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}
Naveen Sharma 7 years, 1 month 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}
1Thank You