At what height above the surface of earth will the value of acceleration due to gravity becomes 64% of its value at the surface of earth radius of Earth R?

Concept:

An object is subjected by an external force it will accelerate and its rate of acceleration is directly proportional to the external force.

∵ F = m a

• Acceleration Due to Gravity (g): When an object is attracted by gravitational force it will be accelerated towards the surface of the heavier mass, such acceleration is known as acceleration due to gravity (g) and it is directly proportional to heavier mass (mass of planet M) and inversely proportional to the square of the distance between them.

Gravitational acceleration, \({\rm{g}} \propto \frac{{\rm{M}}}{{{{\rm{r}}^2}}} \Rightarrow {\rm{g}} = \frac{{{\rm{GM}}}}{{{{\rm{r}}^2}}}\)

And S.I unit of gravitation is given as m/s2.

• As moving upward from the surface of the earth,

\({\bf{g}} \propto \frac{1}{{{{\left( {{\bf{R}} + {\bf{h}}} \right)}^2}}}\)

Here R is the radius of the planet and h is the height from its surface which means that as we move away from the surface its gravitational acceleration will decrease exponentially, it can also be expressed as \({\rm{g'}} = \frac{{\rm{g}}}{{{{\left( {1 + \frac{{\rm{h}}}{{\rm{R}}}} \right)}^2}}} = {\rm{g}}\left( {1 - \frac{{2{\rm{h}}}}{{\rm{R}}}} \right)\) 

This is true if h ≪ R.

• As going inside the surface,

Now as we move below the surface the variation of gravitational acceleration gd can be expressed as \({{\bf{g}}_{\bf{d}}} \propto \left( {{\bf{R}} - {\bf{d}}} \right) \Rightarrow {{\bf{g}}_{\bf{d}}} = {\bf{g}}\left( {1 - \frac{{\bf{d}}}{{\bf{R}}}} \right)\) 

This shows that as we move below the planet's surface it will gradually decrease and become zero at planets core.

Here R is the radius of the planet, g is the acceleration due to gravity on the surface, and gd is the acceleration due to gravity at some depth, and d is the depth from its surface.

Calculation:

Given that radius of earth is 6400 km. 1% decrease in gravity

Acceleration due to gravity above the surface of earth at height h is given

\(g' = g\left( {1 - \frac{{2h}}{{{R_e}}}} \right)\)

Here, g’ = 0.99 g

⇒ \(0.99 = 1 - \frac{{2h}}{{{R_e}}}\)

⇒ \(\frac{{2h}}{{{R_e}}} = 0.01\)

⇒ \(h = \frac{{0.01}}{2}\;{R_e}\)

⇒ \(h = \frac{{0.01}}{2} \times 6400 = 32\;Km\)

h = 32 Km