Answer to Question #240990 in Astronomy | Astrophysics for joel

Question

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.

Answer 1

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 $M$ and the object is at distance $r$ from the central body, then the escape velocity will be

$v_e = \sqrt{\dfrac{2GM}{r}}$ .

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

$v_e = \sqrt{\dfrac{2GM}{r}}, \\ v_e = \sqrt{\dfrac{2\cdot6.67\cdot10^{-11}\,\mathrm{N/kg^2\cdot m^2}\cdot5.97\cdot10^{24}\,\mathrm{kg}}{1.12\cdot10^7\,\mathrm{m}}} ,\\ v_e = 8430\,\mathrm{m/s}.$