The ratio of escape velocity on earth $$($$νe $$)$$ to the escape velocity on a planet $$($$ν$$p)$$ whose radius and mean density are twice as that of earth is [NEET $$2016$$]

Question

The ratio of escape velocity on earth $$($$νe $$)$$ to the escape velocity on a planet $$($$ν$$p)$$ whose radius and mean density are twice as that of earth is [NEET $$2016$$]

Infinity Learn · Accepted Answer

Since, the escape velocity of the earth can be given $$asve=2gR=2GMR2R=R83$$$$π$$Gρ                                        ∵ mass of earth, $$m=$$ρ$$43$$$$π$$$$R3$$⇒  $$ve=R83$$$$π$$Gρ                                     … (i)As$$ it is given that, the radius and mean density of planet are twice as that of the earth. So, escape velocity at planet will $$bevp=2R83$$$$π$$$$G2p$$                                      …$$(ii)On$$ dividing Eq. $$(i) by Eq. (ii), we $$getvevp=R83$$$$π$$Gρ$$2R83$$$$π$$$$G2$$ρ⇒$$vevp=122$$

The ratio of escape velocity on earth ( $ν_{e}$ ) to the escape velocity on a planet ( $ν_{p}$ ) whose radius and mean density are twice as that of earth is [NEET 2016]

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The ratio of escape velocity on earth (νe ) to the escape velocity on a planet (νp) whose radius and mean density are twice as that of earth is [NEET 2016]

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Ready to Test Your Skills?

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The ratio of escape velocity on earth ( $ν_{e}$ ) to the escape velocity on a planet ( $ν_{p}$ ) whose radius and mean density are twice as that of earth is [NEET 2016]