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Escape Speed, Orbital Velocity and Energy of an Orbiting Satellite

Throw something fast enough sideways and it keeps falling around the Earth without landing: that speed is the orbital velocity, v₀ = √(GM/r), about 7.9 km/s just above the surface. Throw it faster, at the escape speed vₑ = √(2GM/R) = √(2gR) ≈ 11.2 km/s, and it leaves Earth for ever. vₑ = √2 × v₀. A satellite in a circular orbit has KE = GMm/2r, PE = −GMm/r and total energy E = −GMm/2r. The total is negative, so the satellite is bound. Higher orbits are slower and take longer: T = 2π√(r³/GM).

🎬 Step-by-step story

  1. A small satellite is launched sideways from just above the Earth at a low speed. Its path bends down and it crashes back. Too slow.
  2. Now launch at about 7.9 km/s. It still falls toward the Earth, but the ground curves away just as fast. It goes round in a circle: an orbit.
  3. Now launch at 11.2 km/s. The path opens out and the satellite flies away and never comes back. That is the escape speed.
  4. A satellite in a circular orbit. Orange bar = kinetic energy (up). Purple bar = potential energy (down, twice as big). Green bar = total, negative and equal in size to KE.
  5. A higher orbit, at twice the radius. The satellite moves slower and takes longer for one round. Press the button for the worked example.
  6. Your turn. Pick a launch speed, press Launch, and see: fall back, orbit or escape.

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

If the satellite is falling all the time, why doesn't it hit the ground?

It moves sideways so fast that while it falls a little, the curved ground drops away by the same amount. Compare step 1 (too slow, crashes) with step 2 (just right, circles).

Does direction matter for escape speed?

No (ignoring air). Escape depends on energy, and KE = ½mv² has no direction. Launching eastward helps only because Earth's spin adds speed.

Why is the satellite's total energy negative even though it is moving?

Its PE (−GMm/r) is twice as large as its KE (GMm/2r). The total is −GMm/2r. See the green bar below zero in step 4.

A higher orbit is slower. So why does it need more energy?

Going up, KE falls a little, but PE rises twice as much. The total energy goes up (becomes less negative), so you must supply energy.

Why is 7.9 km/s not enough to escape but enough to orbit?

At 7.9 km/s total energy is −GMm/2R, still negative, so it stays bound in a circle. You need √2 times the speed (11.2 km/s) to make total energy zero.

Orbital velocity: falling but never landing

A satellite in orbit is always falling toward the Earth. But it also moves sideways so fast that the Earth's surface curves away beneath it. So it never lands.

Deriving v₀

  1. For a circular orbit of radius r = R + h, gravity provides the centripetal force: GMm/r² = m v₀²/r.
  2. So v₀ = √(GM/r) = √(GM/(R + h)).
  3. Just above the surface (h ≪ R): v₀ = √(GM/R) = √(gR) = √(9.8 × 6.4 × 10⁶) ≈ 7.9 km/s.

v₀ does not depend on the satellite's mass. Higher orbit (bigger r) means lower speed.

Time period

T = 2πr / v₀ = 2π√(r³/GM). Near the surface T ≈ 84 minutes. So T² ∝ r³ — Kepler's third law again.

Escape speed: leaving for ever

Escape speed is the smallest speed with which a body must be thrown from the surface so that it never comes back (it just reaches infinity with zero speed).

Deriving vₑ from energy

  1. At the surface: KE = ½ m vₑ², PE = −GMm/R.
  2. At infinity, just escaping: KE = 0, PE = 0. Total = 0.
  3. Energy is conserved: ½ m vₑ² − GMm/R = 0.
  4. vₑ = √(2GM/R) = √(2gR) = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km/s.

Energy of an orbiting satellite

For a satellite of mass m in a circular orbit of radius r:

So E = −KE and U = −2 KE. The total is negative: the satellite is bound to the Earth.

Binding energy

The energy you must add to free the satellite from its orbit = +GMm/2r. For an orbit at r, going to a higher orbit needs extra energy, even though the satellite ends up moving slower (KE falls, but U rises twice as much).

Kinds of satellites and weightlessness

Why do astronauts float?

The astronaut and the station fall together with the same acceleration. The floor does not push up on the astronaut, so the weighing machine reads zero. Gravity is still there; the apparent weight is zero. This is weightlessness.

Try it: the sideways throw

Throw a ball sideways gently, then harder, then harder again, from the same height. Each time it lands farther away. Newton imagined a cannon on a very tall mountain: throw fast enough and the ball lands 'beyond the curve' of the Earth — it orbits. In the 3D free play, try 7.0, 7.9, 9.0, 11.2 and 12 km/s and predict before you press Launch: fall back, circle, stretched ellipse or escape?

Key formulas and definitions

Worked examples

1. Find the orbital speed just above Earth's surface (g = 9.8 m/s², R = 6.4 × 10⁶ m).

v₀ = √(gR) = √(6.27 × 10⁷) ≈ 7.92 × 10³ m/s ≈ 7.9 km/s.

2. Find the escape speed from Earth.

vₑ = √(2gR) = √(1.254 × 10⁸) ≈ 1.12 × 10⁴ m/s = 11.2 km/s.

3. Find the orbital speed at a height h = R.

r = 2R. v₀ = √(gR²/2R) = √(gR/2) = 7.9/√2 ≈ 5.6 km/s.

4. A planet has twice Earth's mass and twice its radius. Find its escape speed (Earth: 11.2 km/s).

vₑ ∝ √(M/R) = √(2/2) = 1. So vₑ = 11.2 km/s.

5. Find the period of a satellite orbiting close to Earth's surface.

T = 2π√(R/g) = 2π√(6.4 × 10⁶/9.8) = 2π × 808 ≈ 5077 s ≈ 84.6 min.

6. A 500 kg satellite orbits at r = 2R. Find its KE, PE and total energy (g = 9.8, R = 6.4 × 10⁶ m).

GMm/r = gR²m/2R = mgR/2 = 500 × 9.8 × 6.4 × 10⁶/2 = 1.57 × 10¹⁰ J. KE = 7.84 × 10⁹ J, U = −1.57 × 10¹⁰ J, E = −7.84 × 10⁹ J.

7. How much energy is needed to move that 500 kg satellite from r = 2R to r = 4R?

E = −mgR²/2r. E(2R) = −mgR/4, E(4R) = −mgR/8. ΔE = mgR/8 = 500 × 9.8 × 6.4 × 10⁶/8 = 3.92 × 10⁹ J.

8. Find the radius of a geostationary orbit (T = 86400 s, GM = 4 × 10¹⁴ m³/s²).

r³ = GMT²/4π² = 4 × 10¹⁴ × 7.46 × 10⁹ / 39.48 = 7.56 × 10²². r ≈ 4.23 × 10⁷ m ≈ 42,300 km, so height ≈ 35,900 km.

Common mistakes

Practice quiz

1. Escape speed from Earth is about:
2. The relation between escape speed and orbital speed near the surface is:
3. Total energy of a satellite in orbit of radius r is:
4. As the orbit radius increases, the orbital speed:
5. The period of a geostationary satellite is:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What is the formula for escape velocity?

vₑ = √(2GM/R) = √(2gR) ≈ 11.2 km/s for Earth.

What is the relation between escape velocity and orbital velocity?

Near the surface, vₑ = √2 × v₀, so 11.2 km/s ≈ 1.414 × 7.9 km/s.

What is the total energy of a satellite in orbit?

E = −GMm/2r. It equals −KE and is half of the PE.

Where this is taught

CBSE (India)Class 11Gravitation
USA (Common Core, NGSS, AP)Grade 11Energy and Momentum of Rotating Systems
USA (Common Core, NGSS, AP)Grade 12Energy and Momentum of Rotating Systems
South Korea고등학교 2학년Space-time and motion
China高一Compulsory 2 Ch.7 Gravitation and spaceflight

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