One equation links where you are (r r r how fast you're going (v v v a a a conservation of energy rewritten in orbit-friendly variables. Master why energy depends only on a a a
Definition Vis-viva ("living force")For a body of mass m m m M M M v v v r r r v 2 = G M ( 2 r − 1 a ) v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right) v 2 = GM ( r 2 − a 1 ) a a a semi-major axis of the orbit and G M = μ GM = \mu GM = μ standard gravitational parameter .
Circle: r = a ⇒ v 2 = G M / r r = a \Rightarrow v^2 = GM/r r = a ⇒ v 2 = GM / r Ellipse: a a a v v v r r r Parabola (escape): a → ∞ ⇒ v 2 = 2 G M / r a \to \infty \Rightarrow v^2 = 2GM/r a → ∞ ⇒ v 2 = 2 GM / r Hyperbola: a < 0 ⇒ v 2 > 2 G M / r a < 0 \Rightarrow v^2 > 2GM/r a < 0 ⇒ v 2 > 2 GM / r
Intuition The key insightGravity is conservative, so the total orbital energy E E E along the orbit. It turns out E = − G M m 2 a E = -\dfrac{GMm}{2a} E = − 2 a GM m size , not its shape (eccentricity) or your current position. Bigger orbit ⇒ less negative energy ⇒ closer to escape.
The specific energy (per unit mass) of an orbiting body:
ε = 1 2 v 2 ⏟ kinetic − G M r ⏟ potential \varepsilon = \underbrace{\frac{1}{2}v^2}_{\text{kinetic}} \;-\; \underbrace{\frac{GM}{r}}_{\text{potential}} ε = kinetic 2 1 v 2 − potential r GM
Why this step? Gravitational PE per unit mass is − G M / r -GM/r − GM / r 1 2 v 2 \tfrac12 v^2 2 1 v 2
At the apsides (perigee and apogee) the velocity is purely tangential (perpendicular to r r r v = v p v = v_p v = v p v = v a v = v_a v = v a r p = a ( 1 − e ) , r a = a ( 1 + e ) r_p = a(1-e), \qquad r_a = a(1+e) r p = a ( 1 − e ) , r a = a ( 1 + e )
Why this step? Energy is the same everywhere, but the two apsides give two easy equations we can solve together. Choosing them removes the radial velocity component.
\varepsilon = \frac12 v_p^2 - \frac{GM}{r_p} = \frac12 v_a^2 - \frac{GM}{r_a} \tag{1}
At the apsides v ⃗ ⊥ r ⃗ \vec v \perp \vec r v ⊥ r h = r_p v_p = r_a v_a \;\Rightarrow\; v_a = v_p\frac{r_p}{r_a} \tag{2}
Why this step? Angular momentum is also conserved. With two conserved quantities (energy + L L L v p , v a v_p, v_a v p , v a
Substitute (2) into (1):
1 2 v p 2 − G M r p = 1 2 v p 2 r p 2 r a 2 − G M r a \frac12 v_p^2 - \frac{GM}{r_p} = \frac12 v_p^2\frac{r_p^2}{r_a^2} - \frac{GM}{r_a} 2 1 v p 2 − r p GM = 2 1 v p 2 r a 2 r p 2 − r a GM
Group the v p 2 v_p^2 v p 2 G M GM GM 1 2 v p 2 ( 1 − r p 2 r a 2 ) = G M ( 1 r p − 1 r a ) \frac12 v_p^2\left(1 - \frac{r_p^2}{r_a^2}\right) = GM\left(\frac{1}{r_p} - \frac{1}{r_a}\right) 2 1 v p 2 ( 1 − r a 2 r p 2 ) = GM ( r p 1 − r a 1 )
Factor the left bracket as a difference of squares and the right side over a common denominator:
1 2 v p 2 ⋅ ( r a − r p ) ( r a + r p ) r a 2 = G M ⋅ r a − r p r p r a \frac12 v_p^2 \cdot \frac{(r_a - r_p)(r_a + r_p)}{r_a^2} = GM\cdot\frac{r_a - r_p}{r_p r_a} 2 1 v p 2 ⋅ r a 2 ( r a − r p ) ( r a + r p ) = GM ⋅ r p r a r a − r p
Cancel ( r a − r p ) (r_a - r_p) ( r a − r p ) 1 2 v p 2 r a + r p r a 2 = G M r p r a \frac12 v_p^2\,\frac{r_a + r_p}{r_a^2} = \frac{GM}{r_p r_a} 2 1 v p 2 r a 2 r a + r p = r p r a GM
So
v p 2 = 2 G M r a r p ( r a + r p ) v_p^2 = \frac{2GM\, r_a}{r_p (r_a + r_p)} v p 2 = r p ( r a + r p ) 2 GM r a
Why this step? Pure algebra, but the difference-of-squares cancellation is what makes it collapse to something clean.
Since r p = a ( 1 − e ) r_p = a(1-e) r p = a ( 1 − e ) r a = a ( 1 + e ) r_a = a(1+e) r a = a ( 1 + e ) 2 a 2a 2 a v p 2 = 2 G M r a r p ⋅ 2 a = G M r a a r p v_p^2 = \frac{2GM\,r_a}{r_p \cdot 2a} = \frac{GM\,r_a}{a\,r_p} v p 2 = r p ⋅ 2 a 2 GM r a = a r p GM r a
Now compute ε \varepsilon ε ε = 1 2 v p 2 − G M r p = G M r a 2 a r p − G M r p = G M r p ( r a 2 a − 1 ) \varepsilon = \frac12 v_p^2 - \frac{GM}{r_p} = \frac{GM\,r_a}{2a\,r_p} - \frac{GM}{r_p} = \frac{GM}{r_p}\left(\frac{r_a}{2a} - 1\right) ε = 2 1 v p 2 − r p GM = 2 a r p GM r a − r p GM = r p GM ( 2 a r a − 1 )
Use r a = 2 a − r p r_a = 2a - r_p r a = 2 a − r p ε = G M r p ( 2 a − r p 2 a − 1 ) = G M r p ( − r p 2 a ) = − G M 2 a \varepsilon = \frac{GM}{r_p}\left(\frac{2a - r_p}{2a} - 1\right) = \frac{GM}{r_p}\left(-\frac{r_p}{2a}\right) = -\frac{GM}{2a} ε = r p GM ( 2 a 2 a − r p − 1 ) = r p GM ( − 2 a r p ) = − 2 a GM
This is the big result: ε = − G M 2 a \boxed{\varepsilon = -\dfrac{GM}{2a}} ε = − 2 a GM a a a
Since ε \varepsilon ε any point r r r v v v 1 2 v 2 − G M r = − G M 2 a \frac12 v^2 - \frac{GM}{r} = -\frac{GM}{2a} 2 1 v 2 − r GM = − 2 a GM
Multiply by 2 and rearrange:
v 2 = G M ( 2 r − 1 a ) \boxed{v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)} v 2 = GM ( r 2 − a 1 )
Worked example 1. Speed of a circular orbitFor a circle, r = a r=a r = a Why? A circle's "semi-major axis" is its radius.
v 2 = G M ( 2 r − 1 r ) = G M r ⇒ v = G M r v^2 = GM\left(\frac{2}{r}-\frac{1}{r}\right) = \frac{GM}{r} \Rightarrow v=\sqrt{\frac{GM}{r}} v 2 = GM ( r 2 − r 1 ) = r GM ⇒ v = r GM r = 6.78 × 10 6 r = 6.78\times10^6 r = 6.78 × 1 0 6 G M ⊕ = 3.986 × 10 14 GM_\oplus = 3.986\times10^{14} G M ⊕ = 3.986 × 1 0 14 v ≈ 7.67 v \approx 7.67 v ≈ 7.67
Worked example 2. Escape speedEscape ⇒ barely reach infinity with zero speed ⇒ parabolic ⇒ a → ∞ a\to\infty a → ∞ 1 / a → 0 1/a\to 0 1/ a → 0 Why? Infinite orbit means ε = 0 \varepsilon = 0 ε = 0 v e s c 2 = 2 G M r ⇒ v e s c = 2 v c i r c v_{esc}^2 = \frac{2GM}{r} \Rightarrow v_{esc} = \sqrt{2}\,v_{circ} v esc 2 = r 2 GM ⇒ v esc = 2 v c i r c 2 × \sqrt2 \times 2 ×
Worked example 3. Apogee speed of an elliptical transfer orbitOrbit with r p = 7000 r_p = 7000 r p = 7000 r a = 42000 r_a = 42000 r a = 42000 a = ( r p + r a ) / 2 = 24500 a = (r_p+r_a)/2 = 24500 a = ( r p + r a ) /2 = 24500 Why a a a Vis-viva needs a a a 2 a 2a 2 a r = r a = 42000 r = r_a = 42000 r = r a = 42000 G M = 398600 km 3 / s 2 GM = 398600\ \text{km}^3/\text{s}^2 GM = 398600 km 3 / s 2 v a 2 = 398600 ( 2 42000 − 1 24500 ) v_a^2 = 398600\left(\frac{2}{42000}-\frac{1}{24500}\right) v a 2 = 398600 ( 42000 2 − 24500 1 ) = 398600 ( 4.762 × 10 − 5 − 4.082 × 10 − 5 ) = 398600 ( 6.80 × 10 − 6 ) = 398600(4.762\times10^{-5} - 4.082\times10^{-5}) = 398600(6.80\times10^{-6}) = 398600 ( 4.762 × 1 0 − 5 − 4.082 × 1 0 − 5 ) = 398600 ( 6.80 × 1 0 − 6 ) v a ≈ 1.65 v_a \approx 1.65 v a ≈ 1.65
r r r a a a Why it feels right: for a circle r = a r=a r = a Fix: r r r instantaneous distance (changes every second); a a a fixed property of the whole orbit. Vis-viva needs both .
Common mistake Forgettinga < 0 a<0 a < 0 Why it feels right: "a a a Fix: For unbound (hyperbolic) orbits a a a negative by convention, which correctly makes v 2 = G M ( 2 / r − 1 / a ) > 2 G M / r v^2 = GM(2/r - 1/a) > 2GM/r v 2 = GM ( 2/ r − 1/ a ) > 2 GM / r
G M GM GM G ( M + m ) G(M+m) G ( M + m ) Why it feels right: Newton's exact two-body uses the reduced form.
Fix: For m ≪ M m \ll M m ≪ M G M GM GM G ( M + m ) G(M+m) G ( M + m )
Recall Feynman: explain to a 12-year-oldImagine a ball on a stretchy string going around you in an oval. When it swings in close it whips around fast ; when it drifts far out it goes slow — but the total "oomph" (speed-energy + height-energy) never changes. The vis-viva equation is just a recipe: tell me how far the ball is right now and how big its whole loop is, and I'll tell you exactly how fast it's moving. The "size of the loop" is the only thing that sets the total oomph.
Mnemonic Remember the formula shape"Two over here, one over all" : 2 r \frac{2}{r} r 2 1 a \frac{1}{a} a 1 G M GM GM 2 rides with r , the 1 rides with a .
What does a a a The semi-major axis of the orbit (a fixed property of the whole orbit).
What does r r r The instantaneous distance from the central body's center to the orbiting body.
State the specific orbital energy in terms of a a a ε = − G M / ( 2 a ) \varepsilon = -GM/(2a) ε = − GM / ( 2 a ) — depends only on orbit size.
Derive escape speed from vis-viva. Set
a → ∞ a\to\infty a → ∞ so
1 / a → 0 1/a\to0 1/ a → 0 :
v 2 = 2 G M / r v^2 = 2GM/r v 2 = 2 GM / r , hence
v e s c = 2 G M / r v_{esc}=\sqrt{2GM/r} v esc = 2 GM / r .
Vis-viva for a circular orbit reduces to? v = G M / r v=\sqrt{GM/r} v = GM / r , since
r = a r=a r = a .
Why is energy the same at perigee and apogee? Gravity is conservative; total mechanical energy is conserved everywhere on the orbit.
What two conservation laws are used in the derivation? Conservation of energy and conservation of angular momentum.
For a hyperbolic orbit, what is the sign of a a a Negative, giving
v 2 > 2 G M / r v^2 > 2GM/r v 2 > 2 GM / r (excess hyperbolic speed).
Relationship between escape and circular speed at the same radius? v e s c = 2 v c i r c v_{esc} = \sqrt{2}\,v_{circ} v esc = 2 v c i r c .
What is G M GM GM The standard gravitational parameter,
μ \mu μ .
E equals minus GM over 2a
Intuition Hinglish mein samjhoDekho, vis-viva equation ka matlab simple hai: kisi bhi orbit me, agar tumhe pata hai ki abhi tum central body se kitni door ho (ye hai r r r a a a v v v v 2 = G M ( 2 / r − 1 / a ) v^2 = GM(2/r - 1/a) v 2 = GM ( 2/ r − 1/ a ) r r r a a a
Iske peeche ka asli jaadu hai energy conservation . Gravity ek conservative force hai, isliye total energy (kinetic 1 2 v 2 \frac12 v^2 2 1 v 2 − G M / r -GM/r − GM / r ε = − G M / ( 2 a ) \varepsilon = -GM/(2a) ε = − GM / ( 2 a ) size par depend karti hai, shape ya position par nahi. Bas isi ko rearrange karo to vis-viva mil jaata hai.
Practical me ye bahut kaam ka hai. Satellite launch karna ho, Hohmann transfer se Mars bhejna ho, ya escape velocity nikalni ho — sab vis-viva se. Jaise escape ke liye orbit infinitely badi (a → ∞ a \to \infty a → ∞ 1 / a 1/a 1/ a v = 2 G M / r v = \sqrt{2GM/r} v = 2 GM / r r = a r=a r = a v = G M / r v=\sqrt{GM/r} v = GM / r
