Gravity

Escape Velocity

We have looked at the orbital speed of 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 a constant,

F_g = m g = constant

we found the gravitational potential energy to be

U_g = m g h

Remember, the potential energy is the opposite of the work done by the force of gravity as we move an object from some reference point to its present position y = h.

Now we want to use the general form of the force of gravity,

The minus sign is important. The minus sign means this is an attractive force, acting in the minus-r-direction. The force acts in the direction of decreasing distance r. We have often -- or usually -- just written the absolute value of the force F_{g. But now we need to be more careful}

The potential energy is the opposite of the work done by the force of gravity as we move an object from some reference point to 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 very important. We will handle the limits in a moment.

where r_ref is the "reference position" for which the potential energy is defined to be zero.

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 be zero.

Notice that the potential energy varies inversely as the distance between the two masses. The force varies inversely as the square of 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 energy mean? We must do work to 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 "escape speed". ]

Initially, at radius r and moving with speed v_esc, the total energy of the object is

E_init = - ^GMm/_r + (¹/₂) m v_esc²

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 = 0

And, from Energy Conservation, we know the total energy remains the same, so

E_init = - ^GMm/_r + (¹/₂) m v_esc² = 0 = E_final

- ^GMm/_r + (¹/₂) m v_esc² = 0

(¹/₂) m v_esc2 = ^GMm/_r

(¹/₂) v_esc2 = ^GM/_r

v_esc² = ^{2 GM}/_r

v_esc = SQRT[ ^{2 GM}/_r ]

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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 orbital speed was about

v_orb = 28 000 ^km/_h

or

v_orb = 17 000 ^mi/_h

Satellites			Summary
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(c) Doug Davis, 2001; all rights reserved