Answer to Question #167795 in Physics for Tithi

Question #167795

Show that the escape velocity from the surface of earth is √2 times the velocity of projection of an artificial satellite orbiting close around the earth.


1
Expert's answer
2021-03-04T11:57:01-0500

The escape velocity from the surface of earth is


ve=2GMRv_e=\sqrt{\frac{2GM}{R}}

For orbiting around the earth:


v=GMR+hv=\sqrt{\frac{GM}{R+h}}

For a satellite very close to Earth


v=GMR+0=GMRv=\sqrt{\frac{GM}{R+0}}=\sqrt{\frac{GM}{R}}

Thus


vev=2\frac{v_e}{v}=\sqrt{2}


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