Question #240990

What is the minimum required velocity in km/s of the satellite to escape from the gravitational attraction of the earth? Assumed a circular orbit with an altitude of 4,830 km.


1
Expert's answer
2021-09-23T08:30:33-0400

The minimum velocity is the so called escape velocity. If this velocity is reached, the motion is parabolic. If the central body has the mass MM and the object is at distance rr from the central body, then the escape velocity will be

ve=2GMrv_e = \sqrt{\dfrac{2GM}{r}} .

In our case r=h+R=4830+6378=11208km=1.12107m.r = h + R_{\oplus} = 4830 + 6378 = 11208\,\mathrm{km} = 1.12\cdot10^7\,\mathrm{m}.

ve=2GMr,ve=26.671011N/kg2m25.971024kg1.12107m,ve=8430m/s.v_e = \sqrt{\dfrac{2GM}{r}}, \\ v_e = \sqrt{\dfrac{2\cdot6.67\cdot10^{-11}\,\mathrm{N/kg^2\cdot m^2}\cdot5.97\cdot10^{24}\,\mathrm{kg}}{1.12\cdot10^7\,\mathrm{m}}} ,\\ v_e = 8430\,\mathrm{m/s}.


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