JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 22)
The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity ve such that $${v_i} = \sqrt {{x \over y}} \times {v_e}$$. The value of x will be ____________.
Answer
10
Explanation
Here R = radius of the earth
From energy conservation
$${{ - G{m_e}m} \over R} + {1 \over 2}m{v_i}^2 = {{ - G{m_e}m} \over {11R}} + 0$$
$${1 \over 2}m{v_i}^2 = {{10} \over {11}}{{G{m_e}m} \over R}$$
$${v_i} = \sqrt {{{20} \over {11}}{{G{m_e}} \over R}} $$
$${v_i} = \sqrt {{{10} \over {11}}} {v_e}$$
{$$ \because $$ escape velocity $${v_e} = \sqrt {{{2G{m_e}} \over R}} $$}
Then the value of x = 10
From energy conservation
$${{ - G{m_e}m} \over R} + {1 \over 2}m{v_i}^2 = {{ - G{m_e}m} \over {11R}} + 0$$
$${1 \over 2}m{v_i}^2 = {{10} \over {11}}{{G{m_e}m} \over R}$$
$${v_i} = \sqrt {{{20} \over {11}}{{G{m_e}} \over R}} $$
$${v_i} = \sqrt {{{10} \over {11}}} {v_e}$$
{$$ \because $$ escape velocity $${v_e} = \sqrt {{{2G{m_e}} \over R}} $$}
Then the value of x = 10
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