Escape velocity of a Body from Earth is 11 km/s. Assuming the mass and radius of Earth to be about 81 and 4 times the mass and radius of the Moon

Question

Calculate the escape velocity from the surface of the Moon.

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Answer 1

[Note that G is the Gravitational Constant equal to 6.6725985E-11(N•m²per kg²). Take Earth Mass/Moon Mass as 81.25 and Earth Radius/Moon Radius as 3.66091954.]

Let the body be shot "straight up" from Earth's surface with velocity v_initial. Use energy considerations to find the minimum velocity that will break the hold of Earth's gravity on the object.

Next, write Total Energy = 0.5m_objectv_initial² − GM_Earthm_object÷R_Earth which in turn equals

0.5m_objectv_escape² − GM_Earthm_object÷R_escape.

This gives the Total Energy of the object at any point when its velocity and distance from Earth's center are known. At Earth's surface, with velocity = v_initial, the distance from Earth's center is simply the radius of the Earth. At maximum altitude, final velocity of the object will be 0 and final radius can be taken as infinity (or ∞).

Total Energy is conserved so obtain 0.5m_objectv_initial² − GM_Earthm_object÷R_Earth equals

0.5m_objectv_final² − GM_Earthm_object÷R_escape. This reduces to v_initial² = 0² + 2GM_Earth(1/R_Earth − 1/∞) or

v_initial (now the same as v_escape when on the "edge" of outer space) equals √(2GM_Earth/R_Earth).

For Earth's Escape Velocity, write v_escape = √(2GM_Earth/R_Earth). With M_Earth at 5.98E24 kg and R_Earth at 6.37E6 meters, v_escape comes to 11192.91924 meters per second or 11.2 km/s.

For Lunar Escape Velocity, one would multiply 11192.91924 meters per second by √[(1/81.25)÷(1/3.66091954)] to find 2375.89245 meters per second or 2.4 km/s.