A spacecraft escapes Earth's gravity from the surface. What is the escape speed in terms of G, M_E, and R_E?

A spacecraft escapes Earth's gravity from the surface. What is the escape speed in terms of G, M_E, and R_E?

• A. v_escape = sqrt(G M_E / R_E)
• B. v_escape = sqrt(2 G M_E / R_E)
• C. v_escape = sqrt(G M_E / (2 R_E))
• D. v_escape = 2 sqrt(G M_E / R_E)

Answer: (B)

To escape from Earth from the surface, you need just enough kinetic energy that the spacecraft’s total energy becomes zero as it reaches infinity. Using energy conservation, the initial energy is (1/2) m v^2 minus GM_E m / R_E, and for a threshold escape, the total energy equals zero. Setting (1/2) m v^2 = GM_E m / R_E and solving for v gives v = sqrt(2 GM_E / R_E). This speed is independent of the spacecraft’s mass and is larger than the circular orbital speed, which is sqrt(GM_E / R_E), by a factor of sqrt(2). The circular speed is not enough to escape, while the other expressions either miss the factor of sqrt(2) or overshoot by a larger amount.

To escape from Earth from the surface, you need just enough kinetic energy that the spacecraft’s total energy becomes zero as it reaches infinity. Using energy conservation, the initial energy is (1/2) m v^2 minus GM_E m / R_E, and for a threshold escape, the total energy equals zero. Setting (1/2) m v^2 = GM_E m / R_E and solving for v gives v = sqrt(2 GM_E / R_E).

This speed is independent of the spacecraft’s mass and is larger than the circular orbital speed, which is sqrt(GM_E / R_E), by a factor of sqrt(2). The circular speed is not enough to escape, while the other expressions either miss the factor of sqrt(2) or overshoot by a larger amount.