Orbital Velocity at Different Dimensions?

Ideas about how a world with more than three spatial dimensions would work - what laws of physics would be needed, how things would be built, how people would do things and so on.

Orbital Velocity at Different Dimensions?

Postby gonegahgah » Sat Apr 24, 2021 8:01 pm

Could somebody help me out with providing the formulas for orbital velocity under different dimensions, please?

2D?
3D?
4D?
gonegahgah
Tetronian
 
Posts: 490
Joined: Sat Nov 05, 2011 3:27 pm
Location: Queensland, Australia

Re: Orbital Velocity at Different Dimensions?

Postby mr_e_man » Tue Apr 27, 2021 3:07 pm

I'll assume that the orbit is circular, and that the gravitational acceleration has the form G*M/rn-1, where G is a universal constant, M is the attractor's mass, r is the distance from the centre of the attractor, and n is the dimension. The required inward acceleration for uniform circular motion is v2/r, where v is speed. Equating these:

v2/r = GM/rn-1

v = √(GM/rn-2).

This shows that in 2D the orbital speed doesn't depend on the radius. In 3D or higher, the speed decreases with increasing radius. (And in 1D you can't have a circular orbit.)

For an orbit around a planet of radius R, you don't need to know G and M directly; you can instead measure the gravity g at the surface:

g = GM/Rn-1

GM = g Rn-1

v = √(g Rn-1/rn-2).

(Of course r > R since the orbit is above the surface.)

Our own 3D Earth has radius R=6.37*103km, and gravity g=9.8*10-3km/s2. For a satellite, say, 200 km above the surface, r=6.57*103km, and our formula gives v=√(g R2/r)=7.8 km/s. For a farther satellite at r=2R, our formula gives v=√(g R/2)=5.6 km/s.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian
 
Posts: 514
Joined: Tue Sep 18, 2018 4:10 am

Re: Orbital Velocity at Different Dimensions?

Postby gonegahgah » Tue Apr 27, 2021 9:08 pm

Awesome. Thank you Mr E for those.
I was primarily after the first equation and thank you for the second as well :)

That provides me:

v(2D) = √(G m)
v(3D) = √(G m/r)
v(4D) = √(G m/r2)

How does that affect escape velocity?
gonegahgah
Tetronian
 
Posts: 490
Joined: Sat Nov 05, 2011 3:27 pm
Location: Queensland, Australia

Re: Orbital Velocity at Different Dimensions?

Postby 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 vv), 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=√(rr) 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?)
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian
 
Posts: 514
Joined: Tue Sep 18, 2018 4:10 am


Return to Higher Spatial Dimensions

Who is online

Users browsing this forum: No registered users and 1 guest