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Discus the variation of acceleration due …

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Discus the variation of acceleration due to gravity with altitude . How does the expression modify when h<<R?

  • 1 answers

Naveen Sharma 7 years ago

Ans. Let g be the value of acceleration due to gravity at the surface of earth and g' at a height h above the surface of earth. If the earth is considered as a sphere of homogeneous composition, then g at any point on the surface of the earth is given by:

{tex}g = {GM\over R^2}{/tex}  …………(1)

Where M mass of earth, R is radius of Earth.

Now, we ll determine Value of g at the distance h from the surface of earth.

Then distance b/w center of earth and point = (R+h)

{tex}g' = {GM\over (R+h)^2}{/tex} ……………(2)

divide eq (2) by (1)

{tex}{g'\over g} ={ { GM\over (R+h)^2} \over {GM\over R^2}}{/tex}

{tex}=> {g'\over g}= {R^2\over (R+h)^2}{/tex}

{tex}=> {g'\over g}= {R^2\over R^2 (1+{h\over R})^2} {/tex}

{tex}=> {g'\over g}= {1\over (1+{h\over R})^2}{/tex}

{tex}=> {g'\over g}= {(1+{h\over R})^{-2}}{/tex}

Expanding using Binomial Theorem and neglecting terms with higher power of R, we get

{tex}=> {g'\over g}=(1-2{h\over R}){/tex}

 

So, The value of acceleration due to gravity 'g' decreases with altitude 'h'.

When h<< R then expression is

{tex}=> {g'\over g}=(1-2h){/tex}

 

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