Question

find the expression for percentage change in gravity with height h from the earth's surface​

Answer 1

Answer:

Force of gravity acting on a body of mass m on the earth surface = F = GMm/R^2 ____________ (1)

Where R is the radius of the earth (considering it a homogenous sphere)

and M here is the total mass of earth concentrated at its centre. G is the gravitational constant.

Now, from Newton’s 2nd Law of Motion,

F = mg.___________________(2)

Just to recapitulate, as a body falls downwards it is continuously acted upon by a force of gravity. The body thus possesses an acceleration, called Acceleration Due to gravity(g).

From equation 1 and 2 we can write

mg = (GMm) / R^2

=> g = GM / R^2 _______________ (3)

Another expression of g here:

If mean density of earth is p then mass of the earth = M = volume X density = (4/3) Pi R^3 p

(Pi = 22/7)

g = G.( (4/3) Pi R^3 p) / R^2

g = (4/3) Pi R p G ________________ (4)

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Answer 2

Question:-

Obtain an expression for percentage change in acceleration due to gravity with height 'h' from the earth's surface​.

Solution:-

Let the acceleration due to gravity at the height 'h' from the surface of the earth be $\sf{g',}$ so that the percentage change in acceleration due to gravity will be,

$\longrightarrow\sf{\delta g=\dfrac{g'-g}{g}\times100}$

We know the acceleration due to gravity at the surface of earth is,

$\longrightarrow\sf{g=\dfrac{GM}{R^2}\quad\quad\dots(1)}$

where,

$\sf{G=}$ gravitational constant

$\sf{M=}$ mass of earth

$\sf{R=}$ radius of earth

As we go away from earth's surface through a height 'h', the mass of earth does not change but the distance from the center of earth changes from $\sf{R}$ to $\sf{R+h.}$

Thus the acceleration due to gravity at the height will be,

$\longrightarrow\sf{g'=\dfrac{GM}{(R+h)^2}\quad\quad\dots(2)}$

On dividing (2) by (1), we get,

$\longrightarrow\sf{\dfrac{g'}{g}=\dfrac{\left(\dfrac{GM}{(R+h)^2}\right)}{\left(\dfrac{GM}{R^2}\right)}}$

$\longrightarrow\sf{\dfrac{g'}{g}=\dfrac{R^2}{(R+h)^2}}$

$\longrightarrow\sf{\dfrac{g'}{g}=\left(\dfrac{R}{R+h}\right)^2}$

$\longrightarrow\sf{g'=g\left(\dfrac{R}{R+h}\right)^2}$

Subtracting 'g' from both sides,

$\longrightarrow\sf{g'-g=g\left(\dfrac{R}{R+h}\right)^2-g}$

$\longrightarrow\sf{g'-g=g\left[\left(\dfrac{R}{R+h}\right)^2-1\right]}$

$\longrightarrow\sf{\dfrac{g'-g}{g}=\left(\dfrac{R}{R+h}\right)^2-1}$

$\longrightarrow\sf{\dfrac{g'-g}{g}=\dfrac{R^2-(R+h)^2}{(R+h)^2}}$

$\longrightarrow\sf{\dfrac{g'-g}{g}=\dfrac{(R+R+h)(R-R-h)}{(R+h)^2}}$

$\longrightarrow\sf{\dfrac{g'-g}{g}=-\dfrac{h(2R+h)}{(R+h)^2}}$

To obtain percentage change we have to multiply both sides by 100.

$\longrightarrow\sf{\dfrac{g'-g}{g}\times100=-\dfrac{h(2R+h)}{(R+h)^2}\times100}$

That is,

$\longrightarrow\sf{\underline{\underline{\delta g=-\dfrac{100h(2R+h)}{(R+h)^2}}}}$