GravityEscape Velocity

We have looked at the

orbital speedof satellites in low Earth orbit and in a geo-sychronous orbit. At orbital speed, the satellite continues in orbit. What speed is necessary to "escape" Earth's gravity entirely and to move infinitely far away from Earth? To answer this question, we need to find the gravitational potential energy for very large distances from Earth.

At Earth's surface, where we considered the force of gravity to be aconstant,F

_{g}= m g = constantwe found the gravitational potential energy to be

U

_{g}= m g hRemember, the

potential energyis the opposite of thework done bythe force of gravity as we move an object from somereference pointto its present position y = h.Now we want to use the

generalform of the force of gravity,The minus sign is important. The minus sign means this is an

attractiveforce, acting in theminus-r-direction. The force acts in the direction ofdecreasing distancer. We have often -- or usually -- just written the absolute value of the forceF_{g. But now we need to be more careful}The

potential energyis the opposite of thework done bythe force of gravity as we move an object from somereference pointto its present position r. That is,Now the intrinsic negative sign in the force

F_{g is important. Along with the negative sign from the definition of potential energy as the negative of the work done by the force, we have}G, M, and m are constants so we can pull them outside the integral sign.

The limits are

veryimportant. We will handle the limits in a moment.where r

_{ref}is the "reference position" for which the potential energy isdefinedto bezero.We will take this reference position to be

infinity. That is, when two masses are infinitly far from each other, we define their gravitational potential energy to bezero.Notice that the potential energy varies inversely as the distance between the two masses. The

forcevaries inversely as thesquareof the distance but the gravitational potential energy varies inversely as the distance itself. It is a one-over-r dependence.Notice, too, the sign. What does a

negative potential energymean? We mustdo workto move the two masses from a distance r apart to being finitly far apart. At an infinite distance part their potential energy is zero.Now, consider a mass m at a distance r from Earth and moving with a speed v

_{esc}. What must be this "escape velocity" v_{esc}so that the object can "escape" Earth and go infinitely far a way? [ Since the direction makes no difference, it would be better and more correct to call v_{esc}the "escapespeed". ]Initially, at radius r and moving with speed v

_{esc}, the total energy of the object isE

_{init}= -^{GMm}/_{r}+ (^{1}/_{2}) m v_{esc}^{2}In the final state, when the object gets far, far away from Earth, it need have no speed at all so its final total energy is zero,

E

_{final}= 0And, from Energy Conservation, we know the total energy remains the same, so

E

_{init}= -^{GMm}/_{r}+ (^{1}/_{2}) m v_{esc}^{2}= 0 = E_{final}-

^{GMm}/_{r}+ (^{1}/_{2}) m v_{esc}^{2}= 0(

^{1}/_{2}) m v_{esc}2 =^{GMm}/_{r}(

^{1}/_{2}) v_{esc}2 =^{GM}/_{r}v

_{esc}^{2}=^{2 GM}/_{r}v

_{esc}= SQRT[^{2 GM}/_{r}]

Now that we have a symbolic solution for the escape velocity, what is a numerical value for escape velocity from Earth's surface?

v

_{esc}= SQRT[^{2 GM}/_{r}]

v

_{esc}= SQRT[^{2 (6.67 x 10 - 11)(5.98 x 1024)}/_{(6.37 x 106)}]^{m}/_{s}

v

_{esc}= 1.12 x 10^{4}^{m}/_{s}

v

_{esc}= 1.12 x 10^{4}^{m}/_{s}[^{1 km}/_{1 000 m}] [^{3 600 s}/_{1 h}] = 4.03 x 10^{4 km}/_{h}

v

_{esc}= 40 300^{km}/_{h }

v

_{esc}= 40 300^{km}/_{h}[^{0.61 mi}/_{1 km}]

v

_{esc}= 25 000^{mi}/_{h}Remember, earlier we found that

orbitalspeed was aboutv

_{orb}= 28 000^{km}/_{h}or

v

_{orb}= 17 000^{mi}/_{h}

SatellitesSummaryReturn to Gravity ToC(c) Doug Davis, 2001; all rights reserved