Gravitation

Acceleration Due to Gravity

Let us carry out a small activity! Throw a ball up in the air. Now let the ball come down on its own. Did you notice one thing? When the ball is going in upwards direction the speed of the ball is less as compared to when it comes down. This is because the acceleration is produced due to the force of gravity. Let us now study about acceleration due to gravity in detail.

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Acceleration Due to Gravity

When the object falls towards the earth due to the earth’s gravitational force, it something we call as free fall of the object.  So during the free fall, the only force acting on the object is the gravitational force of the earth. The acceleration due to gravity is the acceleration produced in the freely falling body due to the influence of the gravitational pull of the earth.

Acceleration due to Gravity

Acceleration due to gravity is denoted by ‘ g ‘ but its values vary. Like, for example, the acceleration due to gravity on the moon is different from that of the earth.

Read the Concept of Acceleration here for better understanding of this topic.

Expression of the Acceleration Due to Gravity

Let us suppose you are standing on the top of the building with a small stone in your hand. So let the mass of the stone be ‘ m ‘. When you throw the stone on the ground, the gravitational force of the earth attracts the stones downwards. The gravitational force acting on the stone is F = mg.

Also, we know that force between two objects is given by the universal law of gravitation. So here one object is the stone and object is the earth.

F = \( \frac{GMm}{d^2} \)

  • M = mass of the earth
  • m = mass of the stone
  • d = distance

⇒ mg =  \( \frac{GMm}{d^2} \), or

 g =   \( \frac{GM}{d^2} \)

Suppose the object is on the surface of the earth or nearby.

Now, in this case, d = R+ h

So, g = \( \frac{GM}{(R+h)^2} \)

Let us calculate the value of g on the earth, i.e h=0. As we know the value of g,

g = \( \frac{GM}{R^2} \)

  • G = 6.7 × 10-11 Nm²/kg²
  • M = 6 × 1024 kg
  • R = 6.4 × 10m

So putting these values we will get the value of acceleration due to gravity.

g = \(\frac{6.7×10^{11}×6×10^{24}} {(6.4×10^{6})^2}\)

g = 9.8 m/s²

This is the value of acceleration due to gravity. The value of this acceleration due to gravity changes from place to place. It is not universal constant.

Learn more about Newtons Universal Law of Gravitation here

Solved Questions For You

Q1. The ratio of the value of gravitational constant between Earth and Moon system and Earth and Sun system is-

  1. > 1
  2. > 1
  3. 1
  4. Cannot be calculated.

Ans: C. In Newton’s law of gravitation, F = \( \frac{GMm}{r^2} \). G is the gravitational constant which has the fixed value 6.7 × 10-11 m³/kg-1 s-2. Therefore the ratio of the value of gravitational constant G between Earth and Moon system and Earth and Sun system is 1.

Q2. Which of the following is/are true about the value of g

  1. The value of g is minimum at the equator
  2. The value of g is maximum at poles.
  3. Average value of g is 9.8 m/s²
  4. All of the above.

Ans: D. At a given place the value of acceleration due to gravity is constant but it varies from one place to another place on the surface of the earth. This is the reason that the earth is not of the perfect shape. It is flattened at the poles and bulged out at the equator.

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