Question

The height above the surface of the earth at which acceleration due to gravity is half the acceleration due to gravity at surface of the earth is ( $R = 6.4 \times {10^6}$ )
(A) $6.4 \times {10^6}$
(B) $2.6 \times {10^6}$
(C) $12.8 \times {10^6}$
(D) $19.2 \times {10^6}$

Answer 1

In order to find the answer to the above question, we have to use the below expression of acceleration due to the gravitation,
$g = \dfrac{{GM}}{{{R^2}}}$
Where, g is acceleration due to gravity
G is gravitational constant
M is mass of the earth
R is the radius of the earth.

Complete answer:
We know that the acceleration due to gravity on any object placed at the surface of earth is given by:
$g = \dfrac{{GM}}{{{R^2}}} - - - - - - {\text{ }}(1)$
Where,g is acceleration due to gravity
G is gravitational constant
M is mass
R is the radius of earth.
Now let us write an expression assuming that we have taken the object to a height h, where g is now half of what it was on the surface, so we can write the expression as,
$\dfrac{g}{2} = \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}} - - - - - - - {\text{ }}\left( 2 \right)$
Dividing equation $2$ by equation $1$ , we will get,
$\dfrac{{(\dfrac{g}{2})}}{{(g)}} = \;\dfrac{{\dfrac{{GM}}{{\left( {R\; + \;h} \right){\;^2}}}}}{{\dfrac{{GM}}{{{R^2}}}}}\;$
Rearranging the above equation,
$\dfrac{g}{2} \times \dfrac{1}{g} = \;\dfrac{{GM}}{{{{(R + h)}^2}}} \times (\dfrac{R}{{GM}})$
$\Rightarrow \dfrac{1}{2} = \dfrac{R}{{{{(R + h)}^2}}}$
Rearranging it,
$2R = {\left( {R\; + \;h} \right)^2}$
Taking square root on both sides will give us,
$1.414\;R = \left( {R\; + \;h} \right)$
$\Rightarrow 1.414R - R = h$
$\Rightarrow 0.414R = h$
Putting in the given value of R, we will get,
$h = 2.6 \times {10^6}$
Hence (b) is the correct option.

Note:
One should be extremely careful with the language of the question, as the height asked here was from the surface of the earth but it can be asked from the center also, in that case we will add the radius of the earth to our answer.