Question

Value of g on the surface of earth is $ 9.8{ m }/{ { s }^{ 2 } }$. At height h=R from the surface the value of g is $ \dfrac { g }{ x }$. Find x.

Answer 1

Hint: Write the formula for acceleration due to gravity on the surface and acceleration due to gravity at height h. And then divide the equations. Arrange the equation in such a way that you get acceleration due to gravity at height h in terms of acceleration due to gravity on the surface.

Formula used:
$ g\quad =\quad \dfrac { GM }{ { R }^{ 2 } }$
$ { g }^{ ' }\quad =\quad \dfrac { GM }{ { \left( R\quad +\quad h \right) }^{ 2 } }$

Complete answer:
Acceleration due to gravity on the surface is given by,
$ g\quad =\quad \dfrac { GM }{ { R }^{ 2 } }$ …(1)

where, g: Acceleration due to gravity
           G: Gravitational constant
            M: Mass of earth
            R: Radius of earth

Acceleration due to gravity at height h is given by,
$ { g }^{ ' }\quad =\quad \dfrac { GM }{ { \left( R\quad +\quad h \right) }^{ 2 } }$...(2)
Dividing equation. (2) by equation. (1) we get,
$ \dfrac { { g }^{ ' } }{ g } =\quad \dfrac { { R }^{ 2 } }{ { \left( R\quad +\quad h \right) }^{ 2 } }$
At h= R,
$ \dfrac { { g }^{ ' } }{ g } =\quad \dfrac { { R }^{ 2 } }{ { \left( R\quad +\quad R \right) }^{ 2 } }$
$ \Rightarrow \dfrac { { g }^{ ' } }{ g } =\quad \dfrac { { R }^{ 2 } }{ { \left( 2R \right) }^{ 2 } }$
$ \Rightarrow \cfrac { { g }^{ ' } }{ g } =\quad \cfrac { 1 }{ 4 }$
$ \Rightarrow { g }^{ ' }=\quad \cfrac { g }{ 4 }$
Therefore, the value of x is 4.

Note:
Acceleration due to gravity g is inversely proportional to radius R and eventually height h. This means as the height increases, acceleration due to gravity decreases. Thus, acceleration due to gravity is smaller at height as compared to that on the surface.