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
And S.I unit of gravitation is given as m/s2.
\({\bf{g}} \propto \frac{1}{{{{\left( {{\bf{R}} + {\bf{h}}} \right)}^2}}}\) 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.
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 Open in App Suggest Corrections 1 |