Question

Obtain an expression for acceleration due to gravity at height h above the earth’s surface.

Answer 1

Hint:The acceleration due to gravity is the acceleration produced by gravity on the body whenever the body is in motion. We know that the acceleration of gravity is maximum at earth’s surface and it is calculated by using the formula $${\displaystyle \frac{GM}{R^2}}$$. This value of acceleration changes as we go upward at height h from the surface.

Formula Used:
$$\text{g = }{\displaystyle \frac{GM}{R^2}}$$
Where, $$\text{g}$$ is the acceleration due to gravity, $$G$$ is gravitational constant, $$M$$ is the mass of earth and $$R$$ is the radius of earth.

Complete step by step answer:
Whenever an object is in motion or at some height then there is some acceleration in the object due to gravity which is called acceleration due to gravity. It is denoted by $$\text{g}$$ and its value can be calculated by using the formula:
$$\text{g = }{\displaystyle \frac{GM}{R^2}}$$
The value of $$\text{g}$$ is maximum at earth’s surface which is calculated as $$9.8\text{ m }{\text{s}}^{-2}$$. Thus each object which is present on the surface of earth is suffered by this value of acceleration. But when the object goes upwards at a height h above the earth’s surface the value of $$9.8\text{ m }{\text{s}}^{-2}$$ changes accordingly. Let’s assume an object present at height h above the earth’s surface.

The acceleration due to gravity at earth’ surface is given by formula:
$$\text{g = }{\displaystyle \frac{GM}{R^2}}$$ ______________$$(1)$$
When the object is present at height then total distance from the centre of earth becomes $$R\text{ + }h$$. Let acceleration due to gravity at this height be represented by $${\text{g}}^{\iota }$$ , then $${\text{g}}^{\iota }$$ will be ,
$${\text{g}}^{\iota }\text{ = }{\displaystyle \frac{GM}{{\left(R\text{ + }h\right)}^2}}$$ ______________$$(2)$$
Now dividing equation $$(1)$$ and equation $$(2)$$ we get the result as,
$$\Rightarrow {\displaystyle \frac{\text{g}}{g^{\iota }}}\text{ = }{\displaystyle \frac{{\displaystyle \frac{GM}{R^2}}}{{\displaystyle \frac{GM}{{\left(R\text{ + }h\right)}^2}}}}$$
It can be simplified as,
$$\Rightarrow {\displaystyle \frac{\text{g}}{g^{\iota }}}\text{ = }{\displaystyle \frac{{\left(R\text{ + }h\right)}^2}{R^2}}$$

Thus acceleration due to gravity at height h can be calculated by formula:
$$\Rightarrow g^{\iota }\text{ = g}{\displaystyle \frac{R^2}{{\left(R\text{ + }h\right)}^2}}$$
Taking R common from the term we have and when $$h\ll R$$ then by using binomial expansion we can expand the above equation as,
$$\Rightarrow g^{\iota }\text{ = g}{\displaystyle \frac{R^2}{R^2{\left(\text{1 + }{\displaystyle \frac{h}{R}}\right)}^2}}$$
$$\Rightarrow g^{\iota }\text{ = g}{\left(\text{1 + }{\displaystyle \frac{h}{R}}\right)}^{-2}$$
It can be expand as,
$$\therefore g^{\iota }\text{ = g}\left(\text{1 - }{\displaystyle \frac{2h}{R}}\right)$$

Therefore the acceleration due to gravity is $g\left(\text{1 - }{\displaystyle \frac{2h}{R}}\right)$ at height $h$ above earth’s surface..

Note:Similarly we can find the acceleration due to gravity when we go in depth into earth. The acceleration at earth's surface is uniform all around. It will vary when the distance from the centre changes. The value of radius and mass of earth can be taken to approximate values. The radius of earth is $$6400\text{ km}$$.