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

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

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

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

Answered by mrW1 last updated on 05/Feb/17

$$Escapevelocityv=\sqrt{\frac{2Gm}{R}}$$$$withG=universalgravitationalconstant$$$$m=Massofplanet$$$$R=Radiusofplanet$$$$$$$$\frac{v_x}{v_{earth}}=\sqrt{\frac{m_x}{m_{earth}}}=\sqrt{\frac{2}{3}}=0.816$$$$EscapevelocityofplanetXis$$$$v_x=0.816v_{earth}=0.816\times 11.2=9.1km/s$$

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

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

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