For a planet having mass equal to mass of the earth but radius is one fourth of radius of the earth. Then escape velocity for this planet will be

Gravitation question

2000 · Q144

Gravitation question

2000 · Q144

View written solutionFree

Correct answer: B

${v_{earth}} = \sqrt {{{2G{M_e}} \over {{R_e}}}}$

${v_{planet}} = \sqrt {{{2G{M_p}} \over {{R_p}}}} = \sqrt {{{2G{M_e}} \over {{R_e}/4}}} = \sqrt {{{8G{M_e}} \over {{R_e}}}}$

${{{v_{planet}}} \over {{v_{earth}}}} = \sqrt {{{8G{M_e}} \over {{R_e}}}} \times \sqrt {{{{R_e}} \over {2G{M_e}}}} = 2$

$\therefore$ ${v_{planet}} = 2 \times {v_{earth}}$

= 2 $\times$ 11.2 = 22.4 km/s

More from Gravitation

Sign in

Gravitation question

2000 · Q144

Gravitation question

2000 · Q144

View written solutionFree

Correct answer: B

${v_{earth}} = \sqrt {{{2G{M_e}} \over {{R_e}}}}$

${v_{planet}} = \sqrt {{{2G{M_p}} \over {{R_p}}}} = \sqrt {{{2G{M_e}} \over {{R_e}/4}}} = \sqrt {{{8G{M_e}} \over {{R_e}}}}$

${{{v_{planet}}} \over {{v_{earth}}}} = \sqrt {{{8G{M_e}} \over {{R_e}}}} \times \sqrt {{{{R_e}} \over {2G{M_e}}}} = 2$

$\therefore$ ${v_{planet}} = 2 \times {v_{earth}}$

= 2 $\times$ 11.2 = 22.4 km/s