Question

what is Newton's law of gravitation drive in experience for acceleration due to gravity

Answer 1

Answer:

Expression for acceleration due to gravity is

$\boxed{\red{\sf\:g\:=\:\dfrac{G\:M}{r^2}}}$

Explanation:

Newton's law of gravitation:

1. Every object in the universe attracts other object with a force.

2. The force is directly proportional to the product of masses of two objects.

3. While it is inversely proportional to the square of the distance between those two objects.

Derivation of expression for acceleration due to gravity:

NOTE: Refer to the attachment for the diagram.

Let a body of mass $\sf\:m$ be released from a distance $\sf\:r$ from the centre of the earth ( O ).

The mass of earth is denoted by M.

According to Newton's law of of gravitation,

The magnitude of the earth's gravitational force is given by -

$\boxed{\red{\sf\:F\:=\:G\:\dfrac{M\:m}{r^2}}}$

Where,

G is the universal gravitational constant.

Now, we know that,

$\displaystyle\sf\:Force\:=\:Mass\:\times\:Acceleration$

The acceleration produced by this force is given by -

$\displaystyle\sf\:Acceleration\:=\:\dfrac{Force}{Mass}\\\\\\\implies\sf\:g\:=\:\dfrac{F}{m}\\\\\\\implies\sf\:g\:=\:\dfrac{G\:M\:\cancel{m}}{r^2\:\times\:\cancel{m}}\\\\\\\implies\boxed{\red{\sf\:g\:=\:\dfrac{G\:M}{r^2}}}$

This is the required formula for the acceleration due to gravity.

This acceleration is always directed towards the centre of the earth.

If the altitude of an object is denoted by h, then r = R + h.

Where,

R is the radius of the earth,

$\boxed{\red{\sf\:g\:=\:\dfrac{G\:M}{\:(\:R\:+\:h\:)^2}}}$

Answer 2

Answer:

Acceleration due to gravity is the acceleration gained by an object due to the gravitational force. Its SI unit is m/s2. It has both magnitude and direction, hence, it’s a vector quantity. Acceleration due to gravity is represented by g. The standard value of g on the surface of the earth at sea level is 9.8 m/s2.