Question

A rocket of mass M is launched vertically from the surface of the Earth with an initial speed V. Assuming the radius of the Earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the Earth is-
A. \[\dfrac{R}{{\dfrac{{gR}}{{{v^2}}} - 1}}\]
B. \[\dfrac{R}{{\dfrac{{gR}}{{2{v^2}}} - 1}}\]
C. \[\dfrac{R}{{\dfrac{{2gR}}{{{v^2}}} - 1}}\]
D. \[R\left( {\dfrac{{gR}}{{{v^2}}} - 1} \right)\]

Answer 1

Hint: When air resistance is neglected then the mechanical energy of the rocket is conserved, i.e. the mechanical energy of the rocket on the surface of the earth will be equal to the mechanical energy of the rocket at the maximum height it reaches. Mechanical energy is the sum of the kinetic energy and the gravitational potential energy.

Formula used:
\[K = \dfrac{{m{v^2}}}{2}\]
where K is the kinetic energy of the body of mass m and moving with velocity v.
\[U = mgh\]
where U is the gravitational potential energy of the body of mass m at height h from the surface of the earth and g is the acceleration due to gravity.

Complete step by step solution:
The mass of the rocket is given as M and the radius of the earth is given as R. We need to find the maximum height the rocket will reach. At maximum height the rocket will be at rest. So there will be no kinetic energy of the rocket at the maximum height. As the friction due to air is neglected so the only force acting on the rocket is the gravitational force due to earth. If the acceleration due to gravity at the surface of the earth is g then the acceleration due to gravity at height h is,
\[{g_h} = \dfrac{{gR}}{{R + h}}\]

Using conservation of mechanical energy,
\[{E_{surface}} = {E_{height}}\]
\[\Rightarrow \dfrac{{m{v^2}}}{2} + mg\left( 0 \right) = \dfrac{{m{{\left( 0 \right)}^2}}}{2} + mg{h_{\max }}\left( {\dfrac{R}{{R + {h_{\max }}}}} \right)\]
\[\Rightarrow \dfrac{{m{v^2}}}{2} = mg{h_{\max }}\left( {\dfrac{R}{{R + {h_{\max }}}}} \right)\]
\[\Rightarrow \dfrac{{{v^2}}}{{2gR}} = \dfrac{{{h_{\max }}}}{{R + {h_{\max }}}}\]
\[\Rightarrow R{v^2} + {v^2}{h_{\max }} = 2gR{h_{\max }}\]
\[\Rightarrow R{v^2} = \left( {2gR - {v^2}} \right){h_{\max }}\]
\[\Rightarrow {h_{\max }} = \dfrac{{R{v^2}}}{{\left( {2gR - {v^2}} \right)}}\]
\[\therefore {h_{\max }} = \dfrac{R}{{\left( {\dfrac{{2gR}}{{{v^2}}} - 1} \right)}}\]
Hence, the maximum height reached by the rocket is \[\dfrac{R}{{\left( {\dfrac{{2gR}}{{{v^2}}} - 1} \right)}}\].

Therefore, the correct option is C.

Note: We should assume that the acceleration due to gravity is not constant with height and the height reached by the rocket is comparable to the radius of the earth. If the height assumed is insignificant to the radius of the earth then the acceleration due to gravity is assumed to be constant with height from the surface of the earth.