Question 443: Gravitation

Question

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

Accepted Answer

$${v_{earth}} = \sqrt {{{2G{M_e}} \over {{R_e}}}} $$
$${v_{planet}} = \sqrt {{{2G{M_p}} \over {{R_p}}}}  = \sqrt {{{2G{M_e}} \over {{R_e}/4}}}  = \sqrt {{{8G{M_e}} \over {{R_e}}}} $$
$${{{v_{planet}}} \over {{v_{earth}}}} = \sqrt {{{8G{M_e}} \over {{R_e}}}}  	imes \sqrt {{{{R_e}} \over {2G{M_e}}}}  = 2$$
$$ 	herefore $$ $${v_{planet}} = 2 	imes {v_{earth}}$$
= 2 $$ 	imes $$ 11.2 = 22.4 km/s

Answer

11.2 km/sec

Answer

22.4 km/sec

Answer

5.6 km/sec

Answer

44.8 km/sec.

Gravitation

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

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