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

2D?

3D?

4D?

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

2D?

3D?

4D?

2D?

3D?

4D?

- gonegahgah
- Tetronian
**Posts:**463**Joined:**Sat Nov 05, 2011 3:27 pm**Location:**Queensland, Australia

I'll assume that the orbit is circular, and that the gravitational acceleration has the form G*M/r^{n-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 v^{2}/r, where v is speed. Equating these:

v^{2}/r = GM/r^{n-1}

v = √(GM/r^{n-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/R^{n-1}

GM = g R^{n-1}

v = √(g R^{n-1}/r^{n-2}).

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

Our own 3D Earth has radius R=6.37*10^{3}km, and gravity g=9.8*10^{-3}km/s^{2}. For a satellite, say, 200 km above the surface, r=6.57*10^{3}km, and our formula gives v=√(g R^{2}/r)=7.8 km/s. For a farther satellite at r=2R, our formula gives v=√(g R/2)=5.6 km/s.

v

v = √(GM/r

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/R

GM = g R

v = √(g R

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

Our own 3D Earth has radius R=6.37*10

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

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ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**404**Joined:**Tue Sep 18, 2018 4:10 am

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/r^{2})

How does that affect escape velocity?

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/r

How does that affect escape velocity?

- gonegahgah
- Tetronian
**Posts:**463**Joined:**Sat Nov 05, 2011 3:27 pm**Location:**Queensland, Australia

Escape speed (not velocity, which would imply a special direction) is gotten by looking at the total energy. Kinetic energy is 1/2 m v^{2}, 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/r^{n-1}) with its direction (-r/r, where r is the position vector, pointing from the attractor to the orbiter) gives the force vector:

F_{G} = - (GMm/r^{n}) r.

Since no forces other than gravity are affecting the orbiter, Newton equates this to ma; that is,

m d^{2}r/dt^{2} = - (GMm/r^{n}) 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/r^{n}) 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/r^{n-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:

E_{G} = - 1/(n-2) GMm/r^{n-2}

1/2 mv^{2} - 1/(n-2) GMm/r^{n-2} = E_{K} + E_{G} = E = constant.

Now, we want to "escape" to r=∞ where E_{G}=0; but since E_{K} 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 mv^{2} - 1/(n-2) GMm/r^{n-2} ≥ 0

v ≥ √( 2/(n-2) GM/r^{n-2} ).

This is the escape speed, for n>2.

But for n=2, the gravitational energy involves a logarithm:

E_{G} = 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:

E_{G} = 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 E_{G}=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?)

F

Since no forces other than gravity are affecting the orbiter, Newton equates this to ma; that is,

m d

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/r

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/r

E

1/2 mv

Now, we want to "escape" to r=∞ where E

1/2 mv

v ≥ √( 2/(n-2) GM/r

This is the escape speed, for n>2.

But for n=2, the gravitational energy involves a logarithm:

E

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:

E

(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 E

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**404**Joined:**Tue Sep 18, 2018 4:10 am

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