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Question Number 10355 by Tawakalitu ayo mi last updated on 05/Feb/17

What is the escape velocity from the surface  of a planet with two third of the earth′s  gravity but the same radius.

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{escape}\:\mathrm{velocity}\:\mathrm{from}\:\mathrm{the}\:\mathrm{surface} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{planet}\:\mathrm{with}\:\mathrm{two}\:\mathrm{third}\:\mathrm{of}\:\mathrm{the}\:\mathrm{earth}'\mathrm{s} \\ $$$$\mathrm{gravity}\:\mathrm{but}\:\mathrm{the}\:\mathrm{same}\:\mathrm{radius}. \\ $$

Commented by Tawakalitu ayo mi last updated on 05/Feb/17

i will reduce it sir.

$$\mathrm{i}\:\mathrm{will}\:\mathrm{reduce}\:\mathrm{it}\:\mathrm{sir}. \\ $$

Answered by mrW1 last updated on 05/Feb/17

Escape velocity v=(√((2Gm)/R))  with G=universal gravitational constant             m=Mass of planet             R=Radius of planet    (v_x /v_(earth) )=(√(m_x /m_(earth) ))=(√(2/3))=0.816  Escape velocity of planet X is  v_x =0.816v_(earth) =0.816×11.2=9.1 km/s

$${Escape}\:{velocity}\:{v}=\sqrt{\frac{\mathrm{2}{Gm}}{{R}}} \\ $$$${with}\:{G}={universal}\:{gravitational}\:{constant} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{m}={Mass}\:{of}\:{planet} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{R}={Radius}\:{of}\:{planet} \\ $$$$ \\ $$$$\frac{{v}_{{x}} }{{v}_{{earth}} }=\sqrt{\frac{{m}_{{x}} }{{m}_{{earth}} }}=\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}=\mathrm{0}.\mathrm{816} \\ $$$${Escape}\:{velocity}\:{of}\:{planet}\:{X}\:{is} \\ $$$${v}_{{x}} =\mathrm{0}.\mathrm{816}{v}_{{earth}} =\mathrm{0}.\mathrm{816}×\mathrm{11}.\mathrm{2}=\mathrm{9}.\mathrm{1}\:{km}/{s} \\ $$

Commented by Tawakalitu ayo mi last updated on 05/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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