# Acceleration of Gravity on Earth

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Acceleration of gravity is the acceleration of earth by which it pulls anything towards it.  A free-falling object has an acceleration of 9.8 m/s^2, downward on earth. This numerical value for the acceleration of a free falling body is called acceleration of gravity on earth.

It is denoted by “g”.

The gravitation of gravity on the surface of the earth at sea level is 9.8 m/s^2. The value of g is dependent on the location. As we move above the surface of the earth the value of g will be decreasing.

As we know that, F(grav) = m*g

And                      F(grav) = G*M*m
d^2

Therefore, Acceleration of gravity, g = G*M
d^2

Where, F(grav) = Gravitational force

G = Gravitational constant

M = Mass of earth

m = mass of object on earth

d = distance from the center of the object to the center of the earth.

Example: A boy drops a stone from a cliff, which reaches the ground in 25 seconds. Calculate the height of the cliff.

Solution:

Initial velocity, v = 0 m/s

Time taken, t = 25 sec

Acceleration due to gravity, a = g = 9.8 m/s^2

Now, we can use the equation for distance (or height)

H = vt + ½ at^2

= 0 + ½ (9.8) (25)^2

= 612.5 m

Example: How long will it take for an apple falling from a 25.4m-tall tree to hit the ground?

Solution:

Height travelled by the apple, h = 25.4 m

Initial velocity, v = 0

Acceleration due to gravity, g = 9.8 m/s^2

Now, we can use the equation for distance (or height)

H = vt + ½ at^2

25.4 = 0 + ½ (9.8) (t)^2

Time, t = 2.27 sec