Earth's radius is 6.4×10⁶ m. Circular orbital speed at surface (ignoring atmosphere) is v = √(GM/R). Calculate it. (G=6.67×10⁻¹¹, M=6×10²⁴ kg).

V = √(6.67×10⁻¹¹ × 6×10²⁴ / 6.4×10⁶) ≈ 7.9 km/s.

If a planet's orbit has semi-major axis a = 4 AU, what is its orbital period? (Earth has a = 1 AU, T = 1 year).

T² ∝ a³ → T = (a/1AU)^(3/2) years.

Escape velocity from Earth is ~11.2 km/s. What is the ratio v_esc / v_circ?

V_esc = √(2GM/R) = √2 × v_circ.

A satellite is in elliptical orbit with periapsis 300 km and apoapsis 2000 km above Earth. Find its orbital period.

A = (r_peri + r_apo)/2 + R_earth. Use Kepler's 3rd law.

Pro 🔒~20 min

Gravitational Fields & Orbital Mechanics

Kepler's laws, satellite orbits, and escape velocity

How it works

Gravitational Fields & Orbital Mechanics demonstrates a key principle: Gravity follows an inverse-square law (F ∝ 1/r²). Gravity follows an inverse-square law (F ∝ 1/r²). The circular orbital velocity v = √(GM/r) is the exact speed for a stable circular orbit — faster means elliptical or escape, slower means the satellite falls. Kepler's 3 laws: (1) orbits are ellipses with the planet at one focus; (2) equal areas in equal times (conservation of angular momentum); (3) T² ∝ a³. Escape velocity is √2 times the circular orbital speed.

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Step-by-step

Set the launch speed and angle.
The satellite launches horizontally from the surface.
At ~7.9 km/s it enters circular orbit.
Increase speed to get elliptical orbits.
At ~11.2 km/s it escapes.
Watch the swept area triangles to verify Kepler's 2nd law.
Change planet mass (Pro) to explore other worlds.

Key formulas

$F = G\frac{m_1 m_2}{r^2}$ Newton's law of universal gravitation
$v_{circ} = \sqrt{\frac{GM}{r}}$ Circular orbital velocity
$v_{esc} = \sqrt{\frac{2GM}{r}}$ Escape velocity
$T^2 = \frac{4\pi^2}{GM}a^3$ Kepler's 3rd law (T² ∝ a³)
$E_{orbit} = -\frac{GMm}{2a}$ Total orbital energy (bound orbit is negative)

Frequently asked questions

Earth's radius is 6.4×10⁶ m. Circular orbital speed at surface (ignoring atmosphere) is v = √(GM/R). Calculate it. (G=6.67×10⁻¹¹, M=6×10²⁴ kg).: V = √(6.67×10⁻¹¹ × 6×10²⁴ / 6.4×10⁶) ≈ 7.9 km/s.
If a planet's orbit has semi-major axis a = 4 AU, what is its orbital period? (Earth has a = 1 AU, T = 1 year).: T² ∝ a³ → T = (a/1AU)^(3/2) years.
Escape velocity from Earth is ~11.2 km/s. What is the ratio v_esc / v_circ?: V_esc = √(2GM/R) = √2 × v_circ.
A satellite is in elliptical orbit with periapsis 300 km and apoapsis 2000 km above Earth. Find its orbital period.: A = (r_peri + r_apo)/2 + R_earth. Use Kepler's 3rd law.