Escape Velocity Calculator
How fast must an object travel to never fall back? Compute escape velocity of a planet, moon, or star — complete with a three-way solver and verified presets.
vₑ = √(2GM/R) · G = 6,67430×10⁻¹¹ m³/kg/s² (CODATA 2018)
Worked example: Earth
- 2GM = 2 × 6,67430×10⁻¹¹ × 5,97237×10²⁴ = 7,972278e+14 m³/s²
- (2GM)/R = 7,972278e+14 / 6,371×10⁶ = 1,251339e+8 m²/s²
- vₑ = √(1,251339e+8) = 11.186,32 m/s = 11,186 km/s
Meaning: 0.00373% of light speed · Mach 32.6 · low orbit velocity 7.910 km/s.
Preset terverifikasi
| Benda langit | Mass (kg) | Radius (km) | vₑ (km/s) |
|---|---|---|---|
| Bumi | 5,97237×10²⁴ | 6.371 | 11,186 |
| Bulan | 7,342×10²² | 1.737,4 | 2,375 |
| Mars | 6,4171×10²³ | 3.389,5 | 5,027 |
| Jupiter | 1,89813×10²⁷ | 69.911 | 60,202 |
| Matahari | 1,98847×10³⁰ | 696.000 | 617,551 |
GM values use IAU 2015 (Earth 398600.4418 · Moon 4902.8000 · Mars 42828.375214 · Jupiter 126686534 · Sun 132712440018 km³/s²); vₑ is recomputed from M and R above, matching Wikipedia’s “Escape velocity” list (Jupiter 60.20 — not 59.5, which uses the old 71,492 km radius convention).
When to orbit, when to escape?
v < vₑ/√2
The object falls back to the surface (if vertical) or its path is insufficient to orbit.
The object falls back to the surface (if vertical) or its path is insufficient to orbit.
vₑ/√2 ≤ v < vₑ
The object orbits: circular at the lower bound, elliptical in between. For Earth: 7.910 – 11.186 km/s.
The object orbits: circular at the lower bound, elliptical in between. For Earth: 7.910 – 11.186 km/s.
v ≥ vₑ
The object escapes forever: a parabola exactly at vₑ, a hyperbola with leftover velocity above it.
The object escapes forever: a parabola exactly at vₑ, a hyperbola with leftover velocity above it.
Frequently asked questions
What is escape velocity?
Escape velocity is the minimum speed an object needs to leave a celestial body’s gravitational influence without further propulsion, never to fall back. It is derived from energy conservation: the initial kinetic energy (½mv²) must equal the gravitational potential energy to overcome (GMm/R), so vₑ = √(2GM/R).
What is Earth’s escape velocity?
11.186 km/s, computed from Earth’s mass of 5.97237×10²⁴ kg and mean radius 6371 km with G = 6.67430×10⁻¹¹ m³/kg/s² (CODATA 2018). That equals 40,271 km/h, or about 0.00373% of light speed.
Does the launched object’s mass affect escape velocity?
No. In the derivation, the object’s mass (m) appears on both sides of the energy equation (½mv² = GMm/R) and cancels out. Escape velocity depends only on the celestial body’s mass (M) and distance from its center (R). A 1-ton rocket and a 1-gram pebble need the same vₑ from the same surface.
What is the difference between orbiting and escaping gravity?
Low circular orbit velocity is v = √(GM/R) = vₑ/√2 — for Earth about 7.910 km/s, not 11.186 km/s. Below vₑ/√2 the object falls back; in between it orbits (ellipse); exactly at vₑ its path is a parabola and it never returns; above vₑ its path is a hyperbola with leftover velocity at infinity.
Why is Jupiter’s escape velocity sometimes written 59.5 km/s and sometimes 60.2 km/s?
A radius-convention difference: 59.5 km/s uses a 71,492 km radius (old convention), while 60.202 km/s uses the 1-bar equatorial radius of 69,911 km with GM = 126686534 km³/s² (IAU 2015). This calculator uses 69,911 km to stay consistent with the modern GM value.
Must a rocket reach 11.186 km/s all at once from the launchpad?
No. The 11.186 km/s figure is an idealized instantaneous energy condition with no air friction. Real rockets climb gradually: they fight gravity and atmospheric drag during the burn, so the total delta-v needed is actually larger. Escape velocity is best applied to ballistic objects or energy analysis, not rocket flight profiles.
What are the escape velocities of the Sun and the Moon?
Sun: 617.551 km/s (about 0.206% of light speed) from a mass of 1.98847×10³⁰ kg and radius 696,000 km. Moon: 2.375 km/s, only 21.2% of Earth’s escape velocity — which is why the Apollo missions escaped the Moon far more easily than Earth.
What happens if the speed is slightly below escape velocity?
The object cannot escape to infinity: it will fall back to the surface (if fired vertically) or enter an elliptical orbit returning to its starting point (if aimed tangentially and fast enough). Only at v ≥ vₑ is its total energy zero or positive, leaving it unbound.
Can escape velocity exceed the speed of light?
By Newton’s formula, yes: for extremely dense bodies, √(2GM/R) can exceed c — that is the rough definition of a black hole’s event horizon. But in that regime Newtonian physics no longer applies and general relativity must be used, so this calculator is only valid for ordinary celestial bodies.
What value of G does this calculator use?
G = 6.67430×10⁻¹¹ m³/kg/s² per CODATA 2018. G is the physical constant hardest to measure precisely, so its last digit may be revised by the next CODATA recommendation; the impact on vₑ is far below 0.01% and insignificant for practical use.
What do the M⊕ and R⊕ units mean in the unit options?
M⊕ is Earth’s mass (5.97237×10²⁴ kg) and R⊕ is Earth’s mean radius (6371 km). Both are just unit shortcuts for entering multiples easily, e.g. Jupiter’s mass ≈ 317.8 M⊕.
Does this calculator account for a planet’s atmosphere and rotation?
No. The result is the ideal escape velocity in vacuum from a non-rotating spherical body. Atmosphere adds drag, while planetary rotation gives an eastward speed bonus (about 0.465 km/s at Earth’s equator). For real mission estimates, both factors must be computed separately.
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