Question

Estimate the value of acceleration due to gravity of the earth when $g = \dfrac{{G{M_e}}}{{R_e^2}}$.

Answer 1

Hint:We have to find the value of acceleration due to gravity of the earth by putting the various constant values in the given formula such as Universal gravitational constant, mass of the earth and radius of the earth.

Formula used:
Acceleration due to gravity of the earth is given as
$g = \dfrac{{G{M_e}}}{{R_e^2}}$
Where, $G$ - Universal gravitational constant, ${M_e}$ - Mass of the earth and \[{R_e}\] - Radius of the earth.

Complete step by step answer:
 The value of acceleration due to gravity of the earth is not constant. It changes when we go upwards, downwards from the surface of the earth. Its value is highest on the surface of the earth i.e. $9.8\,m{s^{ - 2}}$, $0m{s^{ - 2}}$ at the center of the earth and both decrease exponentially as we go down or up from the surface of earth.

Let us write the constant values of the given data, we have
$G = 6.673 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$
$\Rightarrow {M_e} = 6 \times {10^{24}}kg$
\[\Rightarrow {R_e} = 6400km = 64 \times {10^5}m\]
Substituting these values in the given formula, we get
$g = \dfrac{{6.673 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}}}}{{{{\left( {64 \times {{10}^5}} \right)}^2}}}$
$g = \dfrac{{6.673 \times 6 \times {{10}^3}}}{{4096}}$
On solving,
$\Rightarrow g = 9.7749 \times {10^{ - 3}} \times {10^3}$
$\Rightarrow g = 9.7749\,m{s^{ - 2}}$
$\therefore g \approx 9.8\,m{s^{ - 2}} = 980\,cm{s^{ - 2}}$

Hence, the value of acceleration due to gravity of the earth is $9.8\,m{s^{ - 2}} = 980\,cm{s^{ - 2}}$.

Note:We should know all the universal constant values to take in the given formula such as Universal gravitational constant, mass of the earth and radius of the earth with their proper units. Also, we should know the conversion factors such as $1m = 100cm$ & $1km = 1000m$. Do the calculations properly.