M_e}{4\pi R^3_e}=\frac {GmM_e}{R^3_e}r $$

Now, the mass m is located at the surface of the earth, then r = R_{e}. Therefore $$ F = \frac {GmM_e}{R^3_e}R_e \\ \Rightarrow F= \frac {GmM_e}{R^2_e}$$

Now, by Newton’s second law, $$ F = ma = mg $$

Where g is the acceleration due to gravity.

Therefore, $$ mg = \frac {GmM_e}{R^2_e} $$

or $$ g = \frac {GM_e}{R^2_e}$$

The above equation is the formula for calculating acceleration due to gravity on the surface of the earth.

For any planet, the formula of acceleration due to gravity is given by $$ g_p = \frac {GM_p}{R^2_p}$$

Where

- M
_{p}is the mass of the planet - R
_{p}is the radius of the planet

On substituting the values, we get the value of acceleration due to gravity on the surface of the earth

$$ g = \frac {6.67 \times 10^-11 \times 6 \times 10^{24}}{(6.4 \times 10^6)^2} = 9.8~ m s^2 $$

Hence, **the value of g on the surface of the earth is 9.8 m/s ^{2}**.

## Solved Example

**Ques. What will be the relationship between the acceleration due to gravity on the surface of the Earth and on a planet, whose mass and radius are twice that of the Earth?**

**Ans**. We have, acceleration due to gravity on the surface of the earth $$ g_e = \frac {GM_e}{R^2_e}~~~~….(i) $$

Also, the acceleration due to gravity on the surface of any planet $$ g_p = \frac {GM_p}{R^2_p}~~~~….(ii) $$

Given

- Mass of the planet, M
_{p}= 2 x mass of the earth = 2M_{e} - Radius of the planet, R
_{p}= 2 x radius of the earth = 2R_{e}

Therefore, equation (ii) becomes $$ g_p = \frac {G \times 2M_e}{(2R_e)^2} = \frac {2}{4} \frac {GM_e}{R^2_e} $$

From equation (i), we get

$$ g_p = \frac {1}{2}g_e = \frac {g_e}{2} $$

Hence, acceleration due to gravity on the surface of the planet is half of the acceleration due to gravity on the surface of the earth.

**Ques. What will be the relationship between the acceleration due to gravity on the surface of the Earth and on the moon, whose mass is 1/82 times the mass of the Earth and whose radius is 1/3.8 times that of the Earth?**

**Ans**. We have, acceleration due to gravity on the surface of the earth $$ g_e = \frac {GM_e}{R^2_e}~~~~….(i) $$

Also, the acceleration due to gravity on the surface of the moon $$ g_p = \frac {GM_m}{R^2_m}~~~~….(ii) $$

Given

- Mass of the moon, M
_{m}= 1/82 x mass of the earth = M_{e}/82 - Radius of the planet, R
_{m}= 1/3.8 x radius of the earth = R_{e}/3.8

Therefore, equation (ii) becomes $$ g_m = \frac {G \frac {M_e}{82}}{(\frac {R_e}{3.8})^2} = \frac {3.8^2}{82} \frac {GM_e}{R^2_e} $$

From equation (i), we get

$$ g_m = \frac {1}{6}g_e = \frac {g_e}{6} $$

Hence, acceleration due to gravity on the surface of the moon is 1/6 times the acceleration due to gravity on the surface of the earth.

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