In conservation of energy, we equate the initial and final potential and kinetic energies K1+U1=K2+U2,{\displaystyle K_{1}+U_{1}=K_{2}+U_{2},} where K{\displaystyle K} is kinetic energy and U{\displaystyle U} is potential energy.
Kinetic energy is energy of motion, and is equal to 12mv2,{\displaystyle {\frac {1}{2}}mv^{2},} where m{\displaystyle m} is the mass of the rocket and v{\displaystyle 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 −GMmr,{\displaystyle -{\frac {GMm}{r}},} where G{\displaystyle G} is Newton’s gravitational constant, M{\displaystyle M} is the mass of Earth, and r{\displaystyle r} is the distance between the two masses’ centers.
12mv2−GMmr=0{\displaystyle {\frac {1}{2}}mv^{2}-{\frac {GMm}{r}}=0}
12mv2=GMmrv2=2GMrv=2GMr{\displaystyle {\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} 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. } 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.
v=2GMr{\displaystyle 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.
G=6. 67×10−11 N m2 kg−2{\displaystyle 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,[7] X Research source but has proven to be notoriously difficult to measure precisely. G{\displaystyle G} can be written using only base units as 6. 67×10−11 m3 kg−1 s−2,{\displaystyle 6. 67\times 10^{-11}{\rm {\ m^{3}\ kg^{-1}\ s^{-2}}},} since 1 N=1 kg m s−2. {\displaystyle 1{\rm {\ N}}=1{\rm {\ kg\ m\ s^{-2}}}. }[8] X Research source Mass M{\displaystyle M} and radius r{\displaystyle 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.
Search online for a table of masses and radii for other planets or moons.
v=2(6. 67×10−11 m3 kg−1 s−2)(5. 98×1024 kg)(6. 38×106 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=2(6. 67)(5. 98)(6. 38)×107 m2 s−2≈11200 m s−1=11. 2 km 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}}} In the last step, we converted the answer from SI units to km s−1{\displaystyle {\rm {\ km\ s^{-1}}}} by multiplying by the conversion factor 1 km1000 m. {\displaystyle {\frac {\text{1 km}}{\text{1000 m}}}. }
