by mr_e_man » Wed Apr 28, 2021 1:14 am
Escape speed (not velocity, which would imply a special direction) is gotten by looking at the total energy. Kinetic energy is 1/2 m v2, where m is the orbiter's mass. Gravitational energy is an antiderivative of the gravitational force; taking the gradient of the former (and negating) should give the latter. Let's find a formula for this energy. We could analyze this one-dimensionally (moving only radially), but the general n-dimensional motion (actually it's contained in a 2D plane) is no trouble. Combining the force's magnitude (GMm/rn-1) with its direction (-r/r, where r is the position vector, pointing from the attractor to the orbiter) gives the force vector:
FG = - (GMm/rn) r.
Since no forces other than gravity are affecting the orbiter, Newton equates this to ma; that is,
m d2r/dt2 = - (GMm/rn) r.
Moving everything to the left side (you may also cancel m), and taking dot products with the velocity vector v=dr/dt:
m v • dv/dt + (GMm/rn) r • dr/dt = 0.
The first term is d/dt (1/2 m v•v), the derivative of kinetic energy. The second term, assuming n>2, is d/dt (-1/(n-2) GMm/rn-2) (which you can verify using r=√(r•r) along with various rules from calculus). This equation is saying that the derivative of the total energy is 0, so the total energy is constant. It also tells us what the gravitational energy needs to be:
EG = - 1/(n-2) GMm/rn-2
1/2 mv2 - 1/(n-2) GMm/rn-2 = EK + EG = E = constant.
Now, we want to "escape" to r=∞ where EG=0; but since EK is always non-negative (even as it varies over time, particularly when r→∞), this requires the total energy E to be non-negative when r→∞ and therefore (since E is constant) also at the start of the orbit when r is small:
1/2 mv2 - 1/(n-2) GMm/rn-2 ≥ 0
v ≥ √( 2/(n-2) GM/rn-2 ).
This is the escape speed, for n>2.
But for n=2, the gravitational energy involves a logarithm:
EG = GMm ln(r),
so it increases without bound as r→∞. There is no escape from a 2D planet! No matter how fast you're moving initially, you'll eventually get pulled back to the starting radius. You can go as far away as you want, and the force decreases toward 0, but it keeps taking more energy the farther you go.
And for n=1, the magnitude of the force is constant, so the energy increases linearly with distance:
EG = GMm r.
(Actually, regardless of n, the gravitational energy could have an arbitrary added constant. In particular, for n=2, we should absorb that constant into the logarithm using ln(x)-ln(y)=ln(x/y) to get EG=GMm ln(r/R), where R is some length such as the planet's radius, so that we're taking a logarithm of a number, a ratio of two lengths, rather than an actual length. What is the logarithm of a metre?)
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