Question

A man weighs \[80kg\] on the surface of Earth of radius \[R\]. At what height above the surface of the earth his weight will be \[40kg\]?
A. \[\dfrac{R}{2}\]
B. \[\sqrt 2 R\]
C. \[\left( {\sqrt 2 - 1} \right)R\]
D. \[\left( {\sqrt 2 + 1} \right)R\]

Answer 1

Hint:Given is the weight of the man on the surface of Earth. At some distance above the surface of the earth, there will be a change in his weight. This is because weight is considered to be the force due to gravity. But the mass will remain constant and will be the same on or above the surface of the Earth. The mass does change for a given body.

Formula Used:
The weight \[W\]of a body of mass \[M\] is given by:\[W = Mg\]
where, \[g\]is the acceleration due to gravity.
The acceleration due to gravity \[g\] at a distance \[r\] from the centre of Earth is given by: \[g = \dfrac{{G{M_e}}}{{{r^2}}}\]
where, \[G\]is the Gravitational constant, \[{M_e}\] is the mass of Earth and \[r\] is the distance from the centre of Earth.

Complete step by step answer:
The radius from the centre to the surface of the Earth is \[R\]. A man on the surface of the Earth weighs \[80kg\]. The weight \[W\]of a body of mass \[M\] is given by
\[W = Mg\] \[ \to (1)\]
Mass is a measure of an ability to resist the change in acceleration when a net force is applied. It is a constant quantity and cannot be zero for any material body. Therefore, the mass of the man on the surface of the Earth will be the same as that above some height from the surface of the Earth. On the other hand, weight is considered to be the force due to gravity. The weight varies from place to place for the same body and it can also be zero.
The acceleration due to gravity \[g\] at a distance \[r\] from the centre of Earth is given by
\[g = \dfrac{{G{M_e}}}{{{r^2}}}\]
Substituting the value of \[g\]in equation (1)
Therefore, \[W = M\left( {\dfrac{{G{M_e}}}{{{r^2}}}} \right)\] \[ \to (2)\]
A man weighs \[80kg\] on the surface of Earth of radius \[R\]. Let \[{W_1}\] be the weight of that man. According to equation (2),
\[{W_1} = M\left( {\dfrac{{G{M_e}}}{{{R^2}}}} \right)\]
But, \[{W_1} = 80kg\].
\[80kg = M\left( {\dfrac{{G{M_e}}}{{{R^2}}}} \right)\] \[ \to (3)\]
Let \[\left( {R + h} \right)\]be the height above the surface of the earth where his weight will be \[40kg\]where, \[h\]is the height of the man above the surface of Earth. Let that weight be \[{W_2}\]. Therefore, \[{W_1} = 40kg\]. Thus from equation (2)
\[40kg = M\left( {\dfrac{{G{M_e}}}{{{{\left( {R + h} \right)}^2}}}} \right)\] \[ \to (4)\]
Other factors like \[G\], \[{M_e}\] and \[M\]will remain constant.
Dividing equation (3) by equation (4)
\[\dfrac{{80kg}}{{40kg}} = \dfrac{{M\left( {\dfrac{{G{M_e}}}{{{R^2}}}} \right)}}{{M\left( {\dfrac{{G{M_e}}}{{{{\left( {R + h} \right)}^2}}}} \right)}} \\
\Rightarrow \dfrac{2}{1} = \dfrac{{\dfrac{1}{{{R^2}}}}}{{\dfrac{1}{{{{\left( {R + h} \right)}^2}}}}} = \dfrac{{{{\left( {R + h} \right)}^2}}}{{{R^2}}} \\
\Rightarrow \sqrt 2 = \dfrac{{\left( {R + h} \right)}}{R}\]
Evaluating \[h\]
\[\sqrt 2 = \dfrac{{\left( {R + h} \right)}}{R} \\
\Rightarrow \sqrt 2 = 1 + \dfrac{h}{R} \\
\therefore h = R\left( {\sqrt 2 - 1} \right)\]

Hence, option C is the correct answer.

Note:The mass is not always a constant. It increases at higher velocities (the velocities that are equal to the speed of light). But here in this problem, the velocities are considered to be small and therefore the mass is taken to be constant. The mass cannot be zero. Weight of a body is zero at the centre of the Earth but the mass is not zero there.