How do we know that the escape velocity formula is what it is? How was it determined?
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It’s all about energy. Consider an object at infinite distance from a star; define that as zero potential energy. Any closer object will be deeper in the star’s gravity well, so it will have negative potential energy.
But an object may also have kinetic energy, from having a speed relative to that of the star. Kinetic and potential can be converted into one another, but the sum will be constant. So what’s the kinetic energy an object needs to have zero total energy?
Kinetic: 0.5 m v^2 , where v is the speed of the object, and m is its mass.
[Potential](https://en.wikipedia.org/wiki/Gravitational_energy): –GMm/r , where G is the gravitational constant, M is the mass of the star, and r is the distance from the star to the object.
Total: 0.5mv^2 – GMm/r. If that = 0, then
0.5 mv^2 = GMm/r. Solve for v and get
v_esc = sqrt{2GM/r}
So, we’re going to start with some constants. How we got them isn’t important for this topic.
* G: the universal *gravitational constant*
* M: mass of the Earth
* m: mass of the escaping body
At a given distance *r* from the center, the body feels a force pulling it towards the center:
F = G ( (M*m) / r^2 )
The work (force exerted over distance) needed to move the body over a distance *dr* is then:
dW = F * dr = G ( (M*m) / r^2 ) * dr
In order to extend this work from the surface of the planet r0 to arbitrarily far away from Earth, we have to integrate:
>dW = integral from r0 to infinity of:
G ( (M*m) / r^2 ) * dr
The dr term falls away, and we’re left with:
W = G ( (M*m) / r0 )
In order for the body to be able to do this amount of work, it must have exactly this amount of kinetic energy. Recall the formula for kinetic energy, and we have
1/2 mv^2 = G ( (M*m) / r0 )
Here we rearrange terms to isolate the velocity, v, and we arrive at the formula for escape velocity:
v = sqrt (2GM / r0) = sqrt (2gr0)
It’s the velocity that sets the kinetic energy of an object equal and opposite to the gravitational potential energy.
Kinetic energy is mv^(2)/2, gravitational potential energy is mgh. Set them equal and cancel mass, and we get v^(2)=2gh.
Little g is gravitational acceleration on earth, but we can generalize it to be GM/h^(2), where G is the gravitational constant, M is the mass of the object you’re escaping from, and h is the height from the center of mass.
Sub that into the previous formula to get v^(2)=2MG/h, and can root both sides to get v=(2MG/h)^(1/2).
The speed that I leave the room when she says, “Honey, I find it curious that……”. Certainly sub-light, but as close as I can get.
Orbits are reversible. So the easiest way is to start with an object infinitely (or “almost infinitely”) far away and let it fall down to the surface, then see what its speed it when it reaches the object.
Put another way, energy is conserved. To “escape” is to reach infinite distance with no speed, that is, to reach a state where both kinetic and potential energy are zero. (The potential is arbitrary, but by convention, the potential starts at zero at infinite distance and decreases the closer you get. Only the change in potential is physically meaningful, though.) So to compute escape velocity, we need to set potential energy + kinetic energy = 0, the same value it will have when the object has escaped to a very great distance.
Since kinetic energy is 1/2 m v^(2), and gravitational potential energy is -GMm/r, we can compute:
-GMm/r + 1/2 m v^2 = 0
Subtract -GMm/r from both sides:
1/2 m v^2 = GMm/r
Divide through by m:
1/2 v^2 = GM/r
Multiply by 2:
v^2 = 2GM/r
And square root:
v = sqrt(2GM/r)
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