Question 141: Gravitation

Question

A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be :
(Take radius of earth $$=6400 \mathrm{~km}$$ and $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ )

Accepted Answer

Applying conservation of energy
$$ - {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {{1 \over 3}\sqrt {{{2G{m_e}} \over {{R_e}}}} } \right)^2} =  - {{G{M_e}m} \over {{R_e} + h}}$$
$$ - {{G{M_e}m} \over {{R_e}}} + {{G{M_e}m} \over {9{R_e}}} =  - {{G{M_e}m} \over {{R_e} + h}}$$
$${8 \over {9{R_e}}} = {1 \over {{R_e} + h}}$$
$$ \Rightarrow h = {{{R_e}} \over 8} = {{6400} \over 8} = 800$$ km

Answer

800 km

Answer

1600 km

Answer

2133 km

Answer

4800 km

Gravitation

A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be : (Take radius of earth $=6400 \mathrm{~km} and \mathrm{g}=10 \mathrm{~ms}^{-2}$ )

Choose the correct answer: