Answer to Question #169643 in Physics for nasia

Question #169643

Obtain the escape velocity of an atmospheric particle 100 km above

the earth’s surface. Given that ME = 5.98 × 1024 kg, RE = 6370 km, G = 6.67 × 10-11

N.m2

/kg2


1
Expert's answer
2021-03-08T08:17:46-0500

The escape velocity can be found as follows:


vesc=2GMR+h,v_{esc}=\sqrt{\dfrac{2GM}{R+h}},vesc=26.671011 Nm2kg25.981024 kg6370103 m+100103 m=11.1 kms.v_{esc}=\sqrt{\dfrac{2\cdot6.67\cdot10^{-11}\ \dfrac{Nm^2}{kg^2}\cdot5.98\cdot10^{24}\ kg}{6370\cdot10^3\ m+100\cdot10^3\ m}}=11.1\ \dfrac{km}{s}.

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