Question

Find the height above the surface of the earth where the weight of a body becomes half.
(A) $\dfrac{R}{2}$
(B) $\left( {\sqrt 2 - 1} \right)R$
(C) $\dfrac{R}{{\left( {\sqrt 2 + 1} \right)}}$
(D) $\dfrac{R}{{\sqrt 2 }}$

Answer 1

Gravitational Force is an attractive force which depends on the distance between the centers of mass of the bodies between which it is acting. It also depends on the product of their masses.

Complete step by step solution:
According to Newton’s Universal Law of Gravitation, there exists an attractive force amidst any two objects in the universe which is directly proportional to the product of masses of the two objects and inversely proportional to the square of distance between them. So,
${F_G} \propto {M_1}{M_2}$, and
${F_G} \propto \dfrac{1}{{{r^2}}}$
So, ${F_G} \propto \dfrac{{{M_1}{M_2}}}{{{r^2}}}$
$\Rightarrow {F_G} = G\dfrac{{{M_1}{M_2}}}{{{r^2}}}$
Where, ${F_G} =$Gravitational Force,
${M_1},{M_2} =$Masses of the objects,
$r =$Distance between the objects,
$G =$Universal Gravitational Constant$= 6.67 \times {10^{ - 11}}N{m^2}/k{g^2}$.
Now, the weight of any object is actually the gravitational force by which the earth pulls the object towards itself. So for an object of mass $m$, its weight can be calculated by,
$W = G\dfrac{{Mm}}{{{R^2}}}$
Where, $W =$Weight of the object,
$M =$Mass of earth,
$R =$Radius of earth.
We also know that weight can also be calculated by,
$W = mg$,
Where, $g =$Acceleration due to gravity.
The gravitational force caused by the earth is known as gravity. And the acceleration that this gravity produces in any object is known as acceleration due to gravity.
Now it’s inferred that mass of any object is independent of any physical parameters, whereas weight depends on a lot of physical parameters. One of those parameters is distance. As we increase the distance between earth and that object, its weight decreases. So the distance at which its weight becomes half can be calculated by,
$\dfrac{W}{2} = G\dfrac{{Mm}}{{{r^2}}}$
Where, $r =$distance between the center of earth and the object.
$\Rightarrow \dfrac{{mg}}{2} = G\dfrac{{Mm}}{{{r^2}}}$
Since, $g = \dfrac{{GM}}{{{R^2}}}$
So, $\dfrac{{m\left( {\dfrac{{GM}}{{{R^2}}}} \right)}}{2} = G\dfrac{{Mm}}{{{r^2}}}$
$\Rightarrow \dfrac{{GMm}}{{2{R^2}}} = \dfrac{{GMm}}{{{r^2}}}$
$\Rightarrow {r^2} = 2{R^2}$
$\Rightarrow r = \sqrt 2 R$
This is the distance of the object from the center of earth for its weight to become half. To find the distance from the surface of the earth,
$h = r - R$
Where, $h =$Height from the surface where its weight becomes half,
$\Rightarrow h = \sqrt 2 R - R$
$\Rightarrow h = \left( {\sqrt 2 - 1} \right)R$
Now multiplying this equation by $\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 + 1}}$,
$\Rightarrow h = \left( {\sqrt 2 - 1} \right)R \times \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 + 1}}$
$\therefore h = \dfrac{R}{{\sqrt 2 + 1}}$

Therefore, options B and C both are correct answers.

Note: The other factors on which the weight of an object depends are, the mass of the planet on which its weight is being measures, the radius of the planet on which its weight is being measured, the angular velocity of the planet on which its weight is being measured, and the position at which the object is placed which means at the equator, or at the poles.