Understanding Escape Velocity
Define escape velocity. Escape velocity is the velocity of an object required to overcome the gravitational pull of the planet that object is on to escape into space. A larger planet has more mass and requires a much greater escape velocity than a smaller planet with less mass.
Begin with conservation of energy. Conservation of energy states that the total energy of an isolated system remains unchanged. In the derivation below, we will work with an Earth-rocket system and assume that this system is isolated. In conservation of energy, we equate the initial and final potential and kinetic energies K 1 + U 1 = K 2 + U 2 , {\displaystyle K_{1}+U_{1}=K_{2}+U_{2},} K_{{1}}+U_{{1}}=K_{{2}}+U_{{2}}, where K {\displaystyle K} K is kinetic energy and U {\displaystyle U} U is potential energy.
Define kinetic and potential energy. Kinetic energy is energy of motion, and is equal to 1 2 m v 2 , {\displaystyle {\frac {1}{2}}mv^{2},} {\frac {1}{2}}mv^{{2}}, where m {\displaystyle m} m is the mass of the rocket and v {\displaystyle v} v is its velocity. Potential energy is energy that results from where an object is relative to the bodies in the system. In physics, we typically define the potential energy to be 0 at an infinite distance from Earth. Since the gravitational force is attractive, the potential energy of the rocket will always be negative (and smaller the closer it is to Earth). Potential energy in the Earth-rocket system is thus written as − G M m r , {\displaystyle -{\frac {GMm}{r}},} -{\frac {GMm}{r}}, where G {\displaystyle G} G is Newton's gravitational constant, M {\displaystyle M} M is the mass of Earth, and r {\displaystyle r} r is the distance between the two masses' centers.
Substitute these expressions into conservation of energy. When the rocket achieves the minimum velocity required to escape Earth, it will eventually stop at an infinite distance from Earth, so K 2 = 0. {\displaystyle K_{2}=0.} K_{{2}}=0. Then, the rocket will not feel Earth's gravitational pull and will never fall back to Earth, so U 2 = 0 {\displaystyle U_{2}=0} U_{{2}}=0 as well. 1 2 m v 2 − G M m r = 0 {\displaystyle {\frac {1}{2}}mv^{2}-{\frac {GMm}{r}}=0} {\frac {1}{2}}mv^{{2}}-{\frac {GMm}{r}}=0
Solve for v. 1 2 m v 2 = G M m r v 2 = 2 G M r v = 2 G M r {\displaystyle {\begin{aligned}{\frac {1}{2}}mv^{2}&={\frac {GMm}{r}}\\v^{2}&={\frac {2GM}{r}}\\v&={\sqrt {\frac {2GM}{r}}}\end{aligned}}} {\begin{aligned}{\frac {1}{2}}mv^{{2}}&={\frac {GMm}{r}}\\v^{{2}}&={\frac {2GM}{r}}\\v&={\sqrt {{\frac {2GM}{r}}}}\end{aligned}} v {\displaystyle v} v in the above equation is the escape velocity of the rocket - the minimum velocity required to escape the gravitational pull of Earth. Note that the escape velocity is independent of the mass of the rocket m . {\displaystyle m.} m. The mass is reflected in both the potential energy provided by Earth's gravity as well as the kinetic energy provided by the movement of the rocket.
Calculating Escape Velocity
State the equation for escape velocity. v = 2 G M r {\displaystyle v={\sqrt {\frac {2GM}{r}}}} v={\sqrt {{\frac {2GM}{r}}}} The equation assumes the planet you are on is spherical and has constant density. In the real world, the escape velocity depends on where you are at on the surface because a planet bulges at the equator due to its rotation and has slightly varying density due to its composition.
Understand the variables of the equation. G = 6.67 × 10 − 11 N m 2 k g − 2 {\displaystyle G=6.67\times 10^{-11}{\rm {\ N\ m^{2}\ kg^{-2}}}} G=6.67\times 10^{{-11}}{{\rm {\ N\ m^{{2}}\ kg^{{-2}}}}} is Newton's gravitational constant. The value of this constant reflects the fact that gravity is an incredibly weak force. It was determined experimentally by Henry Cavendish in 1798, but has proven to be notoriously difficult to measure precisely. G {\displaystyle G} G can be written using only base units as 6.67 × 10 − 11 m 3 k g − 1 s − 2 , {\displaystyle 6.67\times 10^{-11}{\rm {\ m^{3}\ kg^{-1}\ s^{-2}}},} 6.67\times 10^{{-11}}{{\rm {\ m^{3}\ kg^{{-1}}\ s^{{-2}}}}}, since 1 N = 1 k g m s − 2 . {\displaystyle 1{\rm {\ N}}=1{\rm {\ kg\ m\ s^{-2}}}.} 1{{\rm {\ N}}}=1{{\rm {\ kg\ m\ s^{{-2}}}}}. Mass M {\displaystyle M} M and radius r {\displaystyle r} r are dependent upon the planet you wish to escape from. You must convert to SI units. That is, mass is in kilograms (kg) and distance is in meters (m). If you find values that are in different units, such as miles, convert them to SI.
Determine the mass and radius of the planet you are on. For Earth, assuming that you are at sea level, r = 6.38 × 10 6 m {\displaystyle r=6.38\times 10^{6}{\rm {\ m}}} r=6.38\times 10^{{6}}{{\rm {\ m}}} and M = 5.98 × 10 24 k g . {\displaystyle M=5.98\times 10^{24}{\rm {\ kg}}.} M=5.98\times 10^{{24}}{{\rm {\ kg}}}. Search online for a table of masses and radii for other planets or moons.
Substitute values into the equation. Now that you have the necessary information, you can start solving the equation. v = 2 ( 6.67 × 10 − 11 m 3 k g − 1 s − 2 ) ( 5.98 × 10 24 k g ) ( 6.38 × 10 6 m ) {\displaystyle v={\sqrt {\frac {2(6.67\times 10^{-11}{\rm {\ m^{3}\ kg^{-1}\ s^{-2}}})(5.98\times 10^{24}{\rm {\ kg}})}{(6.38\times 10^{6}{\rm {\ m}})}}}} v={\sqrt {{\frac {2(6.67\times 10^{{-11}}{{\rm {\ m^{{3}}\ kg^{{-1}}\ s^{{-2}}}}})(5.98\times 10^{{24}}{{\rm {\ kg}}})}{(6.38\times 10^{{6}}{{\rm {\ m}}})}}}}
Evaluate. Remember to evaluate your units at the same time and cancel them out as needed to obtain a dimensionally consistent solution. v = 2 ( 6.67 ) ( 5.98 ) ( 6.38 ) × 10 7 m 2 s − 2 ≈ 11200 m s − 1 = 11.2 k m s − 1 {\displaystyle {\begin{aligned}v&={\sqrt {{\frac {2(6.67)(5.98)}{(6.38)}}\times 10^{7}{\rm {\ m^{2}\ s^{-2}}}}}\\&\approx 11200{\rm {\ m\ s^{-1}}}\\&=11.2{\rm {\ km\ s^{-1}}}\end{aligned}}} {\begin{aligned}v&={\sqrt {{\frac {2(6.67)(5.98)}{(6.38)}}\times 10^{{7}}{{\rm {\ m^{{2}}\ s^{{-2}}}}}}}\\&\approx 11200{{\rm {\ m\ s^{{-1}}}}}\\&=11.2{{\rm {\ km\ s^{{-1}}}}}\end{aligned}} In the last step, we converted the answer from SI units to k m s − 1 {\displaystyle {\rm {\ km\ s^{-1}}}} {{\rm {\ km\ s^{{-1}}}}} by multiplying by the conversion factor 1 km 1000 m . {\displaystyle {\frac {\text{1 km}}{\text{1000 m}}}.} {\frac {{\text{1 km}}}{{\text{1000 m}}}}.
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