The acceleration observed in any body due to gravity is called Acceleration due to gravity.

The force due to gravity is given by,

The force due to gravity is given by,

Where m = mass of the body and

g = gravity.The Universal Law of gravitation gives,

Where G is a constant equal to 6.67 $\times$ 10^{-11} N-m^{2}/kg^{2},

Equating both the force formula we get

The above formula can also be called **Gravitational Acceleration Formula**. It is used to find the acceleration due to gravity any where in space. On earth the acceleration due to gravity is 9.8 m/s^{2}.

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Below are given some problems based on acceleration due to gravity using the above formula.

### Solved Examples

**Question 1: **Calculate the acceleration due to gravity for a planet having mass 4 $\times$ 10^{23} Kg and is 3 $\times$ 10^{6} m away from sun?

** Solution: **

Given: Mass of planet M = 4 $\times$ 106 Kg,

Radius r = 3 $\times$ 106 m,

Gravitational Constant G = 6.67 $\times$ 10-11 N-m2/kg2

Acceleration due to gravity g = $\frac{GM}{r^{2}}$

= $\frac{6.67 \times 10^{-11} \times 4 \times 10^{23}}{3 \times 10^{6} m}$

= 2.964 m/s^{2}.

**Question 2: **Calculate the acceleration due to gravity for the planet pluto if it is having mass of 1.22 $\times$ 10^{22} Kg and radius is 1.15 $\times$ 10^{6} m.

** Solution: **

Given: Mass M = 1.22 $\times$ 10^{22} Kg and

Radius r = 1.15 $\times$ 10^{6} m

The acceleration due to gravity g = $\frac{GM}{r^{2}}$

= $\frac{6.67 \times 10^{-11} \times 1.22 \times 10^{22}}{(1.15 \times 10^{6})^{2}}$

= 0.61m/s^{2}.

Given: Mass of planet M = 4 $\times$ 106 Kg,

Radius r = 3 $\times$ 106 m,

Gravitational Constant G = 6.67 $\times$ 10-11 N-m2/kg2

Acceleration due to gravity g = $\frac{GM}{r^{2}}$

= $\frac{6.67 \times 10^{-11} \times 4 \times 10^{23}}{3 \times 10^{6} m}$

= 2.964 m/s

Given: Mass M = 1.22 $\times$ 10

Radius r = 1.15 $\times$ 10

The acceleration due to gravity g = $\frac{GM}{r^{2}}$

= $\frac{6.67 \times 10^{-11} \times 1.22 \times 10^{22}}{(1.15 \times 10^{6})^{2}}$

= 0.61m/s

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